mirror of
https://github.com/halfrost/LeetCode-Go.git
synced 2025-07-05 08:27:30 +08:00
Add solution 1654、1655
This commit is contained in:
@ -5,52 +5,52 @@ import (
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"testing"
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"testing"
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)
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)
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type question1649 struct {
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type question1653 struct {
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para1649
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para1653
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ans1649
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ans1653
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}
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}
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// para 是参数
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// para 是参数
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// one 代表第一个参数
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// one 代表第一个参数
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type para1649 struct {
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type para1653 struct {
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s string
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s string
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}
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}
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// ans 是答案
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// ans 是答案
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// one 代表第一个答案
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// one 代表第一个答案
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type ans1649 struct {
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type ans1653 struct {
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one int
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one int
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}
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}
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func Test_Problem1649(t *testing.T) {
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func Test_Problem1653(t *testing.T) {
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qs := []question1649{
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qs := []question1653{
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{
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{
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para1649{"aababbab"},
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para1653{"aababbab"},
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ans1649{2},
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ans1653{2},
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},
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},
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{
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{
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para1649{"bbaaaaabb"},
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para1653{"bbaaaaabb"},
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ans1649{2},
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ans1653{2},
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},
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},
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{
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{
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para1649{"b"},
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para1653{"b"},
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ans1649{0},
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ans1653{0},
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},
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},
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{
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{
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para1649{"ababaaaabbbbbaaababbbbbbaaabbaababbabbbbaabbbbaabbabbabaabbbababaa"},
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para1653{"ababaaaabbbbbaaababbbbbbaaabbaababbabbbbaabbbbaabbabbabaabbbababaa"},
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ans1649{25},
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ans1653{25},
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},
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},
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}
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}
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fmt.Printf("------------------------Leetcode Problem 1649------------------------\n")
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fmt.Printf("------------------------Leetcode Problem 1653------------------------\n")
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for _, q := range qs {
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for _, q := range qs {
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_, p := q.ans1649, q.para1649
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_, p := q.ans1653, q.para1653
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fmt.Printf("【input】:%v 【output】:%v \n", p, minimumDeletions(p.s))
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fmt.Printf("【input】:%v 【output】:%v \n", p, minimumDeletions(p.s))
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}
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}
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fmt.Printf("\n\n\n")
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fmt.Printf("\n\n\n")
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@ -0,0 +1,29 @@
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package leetcode
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func minimumJumps(forbidden []int, a int, b int, x int) int {
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visited := make([]bool, 6000)
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for i := range forbidden {
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visited[forbidden[i]] = true
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}
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queue, res := [][2]int{{0, 0}}, -1
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for len(queue) > 0 {
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length := len(queue)
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res++
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for i := 0; i < length; i++ {
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cur, isBack := queue[i][0], queue[i][1]
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if cur == x {
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return res
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}
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if isBack == 0 && cur-b > 0 && !visited[cur-b] {
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visited[cur-b] = true
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queue = append(queue, [2]int{cur - b, 1})
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}
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if cur+a < len(visited) && !visited[cur+a] {
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visited[cur+a] = true
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queue = append(queue, [2]int{cur + a, 0})
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}
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}
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queue = queue[length:]
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}
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return -1
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}
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@ -0,0 +1,60 @@
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package leetcode
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import (
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"fmt"
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"testing"
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)
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type question1654 struct {
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para1654
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ans1654
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}
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// para 是参数
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// one 代表第一个参数
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type para1654 struct {
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forbidden []int
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a int
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b int
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x int
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}
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// ans 是答案
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// one 代表第一个答案
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type ans1654 struct {
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one int
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}
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func Test_Problem1654(t *testing.T) {
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qs := []question1654{
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{
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para1654{[]int{14, 4, 18, 1, 15}, 3, 15, 9},
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ans1654{3},
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},
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{
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para1654{[]int{8, 3, 16, 6, 12, 20}, 15, 13, 11},
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ans1654{-1},
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},
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{
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para1654{[]int{1, 6, 2, 14, 5, 17, 4}, 16, 9, 7},
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ans1654{2},
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},
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{
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para1654{[]int{1998}, 1999, 2000, 2000},
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ans1654{3998},
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},
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}
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fmt.Printf("------------------------Leetcode Problem 1654------------------------\n")
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for _, q := range qs {
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_, p := q.ans1654, q.para1654
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fmt.Printf("【input】:%v 【output】:%v \n", p, minimumJumps(p.forbidden, p.a, p.b, p.x))
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}
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fmt.Printf("\n\n\n")
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}
|
101
leetcode/1654.Minimum-Jumps-to-Reach-Home/README.md
Normal file
101
leetcode/1654.Minimum-Jumps-to-Reach-Home/README.md
Normal file
@ -0,0 +1,101 @@
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# [1654. Minimum Jumps to Reach Home](https://leetcode.com/problems/minimum-jumps-to-reach-home/)
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## 题目
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A certain bug's home is on the x-axis at position `x`. Help them get there from position `0`.
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The bug jumps according to the following rules:
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- It can jump exactly `a` positions **forward** (to the right).
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- It can jump exactly `b` positions **backward** (to the left).
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- It cannot jump backward twice in a row.
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- It cannot jump to any `forbidden` positions.
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The bug may jump forward **beyond** its home, but it **cannot jump** to positions numbered with **negative** integers.
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Given an array of integers `forbidden`, where `forbidden[i]` means that the bug cannot jump to the position `forbidden[i]`, and integers `a`, `b`, and `x`, return *the minimum number of jumps needed for the bug to reach its home*. If there is no possible sequence of jumps that lands the bug on position `x`, return `1.`
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**Example 1:**
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```
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Input: forbidden = [14,4,18,1,15], a = 3, b = 15, x = 9
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Output: 3
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Explanation: 3 jumps forward (0 -> 3 -> 6 -> 9) will get the bug home.
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```
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**Example 2:**
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```
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Input: forbidden = [8,3,16,6,12,20], a = 15, b = 13, x = 11
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Output: -1
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```
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**Example 3:**
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```
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Input: forbidden = [1,6,2,14,5,17,4], a = 16, b = 9, x = 7
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Output: 2
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Explanation: One jump forward (0 -> 16) then one jump backward (16 -> 7) will get the bug home.
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```
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**Constraints:**
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- `1 <= forbidden.length <= 1000`
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- `1 <= a, b, forbidden[i] <= 2000`
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- `0 <= x <= 2000`
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- All the elements in `forbidden` are distinct.
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- Position `x` is not forbidden.
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## 题目大意
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有一只跳蚤的家在数轴上的位置 x 处。请你帮助它从位置 0 出发,到达它的家。
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跳蚤跳跃的规则如下:
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- 它可以 往前 跳恰好 a 个位置(即往右跳)。
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- 它可以 往后 跳恰好 b 个位置(即往左跳)。
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- 它不能 连续 往后跳 2 次。
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- 它不能跳到任何 forbidden 数组中的位置。
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跳蚤可以往前跳 超过 它的家的位置,但是它 不能跳到负整数 的位置。给你一个整数数组 forbidden ,其中 forbidden[i] 是跳蚤不能跳到的位置,同时给你整数 a, b 和 x ,请你返回跳蚤到家的最少跳跃次数。如果没有恰好到达 x 的可行方案,请你返回 -1 。
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## 解题思路
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- 给出坐标 x ,可以往前跳的步长 a,往后跳的步长 b。要求输出能跳回家的最少跳跃次数。
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- 求最少跳跃次数,思路用 BFS 求解,最先到达坐标 x 的方案即是最少跳跃次数。对`forbidden` 的处理是把记忆化数组里面把他们标记为 true。禁止连续往后跳 2 次的限制,要求我们在 BFS 入队的时候再记录一下跳跃方向,每次往后跳的时候判断前一跳是否是往后跳,如果是往后跳,此次就不能往后跳了。
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## 代码
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```go
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package leetcode
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func minimumJumps(forbidden []int, a int, b int, x int) int {
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visited := make([]bool, 6000)
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for i := range forbidden {
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visited[forbidden[i]] = true
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|
}
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queue, res := [][2]int{{0, 0}}, -1
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for len(queue) > 0 {
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length := len(queue)
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res++
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for i := 0; i < length; i++ {
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cur, isBack := queue[i][0], queue[i][1]
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if cur == x {
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return res
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|
}
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if isBack == 0 && cur-b > 0 && !visited[cur-b] {
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visited[cur-b] = true
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queue = append(queue, [2]int{cur - b, 1})
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|
}
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if cur+a < len(visited) && !visited[cur+a] {
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visited[cur+a] = true
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queue = append(queue, [2]int{cur + a, 0})
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}
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}
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queue = queue[length:]
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}
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return -1
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}
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```
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@ -0,0 +1,30 @@
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package leetcode
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func canDistribute(nums []int, quantity []int) bool {
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freq := make(map[int]int)
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for _, n := range nums {
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freq[n]++
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}
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return dfs(freq, quantity)
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}
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|
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func dfs(freq map[int]int, quantity []int) bool {
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if len(quantity) == 0 {
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return true
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|
}
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|
visited := make(map[int]bool)
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|
for i := range freq {
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|
if visited[freq[i]] {
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|
continue
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|
}
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|
visited[freq[i]] = true
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|
if freq[i] >= quantity[0] {
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|
freq[i] -= quantity[0]
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|
if dfs(freq, quantity[1:]) {
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|
return true
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|
}
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|
freq[i] += quantity[0]
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|
}
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|
}
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|
return false
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|
}
|
@ -0,0 +1,63 @@
|
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|
package leetcode
|
||||||
|
|
||||||
|
import (
|
||||||
|
"fmt"
|
||||||
|
"testing"
|
||||||
|
)
|
||||||
|
|
||||||
|
type question1655 struct {
|
||||||
|
para1655
|
||||||
|
ans1655
|
||||||
|
}
|
||||||
|
|
||||||
|
// para 是参数
|
||||||
|
// one 代表第一个参数
|
||||||
|
type para1655 struct {
|
||||||
|
nums []int
|
||||||
|
quantity []int
|
||||||
|
}
|
||||||
|
|
||||||
|
// ans 是答案
|
||||||
|
// one 代表第一个答案
|
||||||
|
type ans1655 struct {
|
||||||
|
one bool
|
||||||
|
}
|
||||||
|
|
||||||
|
func Test_Problem1655(t *testing.T) {
|
||||||
|
|
||||||
|
qs := []question1655{
|
||||||
|
|
||||||
|
{
|
||||||
|
para1655{[]int{1, 2, 3, 4}, []int{2}},
|
||||||
|
ans1655{false},
|
||||||
|
},
|
||||||
|
|
||||||
|
{
|
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|
para1655{[]int{1, 2, 3, 3}, []int{2}},
|
||||||
|
ans1655{true},
|
||||||
|
},
|
||||||
|
|
||||||
|
{
|
||||||
|
para1655{[]int{1, 1, 2, 2}, []int{2, 2}},
|
||||||
|
ans1655{true},
|
||||||
|
},
|
||||||
|
|
||||||
|
{
|
||||||
|
para1655{[]int{1, 1, 2, 3}, []int{2, 2}},
|
||||||
|
ans1655{false},
|
||||||
|
},
|
||||||
|
|
||||||
|
{
|
||||||
|
para1655{[]int{1, 1, 1, 1, 1}, []int{2, 3}},
|
||||||
|
ans1655{true},
|
||||||
|
},
|
||||||
|
}
|
||||||
|
|
||||||
|
fmt.Printf("------------------------Leetcode Problem 1655------------------------\n")
|
||||||
|
|
||||||
|
for _, q := range qs {
|
||||||
|
_, p := q.ans1655, q.para1655
|
||||||
|
fmt.Printf("【input】:%v 【output】:%v \n", p, canDistribute(p.nums, p.quantity))
|
||||||
|
}
|
||||||
|
fmt.Printf("\n\n\n")
|
||||||
|
}
|
112
leetcode/1655.Distribute-Repeating-Integers/README.md
Normal file
112
leetcode/1655.Distribute-Repeating-Integers/README.md
Normal file
@ -0,0 +1,112 @@
|
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|
# [1655. Distribute Repeating Integers](https://leetcode.com/problems/distribute-repeating-integers/)
|
||||||
|
|
||||||
|
|
||||||
|
## 题目
|
||||||
|
|
||||||
|
You are given an array of `n` integers, `nums`, where there are at most `50` unique values in the array. You are also given an array of `m` customer order quantities, `quantity`, where `quantity[i]` is the amount of integers the `ith` customer ordered. Determine if it is possible to distribute `nums` such that:
|
||||||
|
|
||||||
|
- The `ith` customer gets **exactly** `quantity[i]` integers,
|
||||||
|
- The integers the `ith` customer gets are **all equal**, and
|
||||||
|
- Every customer is satisfied.
|
||||||
|
|
||||||
|
Return `true` *if it is possible to distribute* `nums` *according to the above conditions*.
|
||||||
|
|
||||||
|
**Example 1:**
|
||||||
|
|
||||||
|
```
|
||||||
|
Input: nums = [1,2,3,4], quantity = [2]
|
||||||
|
Output: false
|
||||||
|
Explanation: The 0th customer cannot be given two different integers.
|
||||||
|
```
|
||||||
|
|
||||||
|
**Example 2:**
|
||||||
|
|
||||||
|
```
|
||||||
|
Input: nums = [1,2,3,3], quantity = [2]
|
||||||
|
Output: true
|
||||||
|
Explanation: The 0th customer is given [3,3]. The integers [1,2] are not used.
|
||||||
|
```
|
||||||
|
|
||||||
|
**Example 3:**
|
||||||
|
|
||||||
|
```
|
||||||
|
Input: nums = [1,1,2,2], quantity = [2,2]
|
||||||
|
Output: true
|
||||||
|
Explanation: The 0th customer is given [1,1], and the 1st customer is given [2,2].
|
||||||
|
```
|
||||||
|
|
||||||
|
**Example 4:**
|
||||||
|
|
||||||
|
```
|
||||||
|
Input: nums = [1,1,2,3], quantity = [2,2]
|
||||||
|
Output: false
|
||||||
|
Explanation: Although the 0th customer could be given [1,1], the 1st customer cannot be satisfied.
|
||||||
|
```
|
||||||
|
|
||||||
|
**Example 5:**
|
||||||
|
|
||||||
|
```
|
||||||
|
Input: nums = [1,1,1,1,1], quantity = [2,3]
|
||||||
|
Output: true
|
||||||
|
Explanation: The 0th customer is given [1,1], and the 1st customer is given [1,1,1].
|
||||||
|
```
|
||||||
|
|
||||||
|
**Constraints:**
|
||||||
|
|
||||||
|
- `n == nums.length`
|
||||||
|
- `1 <= n <= 105`
|
||||||
|
- `1 <= nums[i] <= 1000`
|
||||||
|
- `m == quantity.length`
|
||||||
|
- `1 <= m <= 10`
|
||||||
|
- `1 <= quantity[i] <= 105`
|
||||||
|
- There are at most `50` unique values in `nums`.
|
||||||
|
|
||||||
|
## 题目大意
|
||||||
|
|
||||||
|
给你一个长度为 n 的整数数组 nums ,这个数组中至多有 50 个不同的值。同时你有 m 个顾客的订单 quantity ,其中,整数 quantity[i] 是第 i 位顾客订单的数目。请你判断是否能将 nums 中的整数分配给这些顾客,且满足:
|
||||||
|
|
||||||
|
- 第 i 位顾客 恰好 有 quantity[i] 个整数。
|
||||||
|
- 第 i 位顾客拿到的整数都是 相同的 。
|
||||||
|
- 每位顾客都满足上述两个要求。
|
||||||
|
|
||||||
|
如果你可以分配 nums 中的整数满足上面的要求,那么请返回 true ,否则返回 false 。
|
||||||
|
|
||||||
|
## 解题思路
|
||||||
|
|
||||||
|
- 给定一个数组 nums,订单数组 quantity,要求按照订单满足顾客的需求。如果能满足输出 true,不能满足输出 false。
|
||||||
|
- 用 DFS 记忆化暴力搜索。代码实现不难。(不知道此题为什么是 Hard)
|
||||||
|
|
||||||
|
## 代码
|
||||||
|
|
||||||
|
```go
|
||||||
|
package leetcode
|
||||||
|
|
||||||
|
func canDistribute(nums []int, quantity []int) bool {
|
||||||
|
freq := make(map[int]int)
|
||||||
|
for _, n := range nums {
|
||||||
|
freq[n]++
|
||||||
|
}
|
||||||
|
return dfs(freq, quantity)
|
||||||
|
}
|
||||||
|
|
||||||
|
func dfs(freq map[int]int, quantity []int) bool {
|
||||||
|
if len(quantity) == 0 {
|
||||||
|
return true
|
||||||
|
}
|
||||||
|
visited := make(map[int]bool)
|
||||||
|
for i := range freq {
|
||||||
|
if visited[freq[i]] {
|
||||||
|
continue
|
||||||
|
}
|
||||||
|
visited[freq[i]] = true
|
||||||
|
if freq[i] >= quantity[0] {
|
||||||
|
freq[i] -= quantity[0]
|
||||||
|
if dfs(freq, quantity[1:]) {
|
||||||
|
return true
|
||||||
|
}
|
||||||
|
freq[i] += quantity[0]
|
||||||
|
}
|
||||||
|
}
|
||||||
|
return false
|
||||||
|
}
|
||||||
|
```
|
@ -1,56 +0,0 @@
|
|||||||
package leetcode
|
|
||||||
|
|
||||||
func decrypt(code []int, k int) []int {
|
|
||||||
if k == 0 {
|
|
||||||
for i := 0; i < len(code); i++ {
|
|
||||||
code[i] = 0
|
|
||||||
}
|
|
||||||
return code
|
|
||||||
}
|
|
||||||
count, sum, res := k, 0, make([]int, len(code))
|
|
||||||
if k > 0 {
|
|
||||||
for i := 0; i < len(code); i++ {
|
|
||||||
for j := i + 1; j < len(code); j++ {
|
|
||||||
if count == 0 {
|
|
||||||
break
|
|
||||||
}
|
|
||||||
sum += code[j]
|
|
||||||
count--
|
|
||||||
}
|
|
||||||
if count > 0 {
|
|
||||||
for j := 0; j < len(code); j++ {
|
|
||||||
if count == 0 {
|
|
||||||
break
|
|
||||||
}
|
|
||||||
sum += code[j]
|
|
||||||
count--
|
|
||||||
}
|
|
||||||
}
|
|
||||||
res[i] = sum
|
|
||||||
sum, count = 0, k
|
|
||||||
}
|
|
||||||
}
|
|
||||||
if k < 0 {
|
|
||||||
for i := 0; i < len(code); i++ {
|
|
||||||
for j := i - 1; j >= 0; j-- {
|
|
||||||
if count == 0 {
|
|
||||||
break
|
|
||||||
}
|
|
||||||
sum += code[j]
|
|
||||||
count++
|
|
||||||
}
|
|
||||||
if count < 0 {
|
|
||||||
for j := len(code) - 1; j >= 0; j-- {
|
|
||||||
if count == 0 {
|
|
||||||
break
|
|
||||||
}
|
|
||||||
sum += code[j]
|
|
||||||
count++
|
|
||||||
}
|
|
||||||
}
|
|
||||||
res[i] = sum
|
|
||||||
sum, count = 0, k
|
|
||||||
}
|
|
||||||
}
|
|
||||||
return res
|
|
||||||
}
|
|
@ -1,53 +0,0 @@
|
|||||||
package leetcode
|
|
||||||
|
|
||||||
import (
|
|
||||||
"fmt"
|
|
||||||
"testing"
|
|
||||||
)
|
|
||||||
|
|
||||||
type question1652 struct {
|
|
||||||
para1652
|
|
||||||
ans1652
|
|
||||||
}
|
|
||||||
|
|
||||||
// para 是参数
|
|
||||||
// one 代表第一个参数
|
|
||||||
type para1652 struct {
|
|
||||||
code []int
|
|
||||||
k int
|
|
||||||
}
|
|
||||||
|
|
||||||
// ans 是答案
|
|
||||||
// one 代表第一个答案
|
|
||||||
type ans1652 struct {
|
|
||||||
one []int
|
|
||||||
}
|
|
||||||
|
|
||||||
func Test_Problem1652(t *testing.T) {
|
|
||||||
|
|
||||||
qs := []question1652{
|
|
||||||
|
|
||||||
{
|
|
||||||
para1652{[]int{5, 7, 1, 4}, 3},
|
|
||||||
ans1652{[]int{12, 10, 16, 13}},
|
|
||||||
},
|
|
||||||
|
|
||||||
{
|
|
||||||
para1652{[]int{1, 2, 3, 4}, 0},
|
|
||||||
ans1652{[]int{0, 0, 0, 0}},
|
|
||||||
},
|
|
||||||
|
|
||||||
{
|
|
||||||
para1652{[]int{2, 4, 9, 3}, -2},
|
|
||||||
ans1652{[]int{12, 5, 6, 13}},
|
|
||||||
},
|
|
||||||
}
|
|
||||||
|
|
||||||
fmt.Printf("------------------------Leetcode Problem 1652------------------------\n")
|
|
||||||
|
|
||||||
for _, q := range qs {
|
|
||||||
_, p := q.ans1652, q.para1652
|
|
||||||
fmt.Printf("【input】:%v 【output】:%v \n", p, decrypt(p.code, p.k))
|
|
||||||
}
|
|
||||||
fmt.Printf("\n\n\n")
|
|
||||||
}
|
|
@ -1,49 +0,0 @@
|
|||||||
package leetcode
|
|
||||||
|
|
||||||
func minimumDeletions(s string) int {
|
|
||||||
ai, bi, sum, temp, array := 0, 0, 0, 0, []int{}
|
|
||||||
for ai = 0; ai < len(s); ai++ {
|
|
||||||
if s[ai] == 'a' {
|
|
||||||
break
|
|
||||||
}
|
|
||||||
}
|
|
||||||
if ai != 0 && ai != len(s) {
|
|
||||||
sum += ai
|
|
||||||
}
|
|
||||||
for bi = ai; bi < len(s); bi++ {
|
|
||||||
if s[bi] == 'b' {
|
|
||||||
break
|
|
||||||
}
|
|
||||||
}
|
|
||||||
if s[bi-1] == 'a' {
|
|
||||||
ai = bi - 1
|
|
||||||
}
|
|
||||||
if s[bi-1] == 'b' && bi != len(s) {
|
|
||||||
ai = bi + 1
|
|
||||||
}
|
|
||||||
for j := bi; j < len(s); j++ {
|
|
||||||
if s[j] == 'b' {
|
|
||||||
temp++
|
|
||||||
}
|
|
||||||
if s[j] == 'a' && temp != 0 {
|
|
||||||
array = append(array, temp)
|
|
||||||
temp = 0
|
|
||||||
}
|
|
||||||
}
|
|
||||||
if len(array) == 0 {
|
|
||||||
return sum
|
|
||||||
}
|
|
||||||
dp := make([]int, len(array))
|
|
||||||
dp[0] = min(array[0], len(array))
|
|
||||||
for i := 1; i < len(array); i++ {
|
|
||||||
dp[i] = min(dp[i-1]+array[i], dp[i-1]+len(array)-(i+1)+1)
|
|
||||||
}
|
|
||||||
return sum + dp[len(array)-1]
|
|
||||||
}
|
|
||||||
|
|
||||||
func min(a int, b int) int {
|
|
||||||
if a > b {
|
|
||||||
return b
|
|
||||||
}
|
|
||||||
return a
|
|
||||||
}
|
|
@ -1,57 +0,0 @@
|
|||||||
package leetcode
|
|
||||||
|
|
||||||
import (
|
|
||||||
"fmt"
|
|
||||||
"testing"
|
|
||||||
)
|
|
||||||
|
|
||||||
type question1649 struct {
|
|
||||||
para1649
|
|
||||||
ans1649
|
|
||||||
}
|
|
||||||
|
|
||||||
// para 是参数
|
|
||||||
// one 代表第一个参数
|
|
||||||
type para1649 struct {
|
|
||||||
s string
|
|
||||||
}
|
|
||||||
|
|
||||||
// ans 是答案
|
|
||||||
// one 代表第一个答案
|
|
||||||
type ans1649 struct {
|
|
||||||
one int
|
|
||||||
}
|
|
||||||
|
|
||||||
func Test_Problem1649(t *testing.T) {
|
|
||||||
|
|
||||||
qs := []question1649{
|
|
||||||
|
|
||||||
// {
|
|
||||||
// para1649{"aababbab"},
|
|
||||||
// ans1649{2},
|
|
||||||
// },
|
|
||||||
|
|
||||||
// {
|
|
||||||
// para1649{"bbaaaaabb"},
|
|
||||||
// ans1649{2},
|
|
||||||
// },
|
|
||||||
|
|
||||||
{
|
|
||||||
para1649{"b"},
|
|
||||||
ans1649{0},
|
|
||||||
},
|
|
||||||
|
|
||||||
{
|
|
||||||
para1649{"ababaaaabbbbbaaababbbbbbaaabbaababbabbbbaabbbbaabbabbabaabbbababaa"},
|
|
||||||
ans1649{25},
|
|
||||||
},
|
|
||||||
}
|
|
||||||
|
|
||||||
fmt.Printf("------------------------Leetcode Problem 1649------------------------\n")
|
|
||||||
|
|
||||||
for _, q := range qs {
|
|
||||||
_, p := q.ans1649, q.para1649
|
|
||||||
fmt.Printf("【input】:%v 【output】:%v \n", p, minimumDeletions(p.s))
|
|
||||||
}
|
|
||||||
fmt.Printf("\n\n\n")
|
|
||||||
}
|
|
125
website/content/ChapterFour/1652.Defuse-the-Bomb.md
Normal file
125
website/content/ChapterFour/1652.Defuse-the-Bomb.md
Normal file
@ -0,0 +1,125 @@
|
|||||||
|
# [1652. Defuse the Bomb](https://leetcode.com/problems/defuse-the-bomb/)
|
||||||
|
|
||||||
|
|
||||||
|
## 题目
|
||||||
|
|
||||||
|
You have a bomb to defuse, and your time is running out! Your informer will provide you with a **circular** array `code` of length of `n` and a key `k`.
|
||||||
|
|
||||||
|
To decrypt the code, you must replace every number. All the numbers are replaced **simultaneously**.
|
||||||
|
|
||||||
|
- If `k > 0`, replace the `ith` number with the sum of the **next** `k` numbers.
|
||||||
|
- If `k < 0`, replace the `ith` number with the sum of the **previous** `k` numbers.
|
||||||
|
- If `k == 0`, replace the `ith` number with `0`.
|
||||||
|
|
||||||
|
As `code` is circular, the next element of `code[n-1]` is `code[0]`, and the previous element of `code[0]` is `code[n-1]`.
|
||||||
|
|
||||||
|
Given the **circular** array `code` and an integer key `k`, return *the decrypted code to defuse the bomb*!
|
||||||
|
|
||||||
|
**Example 1:**
|
||||||
|
|
||||||
|
```
|
||||||
|
Input: code = [5,7,1,4], k = 3
|
||||||
|
Output: [12,10,16,13]
|
||||||
|
Explanation: Each number is replaced by the sum of the next 3 numbers. The decrypted code is [7+1+4, 1+4+5, 4+5+7, 5+7+1]. Notice that the numbers wrap around.
|
||||||
|
```
|
||||||
|
|
||||||
|
**Example 2:**
|
||||||
|
|
||||||
|
```
|
||||||
|
Input: code = [1,2,3,4], k = 0
|
||||||
|
Output: [0,0,0,0]
|
||||||
|
Explanation: When k is zero, the numbers are replaced by 0.
|
||||||
|
```
|
||||||
|
|
||||||
|
**Example 3:**
|
||||||
|
|
||||||
|
```
|
||||||
|
Input: code = [2,4,9,3], k = -2
|
||||||
|
Output: [12,5,6,13]
|
||||||
|
Explanation: The decrypted code is [3+9, 2+3, 4+2, 9+4]. Notice that the numbers wrap around again. If k is negative, the sum is of the previous numbers.
|
||||||
|
```
|
||||||
|
|
||||||
|
**Constraints:**
|
||||||
|
|
||||||
|
- `n == code.length`
|
||||||
|
- `1 <= n <= 100`
|
||||||
|
- `1 <= code[i] <= 100`
|
||||||
|
- `(n - 1) <= k <= n - 1`
|
||||||
|
|
||||||
|
## 题目大意
|
||||||
|
|
||||||
|
你有一个炸弹需要拆除,时间紧迫!你的情报员会给你一个长度为 n 的 循环 数组 code 以及一个密钥 k 。为了获得正确的密码,你需要替换掉每一个数字。所有数字会 同时 被替换。
|
||||||
|
|
||||||
|
- 如果 k > 0 ,将第 i 个数字用 接下来 k 个数字之和替换。
|
||||||
|
- 如果 k < 0 ,将第 i 个数字用 之前 k 个数字之和替换。
|
||||||
|
- 如果 k == 0 ,将第 i 个数字用 0 替换。
|
||||||
|
|
||||||
|
由于 code 是循环的, code[n-1] 下一个元素是 code[0] ,且 code[0] 前一个元素是 code[n-1] 。
|
||||||
|
|
||||||
|
给你 循环 数组 code 和整数密钥 k ,请你返回解密后的结果来拆除炸弹!
|
||||||
|
|
||||||
|
## 解题思路
|
||||||
|
|
||||||
|
- 给出一个 code 数组,要求按照规则替换每个字母。
|
||||||
|
- 简单题,按照题意描述循环即可。
|
||||||
|
|
||||||
|
## 代码
|
||||||
|
|
||||||
|
```go
|
||||||
|
package leetcode
|
||||||
|
|
||||||
|
func decrypt(code []int, k int) []int {
|
||||||
|
if k == 0 {
|
||||||
|
for i := 0; i < len(code); i++ {
|
||||||
|
code[i] = 0
|
||||||
|
}
|
||||||
|
return code
|
||||||
|
}
|
||||||
|
count, sum, res := k, 0, make([]int, len(code))
|
||||||
|
if k > 0 {
|
||||||
|
for i := 0; i < len(code); i++ {
|
||||||
|
for j := i + 1; j < len(code); j++ {
|
||||||
|
if count == 0 {
|
||||||
|
break
|
||||||
|
}
|
||||||
|
sum += code[j]
|
||||||
|
count--
|
||||||
|
}
|
||||||
|
if count > 0 {
|
||||||
|
for j := 0; j < len(code); j++ {
|
||||||
|
if count == 0 {
|
||||||
|
break
|
||||||
|
}
|
||||||
|
sum += code[j]
|
||||||
|
count--
|
||||||
|
}
|
||||||
|
}
|
||||||
|
res[i] = sum
|
||||||
|
sum, count = 0, k
|
||||||
|
}
|
||||||
|
}
|
||||||
|
if k < 0 {
|
||||||
|
for i := 0; i < len(code); i++ {
|
||||||
|
for j := i - 1; j >= 0; j-- {
|
||||||
|
if count == 0 {
|
||||||
|
break
|
||||||
|
}
|
||||||
|
sum += code[j]
|
||||||
|
count++
|
||||||
|
}
|
||||||
|
if count < 0 {
|
||||||
|
for j := len(code) - 1; j >= 0; j-- {
|
||||||
|
if count == 0 {
|
||||||
|
break
|
||||||
|
}
|
||||||
|
sum += code[j]
|
||||||
|
count++
|
||||||
|
}
|
||||||
|
}
|
||||||
|
res[i] = sum
|
||||||
|
sum, count = 0, k
|
||||||
|
}
|
||||||
|
}
|
||||||
|
return res
|
||||||
|
}
|
||||||
|
```
|
@ -0,0 +1,90 @@
|
|||||||
|
# [1653. Minimum Deletions to Make String Balanced](https://leetcode.com/problems/minimum-deletions-to-make-string-balanced/)
|
||||||
|
|
||||||
|
|
||||||
|
## 题目
|
||||||
|
|
||||||
|
You are given a string `s` consisting only of characters `'a'` and `'b'`.
|
||||||
|
|
||||||
|
You can delete any number of characters in `s` to make `s` **balanced**. `s` is **balanced** if there is no pair of indices `(i,j)` such that `i < j` and `s[i] = 'b'` and `s[j]= 'a'`.
|
||||||
|
|
||||||
|
Return *the **minimum** number of deletions needed to make* `s` ***balanced***.
|
||||||
|
|
||||||
|
**Example 1:**
|
||||||
|
|
||||||
|
```
|
||||||
|
Input: s = "aababbab"
|
||||||
|
Output: 2
|
||||||
|
Explanation: You can either:
|
||||||
|
Delete the characters at 0-indexed positions 2 and 6 ("aababbab" -> "aaabbb"), or
|
||||||
|
Delete the characters at 0-indexed positions 3 and 6 ("aababbab" -> "aabbbb").
|
||||||
|
```
|
||||||
|
|
||||||
|
**Example 2:**
|
||||||
|
|
||||||
|
```
|
||||||
|
Input: s = "bbaaaaabb"
|
||||||
|
Output: 2
|
||||||
|
Explanation: The only solution is to delete the first two characters.
|
||||||
|
```
|
||||||
|
|
||||||
|
**Constraints:**
|
||||||
|
|
||||||
|
- `1 <= s.length <= 105`
|
||||||
|
- `s[i]` is `'a'` or `'b'`.
|
||||||
|
|
||||||
|
## 题目大意
|
||||||
|
|
||||||
|
给你一个字符串 s ,它仅包含字符 'a' 和 'b' 。你可以删除 s 中任意数目的字符,使得 s 平衡 。我们称 s 平衡的 当不存在下标对 (i,j) 满足 i < j 且 s[i] = 'b' 同时 s[j]= 'a' 。请你返回使 s 平衡 的 最少 删除次数。
|
||||||
|
|
||||||
|
## 解题思路
|
||||||
|
|
||||||
|
- 给定一个字符串,要求删除最少次数,使得字母 a 都排在字母 b 的前面。
|
||||||
|
- 很容易想到的一个解题思路是 DP。定义 `dp[i]` 为字符串下标 [ 0, i ] 这个区间内使得字符串平衡的最少删除次数。当 `s[i] == 'a'` 的时候,有 2 种情况,一种是 `s[i]` 前面全是 `[aa……aa]` 的情况,这个时候只需要把其中的所有的字母 `b` 删除即可。还有一种情况是 `s[i]` 前面有字母 `a` 也有字母 `b`,即 `[aaa……abb……b]`,这种情况就需要考虑 `dp[i-1]` 了。当前字母是 `a`,那么肯定要删除字母 `a`,来维持前面有一段字母 `b` 的情况。当 `s[i] == 'b'` 的时候,不管是 `[aa……aa]` 这种情况,还是 `[aaa……abb……b]` 这种情况,当前字母 `b` 都可以直接附加在后面,也能保证整个字符串是平衡的。所以状态转移方程为 `dp[i+1] = min(dp[i] + 1, bCount), s[i] == 'a'`,`dp[i+1] = dp[i], s[i] == 'b'`。最终答案存在 `dp[n]` 中。由于前后项的递推关系中只用到一次前一项,所以我们还可以优化一下空间,用一个变量保存前一项的结果。优化以后的代码见解法一。
|
||||||
|
- 这一题还有一个模拟的思路。题目要求找到最小删除字数,那么就是要找到一个“临界点”,在这个临界点的左边删除所有的字母 b,在这个临界点的右边删除所有的字母 a。在所有的“临界点”中找到删除最少的次数。代码实现见解法二。
|
||||||
|
|
||||||
|
## 代码
|
||||||
|
|
||||||
|
```go
|
||||||
|
package leetcode
|
||||||
|
|
||||||
|
// 解法一 DP
|
||||||
|
func minimumDeletions(s string) int {
|
||||||
|
prev, res, bCount := 0, 0, 0
|
||||||
|
for _, c := range s {
|
||||||
|
if c == 'a' {
|
||||||
|
res = min(prev+1, bCount)
|
||||||
|
prev = res
|
||||||
|
} else {
|
||||||
|
bCount++
|
||||||
|
}
|
||||||
|
}
|
||||||
|
return res
|
||||||
|
}
|
||||||
|
|
||||||
|
func min(a, b int) int {
|
||||||
|
if a < b {
|
||||||
|
return a
|
||||||
|
}
|
||||||
|
return b
|
||||||
|
}
|
||||||
|
|
||||||
|
// 解法二 模拟
|
||||||
|
func minimumDeletions1(s string) int {
|
||||||
|
aCount, bCount, res := 0, 0, 0
|
||||||
|
for i := 0; i < len(s); i++ {
|
||||||
|
if s[i] == 'a' {
|
||||||
|
aCount++
|
||||||
|
}
|
||||||
|
}
|
||||||
|
res = aCount
|
||||||
|
for i := 0; i < len(s); i++ {
|
||||||
|
if s[i] == 'a' {
|
||||||
|
aCount--
|
||||||
|
} else {
|
||||||
|
bCount++
|
||||||
|
}
|
||||||
|
res = min(res, aCount+bCount)
|
||||||
|
}
|
||||||
|
return res
|
||||||
|
}
|
||||||
|
```
|
101
website/content/ChapterFour/1654.Minimum-Jumps-to-Reach-Home.md
Normal file
101
website/content/ChapterFour/1654.Minimum-Jumps-to-Reach-Home.md
Normal file
@ -0,0 +1,101 @@
|
|||||||
|
# [1654. Minimum Jumps to Reach Home](https://leetcode.com/problems/minimum-jumps-to-reach-home/)
|
||||||
|
|
||||||
|
|
||||||
|
## 题目
|
||||||
|
|
||||||
|
A certain bug's home is on the x-axis at position `x`. Help them get there from position `0`.
|
||||||
|
|
||||||
|
The bug jumps according to the following rules:
|
||||||
|
|
||||||
|
- It can jump exactly `a` positions **forward** (to the right).
|
||||||
|
- It can jump exactly `b` positions **backward** (to the left).
|
||||||
|
- It cannot jump backward twice in a row.
|
||||||
|
- It cannot jump to any `forbidden` positions.
|
||||||
|
|
||||||
|
The bug may jump forward **beyond** its home, but it **cannot jump** to positions numbered with **negative** integers.
|
||||||
|
|
||||||
|
Given an array of integers `forbidden`, where `forbidden[i]` means that the bug cannot jump to the position `forbidden[i]`, and integers `a`, `b`, and `x`, return *the minimum number of jumps needed for the bug to reach its home*. If there is no possible sequence of jumps that lands the bug on position `x`, return `1.`
|
||||||
|
|
||||||
|
**Example 1:**
|
||||||
|
|
||||||
|
```
|
||||||
|
Input: forbidden = [14,4,18,1,15], a = 3, b = 15, x = 9
|
||||||
|
Output: 3
|
||||||
|
Explanation: 3 jumps forward (0 -> 3 -> 6 -> 9) will get the bug home.
|
||||||
|
```
|
||||||
|
|
||||||
|
**Example 2:**
|
||||||
|
|
||||||
|
```
|
||||||
|
Input: forbidden = [8,3,16,6,12,20], a = 15, b = 13, x = 11
|
||||||
|
Output: -1
|
||||||
|
```
|
||||||
|
|
||||||
|
**Example 3:**
|
||||||
|
|
||||||
|
```
|
||||||
|
Input: forbidden = [1,6,2,14,5,17,4], a = 16, b = 9, x = 7
|
||||||
|
Output: 2
|
||||||
|
Explanation: One jump forward (0 -> 16) then one jump backward (16 -> 7) will get the bug home.
|
||||||
|
|
||||||
|
```
|
||||||
|
|
||||||
|
**Constraints:**
|
||||||
|
|
||||||
|
- `1 <= forbidden.length <= 1000`
|
||||||
|
- `1 <= a, b, forbidden[i] <= 2000`
|
||||||
|
- `0 <= x <= 2000`
|
||||||
|
- All the elements in `forbidden` are distinct.
|
||||||
|
- Position `x` is not forbidden.
|
||||||
|
|
||||||
|
## 题目大意
|
||||||
|
|
||||||
|
有一只跳蚤的家在数轴上的位置 x 处。请你帮助它从位置 0 出发,到达它的家。
|
||||||
|
|
||||||
|
跳蚤跳跃的规则如下:
|
||||||
|
|
||||||
|
- 它可以 往前 跳恰好 a 个位置(即往右跳)。
|
||||||
|
- 它可以 往后 跳恰好 b 个位置(即往左跳)。
|
||||||
|
- 它不能 连续 往后跳 2 次。
|
||||||
|
- 它不能跳到任何 forbidden 数组中的位置。
|
||||||
|
|
||||||
|
跳蚤可以往前跳 超过 它的家的位置,但是它 不能跳到负整数 的位置。给你一个整数数组 forbidden ,其中 forbidden[i] 是跳蚤不能跳到的位置,同时给你整数 a, b 和 x ,请你返回跳蚤到家的最少跳跃次数。如果没有恰好到达 x 的可行方案,请你返回 -1 。
|
||||||
|
|
||||||
|
## 解题思路
|
||||||
|
|
||||||
|
- 给出坐标 x ,可以往前跳的步长 a,往后跳的步长 b。要求输出能跳回家的最少跳跃次数。
|
||||||
|
- 求最少跳跃次数,思路用 BFS 求解,最先到达坐标 x 的方案即是最少跳跃次数。对`forbidden` 的处理是把记忆化数组里面把他们标记为 true。禁止连续往后跳 2 次的限制,要求我们在 BFS 入队的时候再记录一下跳跃方向,每次往后跳的时候判断前一跳是否是往后跳,如果是往后跳,此次就不能往后跳了。
|
||||||
|
|
||||||
|
## 代码
|
||||||
|
|
||||||
|
```go
|
||||||
|
package leetcode
|
||||||
|
|
||||||
|
func minimumJumps(forbidden []int, a int, b int, x int) int {
|
||||||
|
visited := make([]bool, 6000)
|
||||||
|
for i := range forbidden {
|
||||||
|
visited[forbidden[i]] = true
|
||||||
|
}
|
||||||
|
queue, res := [][2]int{{0, 0}}, -1
|
||||||
|
for len(queue) > 0 {
|
||||||
|
length := len(queue)
|
||||||
|
res++
|
||||||
|
for i := 0; i < length; i++ {
|
||||||
|
cur, isBack := queue[i][0], queue[i][1]
|
||||||
|
if cur == x {
|
||||||
|
return res
|
||||||
|
}
|
||||||
|
if isBack == 0 && cur-b > 0 && !visited[cur-b] {
|
||||||
|
visited[cur-b] = true
|
||||||
|
queue = append(queue, [2]int{cur - b, 1})
|
||||||
|
}
|
||||||
|
if cur+a < len(visited) && !visited[cur+a] {
|
||||||
|
visited[cur+a] = true
|
||||||
|
queue = append(queue, [2]int{cur + a, 0})
|
||||||
|
}
|
||||||
|
}
|
||||||
|
queue = queue[length:]
|
||||||
|
}
|
||||||
|
return -1
|
||||||
|
}
|
||||||
|
```
|
@ -0,0 +1,112 @@
|
|||||||
|
# [1655. Distribute Repeating Integers](https://leetcode.com/problems/distribute-repeating-integers/)
|
||||||
|
|
||||||
|
|
||||||
|
## 题目
|
||||||
|
|
||||||
|
You are given an array of `n` integers, `nums`, where there are at most `50` unique values in the array. You are also given an array of `m` customer order quantities, `quantity`, where `quantity[i]` is the amount of integers the `ith` customer ordered. Determine if it is possible to distribute `nums` such that:
|
||||||
|
|
||||||
|
- The `ith` customer gets **exactly** `quantity[i]` integers,
|
||||||
|
- The integers the `ith` customer gets are **all equal**, and
|
||||||
|
- Every customer is satisfied.
|
||||||
|
|
||||||
|
Return `true` *if it is possible to distribute* `nums` *according to the above conditions*.
|
||||||
|
|
||||||
|
**Example 1:**
|
||||||
|
|
||||||
|
```
|
||||||
|
Input: nums = [1,2,3,4], quantity = [2]
|
||||||
|
Output: false
|
||||||
|
Explanation: The 0th customer cannot be given two different integers.
|
||||||
|
```
|
||||||
|
|
||||||
|
**Example 2:**
|
||||||
|
|
||||||
|
```
|
||||||
|
Input: nums = [1,2,3,3], quantity = [2]
|
||||||
|
Output: true
|
||||||
|
Explanation: The 0th customer is given [3,3]. The integers [1,2] are not used.
|
||||||
|
```
|
||||||
|
|
||||||
|
**Example 3:**
|
||||||
|
|
||||||
|
```
|
||||||
|
Input: nums = [1,1,2,2], quantity = [2,2]
|
||||||
|
Output: true
|
||||||
|
Explanation: The 0th customer is given [1,1], and the 1st customer is given [2,2].
|
||||||
|
```
|
||||||
|
|
||||||
|
**Example 4:**
|
||||||
|
|
||||||
|
```
|
||||||
|
Input: nums = [1,1,2,3], quantity = [2,2]
|
||||||
|
Output: false
|
||||||
|
Explanation: Although the 0th customer could be given [1,1], the 1st customer cannot be satisfied.
|
||||||
|
```
|
||||||
|
|
||||||
|
**Example 5:**
|
||||||
|
|
||||||
|
```
|
||||||
|
Input: nums = [1,1,1,1,1], quantity = [2,3]
|
||||||
|
Output: true
|
||||||
|
Explanation: The 0th customer is given [1,1], and the 1st customer is given [1,1,1].
|
||||||
|
```
|
||||||
|
|
||||||
|
**Constraints:**
|
||||||
|
|
||||||
|
- `n == nums.length`
|
||||||
|
- `1 <= n <= 105`
|
||||||
|
- `1 <= nums[i] <= 1000`
|
||||||
|
- `m == quantity.length`
|
||||||
|
- `1 <= m <= 10`
|
||||||
|
- `1 <= quantity[i] <= 105`
|
||||||
|
- There are at most `50` unique values in `nums`.
|
||||||
|
|
||||||
|
## 题目大意
|
||||||
|
|
||||||
|
给你一个长度为 n 的整数数组 nums ,这个数组中至多有 50 个不同的值。同时你有 m 个顾客的订单 quantity ,其中,整数 quantity[i] 是第 i 位顾客订单的数目。请你判断是否能将 nums 中的整数分配给这些顾客,且满足:
|
||||||
|
|
||||||
|
- 第 i 位顾客 恰好 有 quantity[i] 个整数。
|
||||||
|
- 第 i 位顾客拿到的整数都是 相同的 。
|
||||||
|
- 每位顾客都满足上述两个要求。
|
||||||
|
|
||||||
|
如果你可以分配 nums 中的整数满足上面的要求,那么请返回 true ,否则返回 false 。
|
||||||
|
|
||||||
|
## 解题思路
|
||||||
|
|
||||||
|
- 给定一个数组 nums,订单数组 quantity,要求按照订单满足顾客的需求。如果能满足输出 true,不能满足输出 false。
|
||||||
|
- 用 DFS 记忆化暴力搜索。代码实现不难。(不知道此题为什么是 Hard)
|
||||||
|
|
||||||
|
## 代码
|
||||||
|
|
||||||
|
```go
|
||||||
|
package leetcode
|
||||||
|
|
||||||
|
func canDistribute(nums []int, quantity []int) bool {
|
||||||
|
freq := make(map[int]int)
|
||||||
|
for _, n := range nums {
|
||||||
|
freq[n]++
|
||||||
|
}
|
||||||
|
return dfs(freq, quantity)
|
||||||
|
}
|
||||||
|
|
||||||
|
func dfs(freq map[int]int, quantity []int) bool {
|
||||||
|
if len(quantity) == 0 {
|
||||||
|
return true
|
||||||
|
}
|
||||||
|
visited := make(map[int]bool)
|
||||||
|
for i := range freq {
|
||||||
|
if visited[freq[i]] {
|
||||||
|
continue
|
||||||
|
}
|
||||||
|
visited[freq[i]] = true
|
||||||
|
if freq[i] >= quantity[0] {
|
||||||
|
freq[i] -= quantity[0]
|
||||||
|
if dfs(freq, quantity[1:]) {
|
||||||
|
return true
|
||||||
|
}
|
||||||
|
freq[i] += quantity[0]
|
||||||
|
}
|
||||||
|
}
|
||||||
|
return false
|
||||||
|
}
|
||||||
|
```
|
@ -552,6 +552,10 @@ headless: true
|
|||||||
- [1647.Minimum-Deletions-to-Make-Character-Frequencies-Unique]({{< relref "/ChapterFour/1647.Minimum-Deletions-to-Make-Character-Frequencies-Unique.md" >}})
|
- [1647.Minimum-Deletions-to-Make-Character-Frequencies-Unique]({{< relref "/ChapterFour/1647.Minimum-Deletions-to-Make-Character-Frequencies-Unique.md" >}})
|
||||||
- [1648.Sell-Diminishing-Valued-Colored-Balls]({{< relref "/ChapterFour/1648.Sell-Diminishing-Valued-Colored-Balls.md" >}})
|
- [1648.Sell-Diminishing-Valued-Colored-Balls]({{< relref "/ChapterFour/1648.Sell-Diminishing-Valued-Colored-Balls.md" >}})
|
||||||
- [1649.Create-Sorted-Array-through-Instructions]({{< relref "/ChapterFour/1649.Create-Sorted-Array-through-Instructions.md" >}})
|
- [1649.Create-Sorted-Array-through-Instructions]({{< relref "/ChapterFour/1649.Create-Sorted-Array-through-Instructions.md" >}})
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|
- [1652.Defuse-the-Bomb]({{< relref "/ChapterFour/1652.Defuse-the-Bomb.md" >}})
|
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|
- [1653.Minimum-Deletions-to-Make-String-Balanced]({{< relref "/ChapterFour/1653.Minimum-Deletions-to-Make-String-Balanced.md" >}})
|
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|
- [1654.Minimum-Jumps-to-Reach-Home]({{< relref "/ChapterFour/1654.Minimum-Jumps-to-Reach-Home.md" >}})
|
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|
- [1655.Distribute-Repeating-Integers]({{< relref "/ChapterFour/1655.Distribute-Repeating-Integers.md" >}})
|
||||||
- [1656.Design-an-Ordered-Stream]({{< relref "/ChapterFour/1656.Design-an-Ordered-Stream.md" >}})
|
- [1656.Design-an-Ordered-Stream]({{< relref "/ChapterFour/1656.Design-an-Ordered-Stream.md" >}})
|
||||||
- [1657.Determine-if-Two-Strings-Are-Close]({{< relref "/ChapterFour/1657.Determine-if-Two-Strings-Are-Close.md" >}})
|
- [1657.Determine-if-Two-Strings-Are-Close]({{< relref "/ChapterFour/1657.Determine-if-Two-Strings-Are-Close.md" >}})
|
||||||
- [1658.Minimum-Operations-to-Reduce-X-to-Zero]({{< relref "/ChapterFour/1658.Minimum-Operations-to-Reduce-X-to-Zero.md" >}})
|
- [1658.Minimum-Operations-to-Reduce-X-to-Zero]({{< relref "/ChapterFour/1658.Minimum-Operations-to-Reduce-X-to-Zero.md" >}})
|
||||||
|
Reference in New Issue
Block a user