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111 lines
3.8 KiB
Markdown
111 lines
3.8 KiB
Markdown
# [1738. Find Kth Largest XOR Coordinate Value](https://leetcode.com/problems/find-kth-largest-xor-coordinate-value/)
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## 题目
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You are given a 2D `matrix` of size `m x n`, consisting of non-negative integers. You are also given an integer `k`.
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The **value** of coordinate `(a, b)` of the matrix is the XOR of all `matrix[i][j]` where `0 <= i <= a < m` and `0 <= j <= b < n` **(0-indexed)**.
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Find the `kth` largest value **(1-indexed)** of all the coordinates of `matrix`.
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**Example 1:**
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```
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Input: matrix = [[5,2],[1,6]], k = 1
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Output: 7
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Explanation: The value of coordinate (0,1) is 5 XOR 2 = 7, which is the largest value.
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```
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**Example 2:**
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```
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Input: matrix = [[5,2],[1,6]], k = 2
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Output: 5
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Explanation:The value of coordinate (0,0) is 5 = 5, which is the 2nd largest value.
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```
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**Example 3:**
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```
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Input: matrix = [[5,2],[1,6]], k = 3
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Output: 4
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Explanation: The value of coordinate (1,0) is 5 XOR 1 = 4, which is the 3rd largest value.
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```
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**Example 4:**
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```
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Input: matrix = [[5,2],[1,6]], k = 4
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Output: 0
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Explanation: The value of coordinate (1,1) is 5 XOR 2 XOR 1 XOR 6 = 0, which is the 4th largest value.
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```
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**Constraints:**
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- `m == matrix.length`
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- `n == matrix[i].length`
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- `1 <= m, n <= 1000`
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- `0 <= matrix[i][j] <= 10^6`
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- `1 <= k <= m * n`
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## 题目大意
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给你一个二维矩阵 matrix 和一个整数 k ，矩阵大小为 m x n 由非负整数组成。矩阵中坐标 (a, b) 的 值 可由对所有满足 0 <= i <= a < m 且 0 <= j <= b < n 的元素 matrix[i][j]（下标从 0 开始计数）执行异或运算得到。请你找出 matrix 的所有坐标中第 k 大的值（k 的值从 1 开始计数）。
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## 解题思路
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- 区间异或结果类比于区间二维前缀和。只不过需要注意 x^x = 0 这一性质。举例：
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通过简单推理，可以得出区间二维前缀和 preSum 的递推式。具体代码见解法二。
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- 上面的解法中，preSum 用二维数组计算的。能否再优化空间复杂度，降低成 O(n)？答案是可以的。通过观察可以发现。preSum 可以按照一行一行来生成。先生成 preSum 前一行，下一行生成过程中会用到前一行的信息，异或计算以后，可以覆盖原数据(前一行的信息)，对之后的计算没有影响。这个优化空间复杂度的方法和优化 DP 空间复杂度是完全一样的思路和方法。
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具体代码见解法一。
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- 计算出了 preSum，还需要考虑如何输出第 k 大的值。有 3 种做法，第一种是排序，第二种是优先队列，第三种是第 215 题中的 O(n) 的 partition 方法。时间复杂度最低的当然是 O(n)。但是经过实际测试，runtime 最优的是排序的方法。所以笔者以下两种方法均采用了排序的方法。
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## 代码
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```go
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package leetcode
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import "sort"
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// 解法一 压缩版的前缀和
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func kthLargestValue(matrix [][]int, k int) int {
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if len(matrix) == 0 || len(matrix[0]) == 0 {
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return 0
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}
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res, prefixSum := make([]int, 0, len(matrix)*len(matrix[0])), make([]int, len(matrix[0]))
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for i := range matrix {
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line := 0
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for j, v := range matrix[i] {
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line ^= v
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prefixSum[j] ^= line
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res = append(res, prefixSum[j])
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}
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}
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sort.Ints(res)
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return res[len(res)-k]
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}
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// 解法二 前缀和
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func kthLargestValue1(matrix [][]int, k int) int {
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nums, prefixSum := []int{}, make([][]int, len(matrix)+1)
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prefixSum[0] = make([]int, len(matrix[0])+1)
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for i, row := range matrix {
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prefixSum[i+1] = make([]int, len(matrix[0])+1)
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for j, val := range row {
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prefixSum[i+1][j+1] = prefixSum[i+1][j] ^ prefixSum[i][j+1] ^ prefixSum[i][j] ^ val
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nums = append(nums, prefixSum[i+1][j+1])
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}
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}
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sort.Ints(nums)
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return nums[len(nums)-k]
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}
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``` |