# [1209. Remove All Adjacent Duplicates in String II](https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/)


## 题目

Given a string `s`, a *k* *duplicate removal* consists of choosing `k` adjacent and equal letters from `s` and removing them causing the left and the right side of the deleted substring to concatenate together.

We repeatedly make `k` duplicate removals on `s` until we no longer can.

Return the final string after all such duplicate removals have been made.

It is guaranteed that the answer is unique.

**Example 1:**

```
Input: s = "abcd", k = 2
Output: "abcd"
Explanation:There's nothing to delete.
```

**Example 2:**

```
Input: s = "deeedbbcccbdaa", k = 3
Output: "aa"
Explanation:
First delete "eee" and "ccc", get "ddbbbdaa"
Then delete "bbb", get "dddaa"
Finally delete "ddd", get "aa"
```

**Example 3:**

```
Input: s = "pbbcggttciiippooaais", k = 2
Output: "ps"
```

**Constraints:**

- `1 <= s.length <= 10^5`
- `2 <= k <= 10^4`
- `s` only contains lower case English letters.

## 题目大意

给你一个字符串 s，「k 倍重复项删除操作」将会从 s 中选择 k 个相邻且相等的字母，并删除它们，使被删去的字符串的左侧和右侧连在一起。你需要对 s 重复进行无限次这样的删除操作，直到无法继续为止。在执行完所有删除操作后，返回最终得到的字符串。本题答案保证唯一。

## 解题思路

- 暴力解法。每增加一个字符，就往前扫描 `k` 位，判断是否存在 `k` 个连续相同的字符。消除了 `k` 个相同字符后，重新组成的字符串还可能再次产出 `k` 位相同的字符，（类似消消乐，`k` 个相同的字符碰到一起就“消除”），还需要继续消除。最差情况要再次扫描一次字符串。时间复杂度 O(n^2)，空间复杂度 O(n)。
- 暴力解法的低效在于重复统计字符频次，如果每个字符的频次统计一次就好了。按照这个思路，利用 stack ，每个栈元素存 2 个值，一个是字符，一个是该字符对应的频次。有了栈顶字符频次信息，就不需要重复往前扫描了。只要栈顶字符频次到达了 `k`，就弹出该字符。如此反复，最终剩下的字符串为所求。时间复杂度 O(n)，空间复杂度 O(n)。

## 代码

```go
package leetcode

// 解法一 stack
func removeDuplicates(s string, k int) string {
	stack, arr := [][2]int{}, []byte{}
	for _, c := range s {
		i := int(c - 'a')
		if len(stack) > 0 && stack[len(stack)-1][0] == i {
			stack[len(stack)-1][1]++
			if stack[len(stack)-1][1] == k {
				stack = stack[:len(stack)-1]
			}
		} else {
			stack = append(stack, [2]int{i, 1})
		}
	}
	for _, pair := range stack {
		c := byte(pair[0] + 'a')
		for i := 0; i < pair[1]; i++ {
			arr = append(arr, c)
		}
	}
	return string(arr)
}

// 解法二 暴力
func removeDuplicates1(s string, k int) string {
	arr, count, tmp := []rune{}, 0, '#'
	for _, v := range s {
		arr = append(arr, v)
		for len(arr) > 0 {
			count = 0
			tmp = arr[len(arr)-1]
			for i := len(arr) - 1; i >= 0; i-- {
				if arr[i] != tmp {
					break
				}
				count++
			}
			if count == k {
				arr = arr[:len(arr)-k]
			} else {
				break
			}
		}
	}
	return string(arr)
}
```