You are given a 0-indexed string blocks of length n, where blocks[i] is either 'W' or 'B', representing the color of the ith block. The characters 'W' and 'B' denote the colors white and black, respectively.
You are also given an integer k, which is the desired number of consecutive black blocks.
In one operation, you can recolor a white block such that it becomes a black block.
Return the minimum number of operations needed such that there is at least one occurrence of k consecutive black blocks.
Example 1:
Input: blocks = "WBBWWBBWBW", k = 7
Output: 3
Explanation:
One way to achieve 7 consecutive black blocks is to recolor the 0th, 3rd, and 4th blocks
so that blocks = "BBBBBBBWBW".
It can be shown that there is no way to achieve 7 consecutive black blocks in less than 3 operations.
Therefore, we return 3.
Example 2:
Input: blocks = "WBWBBBW", k = 2
Output: 0
Explanation:
No changes need to be made, since 2 consecutive black blocks already exist.
Therefore, we return 0.
Constraints:
n == blocks.length
1 <= n <= 100
blocks[i] is either 'W' or 'B'.
1 <= k <= n
得到 K 个黑块的最少涂色次数。
给你一个长度为 n 下标从 0 开始的字符串 blocks ，blocks[i] 要么是 'W' 要么是 'B' ，表示第 i 块的颜色。字符 'W' 和 'B' 分别表示白色和黑色。给你一个整数 k ，表示想要 连续 黑色块的数目。
每一次操作中，你可以选择一个白色块将它 涂成 黑色块。
请你返回至少出现 一次 连续 k 个黑色块的 最少 操作次数。
思路
思路是窗口 size 固定的滑动窗口。首先我们看前 k 个字符里有多少个白色块，白色块的数量就是需要操作的次数，记为 count。接着我们扫描字符串剩余的部分，找到全局最小的 count 即可。
复杂度
时间O(n)
空间O(1)
代码
Java实现
class Solution {public int minimumRecolors(String blocks, int k) {int n = blocks.length();int count = 0;// first kfor (int i = 0; i < k; i++) {if (blocks.charAt(i) == 'W') {count++;}}int res = count;for (int i = k; i < n; i++) {if (blocks.charAt(i) == 'W') {count++;}if (blocks.charAt(i - k) == 'W') {count--;}res = Math.min(res, count);}return res;}
}
