Sliding Window Technique - Part 5
Standard sliding window technique here, my implementation was a little more convoluted this time around (I'm sure there is a clearer way to detect the window) but did the work. Notice that the constraints are very generous, hence even an N^2 solution (2-nested loops) would be able to solve this problem (the sliding window solves it in linear time). Code is down below, cheers, ACC.
Minimum Recolors to Get K Consecutive Black Blocks - LeetCode
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
public class Solution { public int MinimumRecolors(string blocks, int k) { int retVal = Int32.MaxValue; int currentWindowRecolors = 0; int left = 0; int right = -1; for (; ; ) { right++; if (right >= blocks.Length) break; if (right - left + 1 > k) { if (blocks[left] == 'W') currentWindowRecolors--; left++; } if (blocks[right] == 'W') currentWindowRecolors++; if(right >= k - 1) retVal = Math.Min(retVal, currentWindowRecolors); } return retVal; } }
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