Question

2379. Minimum Recolors to Get K Consecutive Black Blocks

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Description

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

Solutions

Solution 1

class Solution:
  def minimumRecolors(self, blocks: str, k: int) -> int:
    ans = cnt = blocks[:k].count('W')
    for i in range(k, len(blocks)):
      cnt += blocks[i] == 'W'
      cnt -= blocks[i - k] == 'W'
      ans = min(ans, cnt)
    return ans

class Solution {
  public int minimumRecolors(String blocks, int k) {
    int cnt = 0;
    for (int i = 0; i < k; ++i) {
      cnt += blocks.charAt(i) == 'W' ? 1 : 0;
    }
    int ans = cnt;
    for (int i = k; i < blocks.length(); ++i) {
      cnt += blocks.charAt(i) == 'W' ? 1 : 0;
      cnt -= blocks.charAt(i - k) == 'W' ? 1 : 0;
      ans = Math.min(ans, cnt);
    }
    return ans;
  }
}

class Solution {
public:
  int minimumRecolors(string blocks, int k) {
    int cnt = count(blocks.begin(), blocks.begin() + k, 'W');
    int ans = cnt;
    for (int i = k; i < blocks.size(); ++i) {
      cnt += blocks[i] == 'W';
      cnt -= blocks[i - k] == 'W';
      ans = min(ans, cnt);
    }
    return ans;
  }
};

func minimumRecolors(blocks string, k int) int {
  cnt := strings.Count(blocks[:k], "W")
  ans := cnt
  for i := k; i < len(blocks); i++ {
    if blocks[i] == 'W' {
      cnt++
    }
    if blocks[i-k] == 'W' {
      cnt--
    }
    if ans > cnt {
      ans = cnt
    }
  }
  return ans
}

function minimumRecolors(blocks: string, k: number): number {
  let cnt = 0;
  for (let i = 0; i < k; ++i) {
    cnt += blocks[i] === 'W' ? 1 : 0;
  }
  let ans = cnt;
  for (let i = k; i < blocks.length; ++i) {
    cnt += blocks[i] === 'W' ? 1 : 0;
    cnt -= blocks[i - k] === 'W' ? 1 : 0;
    ans = Math.min(ans, cnt);
  }
  return ans;
}

impl Solution {
  pub fn minimum_recolors(blocks: String, k: i32) -> i32 {
    let k = k as usize;
    let s = blocks.as_bytes();
    let n = s.len();
    let mut count = 0;
    for i in 0..k {
      if s[i] == b'B' {
        count += 1;
      }
    }
    let mut ans = k - count;
    for i in k..n {
      if s[i - k] == b'B' {
        count -= 1;
      }
      if s[i] == b'B' {
        count += 1;
      }
      ans = ans.min(k - count);
    }
    ans as i32
  }
}

class Solution {
  /**
   * @param String $blocks
   * @param Integer $k
   * @return Integer
   */
  function minimumRecolors($blocks, $k) {
    $cnt = 0;
    for ($i = 0; $i < $k; $i++) {
      if ($blocks[$i] === 'W') {
        $cnt++;
      }
    }
    $min = $cnt;
    for ($i = $k; $i < strlen($blocks); $i++) {
      if ($blocks[$i] === 'W') {
        $cnt++;
      }
      if ($blocks[$i - $k] === 'W') {
        $cnt--;
      }
      $min = min($min, $cnt);
    }
    return $min;
  }
}

#define min(a, b) (((a) < (b)) ? (a) : (b))

int minimumRecolors(char* blocks, int k) {
  int n = strlen(blocks);
  int count = 0;
  for (int i = 0; i < k; i++) {
    count += blocks[i] == 'B';
  }
  int ans = k - count;
  for (int i = k; i < n; i++) {
    count -= blocks[i - k] == 'B';
    count += blocks[i] == 'B';
    ans = min(ans, k - count);
  }
  return ans;
}