Welcome to Subscribe On Youtube

Formatted question description: https://leetcode.ca/all/2379.html

2379. Minimum Recolors to Get K Consecutive Black Blocks

  • Difficulty: Easy.
  • Related Topics: String, Sliding Window.
  • Similar Questions: Max Consecutive Ones III, Maximum Points You Can Obtain from Cards, Maximum Number of Vowels in a Substring of Given Length.

Problem

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

Solution (Java, C++, Python)

  • class Solution {
        public int minimumRecolors(String blocks, int k) {
            int n = blocks.length();
            int ans;
            int i;
            int cur = 0;
            for (i = 0; i < k; i++) {
                if (blocks.charAt(i) == 'W') {
                    cur++;
                }
            }
            ans = cur;
            for (i = k; i < n; i++) {
                if (blocks.charAt(i) == 'W') {
                    cur++;
                }
                if (blocks.charAt(i - k) == 'W') {
                    cur--;
                }
                ans = Math.min(ans, cur);
            }
            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:
        def minimumRecolors(self, blocks: str, k: int) -> int:
            cnt = blocks[:k].count('W')
            ans = cnt
            i, n = k, len(blocks)
            while i < n:
                cnt += blocks[i] == 'W'
                cnt -= blocks[i - k] == 'W'
                ans = min(ans, cnt)
                i += 1
            return ans
    
    ############
    
    # 2379. Minimum Recolors to Get K Consecutive Black Blocks
    # https://leetcode.com/problems/minimum-recolors-to-get-k-consecutive-black-blocks/
    
    class Solution:
        def minimumRecolors(self, blocks: str, k: int) -> int:
            n = len(blocks)
            b = 0
            res = inf
            
            for i, x in enumerate(blocks):
                if i >= k:
                    if blocks[i - k] == "B":
                        b -= 1
                        
                if x == "B":
                    b += 1
                
                if i + 1 >= k:
                    res = min(res, k - b)
            
            return res
    
    
  • 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
        }
    }
    
    

Explain:

nope.

Complexity:

  • Time complexity : O(n).
  • Space complexity : O(n).

All Problems

All Solutions