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.length1 <= n <= 100blocks[i] is either 'W' or 'B'.1 <= k <= n
Solutions
Solution 1: Sliding Window
We observe that what the problem actually asks for is the minimum number of white blocks in a sliding window of size k.
Therefore, we only need to traverse the string blocks, use a variable cnt to count the number of white blocks in the current window, and then use a variable ans to maintain the minimum value.
After the traversal ends, we can get the answer.
The time complexity is O(n), where n is the length of the string blocks. The space complexity is O(1).
PythonJavaC++GoTypeScriptRustPHPC
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(code-box)
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;
}
}(code-box)
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;
}
};(code-box)
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
}(code-box)
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;
}(code-box)
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
}
}(code-box)
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;
}
}(code-box)
#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;
}(code-box)
