comments | difficulty | edit_url | rating | source | tags | ||
---|---|---|---|---|---|---|---|
true |
Easy |
1360 |
Biweekly Contest 85 Q1 |
|
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
We observe that what the problem actually asks for is the minimum number of white blocks in a sliding window of size
Therefore, we only need to traverse the string
After the traversal ends, we can get the answer.
The time complexity is
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;
}