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# Minimum Recolors To Get K Consecutive Black Blocks - Leetcode Solution

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LeetCode: Minimum Recolors To Get K Consecutive Black Blocks Leetcode Solution

Difficulty: Easy

Topics: string sliding-window

Problem Statement:

Given a binary string s (a string consisting only of '0' and '1') and a positive integer k, you need to find the minimum number of recolors required to make k consecutive black blocks in the resulting string. A recolor operation is defined as flipping all the bits of a substring in s. In other words, if s[i]=0 and s[i+m-1]=1, then we can flip s[i] to 1, s[i+1] to 0... and s[i+m-1] to 0, such that s[i:i+m-1] becomes a block of 1s.

Solution:

We can solve this problem using two-pointer approach. We will keep two pointers left and right which will represent the starting and ending index of the current substring and the number of black blocks we have seen so far in this substring respectively. We will start with both pointers pointing to the beginning of the string.

We will use a sliding window approach to expand the substring to the right as long as we have less than k black blocks in this substring. Whenever we encounter a '0' in the string we will increase the count of black blocks.

Once we have k black blocks in the current substring, we can calculate the number of recolors required to make this substring black. For this we just need to check the length of the substring and subtract the count of '1's in it. The count of '1's in the current substring can be calculated using two-pointer approach as well by maintaining a count of '1's in the substring and updating it whenever we move the left or right pointer.

Once we know the number of recolors required for the current substring, we will slide the window by moving the left and right pointers to the right by one position and updating the count of black blocks and '1's in the substring accordingly. We will keep track of the minimum number of recolors required so far.

We will continue this process until we have processed the entire string.

Algorithm:

- Initialize left and right pointers to 0.
- Initialize a count of black blocks and '1's to 0.
- Initialize the minimum number of recolors required so far to the length of the string.
- While the right pointer is less than the length of the string a. If the count of black blocks is less than k i. If s[right] is '0', increment the count of black blocks b. If the count of black blocks is equal to k i. Calculate the number of recolors required for the current substring: 1. Count the number of '1's in the substring 2. Calculate the number of recolors as the length of the substring minus the count of '1's ii. Update the minimum number of recolors required so far. iii. Slide the window by incrementing both left and right pointers by one and updating the count of black blocks and '1's in the substring.
- Return the minimum number of recolors required.

Time Complexity:

The algorithm will take O(n) time to process the entire string, where n is the length of the string. The time complexity of calculating the number of recolors for a substring and sliding the window by one is O(1). Therefore, the overall time complexity of the algorithm is O(n).

Space Complexity:

The algorithm uses constant amount of extra space to keep track of left and right pointers, the count of black blocks and '1's and the minimum number of recolors required so far. Therefore, the space complexity of the algorithm is O(1).

Implementation:

Here is an implementation of the algorithm in Python:

def minRecolor(s: str, k: int) -> int: n = len(s) left = right = 0 count_blk = count_ones = 0 min_recolor = n

```
while right < n:
if count_blk < k:
if s[right] == '0':
count_blk += 1
right += 1
else:
num_ones = count_ones = 0
for i in range(left, right):
if s[i] == '1':
num_ones += 1
min_recolor = min(min_recolor, right - left - num_ones)
if s[left] == '0':
count_blk -= 1
left += 1
while count_blk == k:
num_ones = count_ones = 0
for i in range(left, right):
if s[i] == '1':
num_ones += 1
min_recolor = min(min_recolor, right - left - num_ones)
if s[left] == '0':
count_blk -= 1
left += 1
return min_recolor
```

We first initialize left, right, count_blk, count_ones, and min_recolor to 0. We use a while loop to iterate over the string as long as right is less than the length of the string.

If count_blk is less than k, we check if the character at the right pointer is '0'. If it is '0', we increment the count_blk and move the right pointer one step to the right.

If count_blk is equal to k, we calculate the number of ones in the current substring by iterating over the substring from left to right and counting the '1's. We then calculate the number of recolors as the length of the substring minus the count of '1's, update min_recolor if necessary, and slide the window to the right by incrementing left and right pointers by one and updating the count_blk and count_ones accordingly.

After the while loop, we check if count_blk is equal to k and we need to process the last substring. We repeat the same process of calculating the number of ones and the number of recolors.

Finally, we return the min_recolor.

## Minimum Recolors To Get K Consecutive Black Blocks Solution Code

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