Minimum Suffix Flips - Leetcode Solution

Problem Statement:

Given a binary string s, we can flip every consecutive group of k characters. For example, if we have s = "00110", and k = 3, we can flip the first and second group of characters to get "10010", or the second and third group of characters to get "00101".

Return the minimum number of flips required to convert s into a string with no consecutive groups of the same character.

Example 1:

Input: s = "010110", k = 2 Output: 2 Explanation: We can flip the first and second, or third and fourth segments.

Example 2:

Input: s = "0010111", k = 2 Output: 1

Approach:

The problem can be solved using a greedy approach. We need to start from the beginning of the string and check if the current segment of k characters needs to be flipped. If yes, we can flip it and count it as a flip. We need to continue this process till the end of the string. We can optimize this approach by keeping track of the number of flips done so far, and the current segment that needs to be flipped.

Algorithm:

1. Initialize count variable to 0.
2. Start from the beginning of the string, and for each segment of k characters: a. Check if the segment needs to be flipped. b. If yes, flip the segment and increment the count variable by 1. c. Move to the next segment.
3. Return the count variable.

Code:

Python solution:

class Solution: def minFlips(self, s: str, k: int) -> int: n = len(s) count = 0 for i in range(n-k+1): if s[i:i+k].count('0') == k//2: continue count += 1 temp = s[i:i+k][::-1] for j in range(k): s[i+j] = '1' if temp[j] == '0' else '0' return count

Time Complexity: O(N*K)

Space Complexity: O(N)

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