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Smallest Substring With Identical Characters II

Difficulty: Hard

Problem Description

You are given a binary string s of length n and an integer numOps. You are allowed to perform the following operation on s at most numOps times: Select any index i (where 0 <= i < n) and flip s[i]. If s[i] == '1', change s[i] to '0' and vice versa. You need to minimize the length of the longest substring of s such that all the characters in the substring are identical. Return the minimum length after the operations.

Key Insights

  • The problem requires minimizing the length of the longest substring of identical characters after performing at most numOps flips.
  • A two-pointer or sliding window approach can be effective for determining segments of identical characters.
  • The number of operations allowed (numOps) can be used to convert a segment of one character into another, thereby reducing the longest segment.

Space and Time Complexity

Time Complexity: O(n)
Space Complexity: O(1)

Solution

To solve this problem, we can utilize a two-pointer or sliding window technique to track segments of identical characters. The idea is to maintain a window that counts the number of flips required to convert a section of the string into a homogeneous segment. By adjusting the window size and counting the flips, we can determine the maximum length of identical characters that can be achieved with the allowed flips.

  1. Use two pointers to represent the current segment of the window.
  2. Count the number of flips needed to convert the characters within the window to either '0' or '1'.
  3. If the number of flips exceeds numOps, move the left pointer to reduce the window size.
  4. Continuously track the length of the longest segment of identical characters that can be achieved.

Code Solutions

def min_length_after_operations(s: str, numOps: int) -> int:
    n = len(s)
    left = 0
    max_length = 0
    
    for right in range(n):
        # Count the number of '0's and '1's in the current window
        if s[right] == '0':
            numOps -= 1  # We're considering flipping this '0' to '1'
        
        # If we exceed the allowed flips, shrink the window from the left
        while numOps < 0:
            if s[left] == '0':
                numOps += 1  # We're flipping this '1' back to '0'
            left += 1
        
        # Update the maximum length of the current window
        max_length = max(max_length, right - left + 1)
    
    return n - max_length  # The minimum length of the longest substring
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