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Smallest Substring With Identical Characters I

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 involves modifying a binary string to minimize the length of the longest identical substring.
  • Flipping characters can create larger blocks of identical characters, which can reduce the maximum length of identical substrings.
  • The number of allowed flips (numOps) directly affects how many segments of the string can be merged.
  • The problem can be approached using a sliding window or a two-pointer technique to evaluate potential flips and their impacts on substring lengths.

Space and Time Complexity

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

Solution

The solution can be approached using the sliding window (or two-pointer) technique. The idea is to maintain a window that represents the longest substring of identical characters while keeping track of how many flips are needed to achieve that. We can iterate through the string and expand our window until the number of required flips exceeds numOps. When we exceed the limit, we will shrink the window from the left until we are back within the limit. Finally, we can determine the minimum length of the longest substring of identical characters after performing the allowed flips.

Code Solutions

def minLengthAfterOps(s: str, numOps: int) -> int:
    left = 0
    max_length = 0
    count_0 = 0
    count_1 = 0
    
    for right in range(len(s)):
        if s[right] == '0':
            count_0 += 1
        else:
            count_1 += 1
        
        # If we want to flip all to '0'
        while count_1 > numOps:
            if s[left] == '0':
                count_0 -= 1
            else:
                count_1 -= 1
            left += 1
        
        max_length = max(max_length, right - left + 1)
    
    # Reset for the second case: flipping all to '1'
    left = 0
    count_0 = 0
    count_1 = 0
    
    for right in range(len(s)):
        if s[right] == '1':
            count_1 += 1
        else:
            count_0 += 1
        
        # If we want to flip all to '1'
        while count_0 > numOps:
            if s[left] == '1':
                count_1 -= 1
            else:
                count_0 -= 1
            left += 1
        
        max_length = max(max_length, right - left + 1)
    
    return len(s) - max_length
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