Problem Description

You have a binary string s consisting of only '0's and '1's with length n, and you're allowed to perform up to numOps operations on it.

Each operation allows you to:

• Choose any index i (where 0 <= i < n)
• Flip the character at that position (change '0' to '1' or '1' to '0')

Your goal is to minimize the length of the longest substring where all characters are identical (all '0's or all '1's).

For example:

• In string "111000", the longest substring of identical characters has length 3 (either "111" or "000")
• After flipping the character at index 2 to get "110000", the longest substring now has length 4 ("0000")
• After flipping the character at index 3 to get "110100", the longest substring now has length 2 (either "11" or "00")

The function should return the minimum possible length of such a longest identical substring after performing at most numOps flips optimally.

The solution uses binary search to find the minimum achievable length. For each candidate length m, it checks whether it's possible to ensure no identical substring exceeds length m using at most numOps operations. The check function handles two cases:

• When m = 1: The string needs to alternate between '0' and '1', so it counts how many positions need flipping to achieve either "010101..." or "101010..." pattern
• When m > 1: It counts how many flips are needed within each group of consecutive identical characters to break them into segments of length at most m

Intuition

The key insight is that we want to find the minimum possible value for the maximum length of consecutive identical characters. This is a classic optimization problem where we're minimizing a maximum value - a perfect candidate for binary search.

Why binary search? The answer lies in the monotonic property: if we can achieve a maximum length of m with some number of operations, then we can definitely achieve any length greater than m with the same or fewer operations. Conversely, if we cannot achieve length m, we cannot achieve any length smaller than m either.

For any target maximum length m, we need to figure out the minimum number of flips required to ensure no consecutive identical substring exceeds this length. This breaks down into two distinct strategies:

1. When m = 1: We need a completely alternating pattern like "010101..." or "101010...". We can count how many positions differ from each pattern and choose the one requiring fewer flips. This is optimal because any substring of length 2 or more would contain different characters.

2. When m > 1: We need to break up long runs of identical characters. For a run of k consecutive identical characters, we need to place flips strategically to split it into segments of at most m characters. The optimal way is to place flips every m+1 positions within the run, requiring k // (m+1) flips.

The binary search tries different values of m from 1 to n, checking for each whether it's achievable with at most numOps operations. The smallest achievable m is our answer.

This approach is efficient because instead of trying all possible combinations of flips (which would be exponential), we're leveraging the problem's structure to directly compute the minimum flips needed for any given target length.

Learn more about Binary Search patterns.

Solution Approach

The solution implements a binary search combined with a validation function to find the minimum achievable maximum length of consecutive identical characters.

Binary Search Setup: The main function uses bisect_left to perform binary search on the range [1, n] where n is the length of the string. We're searching for the smallest value m where we can successfully limit all consecutive identical substrings to at most length m.

The Check Function: The check(m) function determines whether it's possible to achieve a maximum consecutive length of m with at most numOps operations.

Case 1: When m = 1

if m == 1:
    t = "01"
    cnt = sum(c == t[i & 1] for i, c in enumerate(s))
    cnt = min(cnt, n - cnt)

• We need the string to alternate completely (no two adjacent characters can be the same)
• We check two patterns: "010101..." and "101010..."
• t[i & 1] generates the alternating pattern: when i is even, i & 1 = 0 gives '0'; when i is odd, i & 1 = 1 gives '1'
• Count mismatches with the "01" pattern starting with '0'
• Since we can also start with '1', the alternative pattern would require n - cnt flips
• Take the minimum of both options

Case 2: When m > 1

else:
    k = 0
    for i, c in enumerate(s):
        k += 1
        if i == len(s) - 1 or c != s[i + 1]:
            cnt += k // (m + 1)
            k = 0

• Iterate through the string to identify consecutive groups of identical characters
• k tracks the current group size
• When we reach the end of a group (character changes or string ends):
  • Calculate flips needed: k // (m + 1)
  • This formula works because placing a flip every m+1 positions optimally breaks a group of size k into segments of at most m
  • Reset k for the next group

Final Check:

return cnt <= numOps

Returns True if the required number of operations doesn't exceed numOps.

Binary Search Execution: bisect_left(range(n), True, lo=1, key=check) finds the leftmost (smallest) value in the range [1, n] where check(m) returns True. This gives us the minimum achievable maximum length of consecutive identical characters.

Example Walkthrough

Let's walk through the solution with s = "111000" and numOps = 2.

Initial Analysis:

• String: "111000" (length n = 6)
• Current longest consecutive identical substring: 3 (both "111" and "000")
• We have 2 operations available

Binary Search Process:

We'll binary search on the range [1, 6] to find the minimum achievable maximum length.

Try m = 3 (midpoint of [1, 6]):

• Check if we can limit all consecutive substrings to length 3
• Current groups: "111" (length 3), "000" (length 3)
• For group "111": 3 // (3+1) = 3 // 4 = 0 flips needed
• For group "000": 3 // (3+1) = 3 // 4 = 0 flips needed
• Total flips needed: 0 ≤ 2 ✓
• Since this works, try smaller values

Try m = 1:

• Need alternating pattern: either "010101" or "101010"
• Compare with "010101": positions 0,1,2 need flipping → 3 flips
• Compare with "101010": positions 3,4,5 need flipping → 3 flips
• Minimum flips needed: 3 > 2 ✗
• Cannot achieve m = 1

Try m = 2:

• Need to break groups longer than 2
• Group "111" (length 3): 3 // (2+1) = 3 // 3 = 1 flip needed
  • One flip at position 1 or 2 would give "101000" or "110000"
• Group "000" (length 3): 3 // (2+1) = 3 // 3 = 1 flip needed
  • One flip at position 3 or 4 would give "110100" or "111010"
• Total flips needed: 2 ≤ 2 ✓
• This works!

Binary Search Conclusion:

• m = 1 doesn't work (needs 3 flips)
• m = 2 works (needs 2 flips)
• Therefore, the minimum achievable maximum length is 2

Verification: After flipping positions 1 and 4: "111000" → "101010"

• All consecutive identical substrings have length ≤ 1 Actually, we can also flip positions 2 and 3: "111000" → "110100"
• Consecutive identical substrings: "11", "01", "00" (max length = 2)

The answer is 2.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n * log n)

The algorithm uses binary search on the range [1, n] where n is the length of string s. The binary search performs O(log n) iterations.

For each iteration of the binary search, the check function is called:

• When m == 1: The function iterates through the string once to count mismatches with alternating patterns "01" or "10", taking O(n) time.
• When m > 1: The function iterates through the string once to count consecutive segments and calculate required operations, taking O(n) time.

Since each check call takes O(n) time and binary search makes O(log n) calls, the overall time complexity is O(n * log n).

Space Complexity: O(1)

The algorithm uses only a constant amount of extra space:

• The check function uses variables cnt, t (when m == 1), and k which require O(1) space.
• The binary search itself only uses a few variables for bounds and mid-point calculations.
• No additional data structures proportional to input size are created.

Therefore, the space complexity is O(1).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Formula for Breaking Consecutive Sequences

The Pitfall: When calculating how many operations are needed to break a consecutive sequence of length k into segments of at most length m, developers often use incorrect formulas like:

• (k - 1) // m - This underestimates the operations needed
• k // m - This doesn't account for the spacing correctly
• (k + m - 1) // m - This is for counting segments, not operations

Why It Happens: The confusion arises from not clearly understanding the relationship between segment placement and the number of flips needed. If you have a string of k consecutive identical characters and want no segment longer than m, you need to place flips strategically.

The Correct Formula: k // (m + 1) is correct because:

• To break a sequence into segments of at most m, you place a flip every m+1 positions
• For example, with "111111" (k=6) and m=2:
  • Place flips at positions 2 and 5: "11011011"
  • This requires 2 flips = 6 // (2+1) = 6 // 3 = 2

Solution: Always use k // (m + 1) for calculating the minimum operations needed to break a consecutive sequence.

2. Mishandling the Alternating Pattern Case (m = 1)

The Pitfall: When m = 1, developers might:

• Only check one alternating pattern ("010101...") without considering the complement
• Use incorrect indexing like i % 2 instead of i & 1 (though both work, bitwise is more efficient)
• Count the matches instead of mismatches

Example of Wrong Implementation:

if m == 1:
    cnt = sum(1 for i, c in enumerate(s) if c != str(i % 2))
    # Missing: doesn't consider the "101010..." pattern

The Correct Approach:

if m == 1:
    pattern = "01"
    # Count mismatches with "010101..." pattern
    cnt = sum(c == pattern[i & 1] for i, c in enumerate(s))
    # Consider both patterns and take minimum
    cnt = min(cnt, n - cnt)

3. Off-by-One Errors in Binary Search Range

The Pitfall: Using incorrect bounds for binary search:

• Starting from 0 instead of 1 (a consecutive sequence must have at least length 1)
• Using n+1 as the upper bound when n is sufficient

Why It Matters:

• Starting from 0 would be meaningless (no consecutive sequence has length 0)
• The maximum possible consecutive length is n (the entire string), so searching beyond n is unnecessary

Solution: Always use lo=1 in the binary search since 1 is the minimum meaningful length for consecutive sequences.

4. Confusing the Binary Search Target

The Pitfall: Misunderstanding what the binary search is looking for. The code uses bisect_left to find where the check function first returns True, but developers might:

• Think we're looking for False values
• Use bisect_right instead, which would give a different result
• Reverse the logic in the check function

The Correct Understanding:

• We want the minimum achievable maximum length
• The check function returns True if we can achieve that maximum length with available operations
• bisect_left finds the smallest value where this is possible

Solution: Ensure the check function returns True when the target is achievable, and use bisect_left to find the minimum such value.