Problem Description

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

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 "111001", the longest substring with identical characters is "111" with length 3
• After some flips, you might reduce this maximum length

The task is to find the minimum possible length of such longest identical substring after performing at most numOps operations optimally.

The solution uses binary search to find this minimum length. For each candidate length m, it checks whether it's possible to achieve that maximum length using at most numOps flips:

• When m = 1, the string should alternate like "010101..." or "101010...", and we count how many flips are needed
• For m > 1, we count how many flips are needed to break up each consecutive group of identical characters into segments of at most length m

The binary search finds the smallest m where the required number of flips doesn't exceed numOps.

Intuition

The key insight is that we're looking for the minimum possible value of the maximum consecutive identical characters. This "minimize the maximum" pattern naturally suggests binary search on the answer.

Think about it this way: if we can achieve a maximum length of m, then we can definitely achieve any length greater than m (we just need fewer or the same number of flips). Conversely, if we cannot achieve length m with our available operations, we cannot achieve any length smaller than m either. This monotonic property makes binary search applicable.

For any target maximum length m, we need to figure out the minimum number of flips required to ensure no consecutive identical substring exceeds length m:

1. Special case when m = 1: This means we want no two adjacent characters to be the same - the string must alternate perfectly like "010101..." or "101010...". We can calculate flips needed for both patterns and take the minimum.

2. General case when m > 1: For each group of consecutive identical characters of length k:

   • If k <= m, no flips needed for this group
   • If k > m, we need to break it up by flipping some characters within it
   • To break a group of length k into segments of at most m, we need k // (m + 1) flips strategically placed

The formula k // (m + 1) comes from the fact that each flip creates a breakpoint, and to divide a segment of length k into parts of at most length m, we need to place flips at intervals. For example, to break "11111" (length 5) into segments of at most 2, we need 5 // 3 = 1 flip in the middle.

By binary searching on the possible values of m from 1 to n, we find the smallest m where the required flips don't exceed numOps.

Learn more about Binary Search patterns.

Solution Approach

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

Binary Search Setup:

• We use bisect_left to search in the range [1, n] where n is the length of the string
• The search finds the smallest value m where check(m) returns True

The check Function: This function determines if we can achieve a maximum consecutive length of m using at most numOps flips.

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 (no two adjacent characters can be the same)
• Two possible patterns: "010101..." or "101010..."
• For pattern "010101...", count mismatches where even indices should be '0' and odd indices should be '1'
• i & 1 gives 0 for even indices and 1 for odd indices
• n - cnt gives the count for the opposite pattern "101010..."
• Take the minimum of both patterns

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 and identify groups of consecutive identical characters
• Variable 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 gives the minimum flips to break a group of size k into segments of at most m
  • Reset k to 0 for the next group

Final Check:

• Return True if cnt <= numOps (we can achieve this maximum length with available operations)
• The binary search finds the minimum m where this condition holds

The time complexity is O(n log n) where each binary search iteration takes O(n) to validate, and we have O(log n) iterations.

Example Walkthrough

Let's walk through the solution with a concrete example:

• String s = "111001"
• numOps = 2 (we can perform up to 2 flips)

Initial Analysis:

• Current consecutive groups: "111" (length 3), "00" (length 2), "1" (length 1)
• Maximum consecutive length = 3

Binary Search Process:

We'll binary search on the answer range [1, 6] (string length is 6).

Iteration 1: Check if m = 3 is achievable

• For each group:
  • "111" (length 3): needs 3 // (3+1) = 0 flips
  • "00" (length 2): needs 2 // (3+1) = 0 flips
  • "1" (length 1): needs 1 // (3+1) = 0 flips
• Total flips needed: 0 ≤ 2 ✓
• Since we can achieve m=3, try smaller values

Iteration 2: Check if m = 2 is achievable

• For each group:
  • "111" (length 3): needs 3 // (2+1) = 1 flip
  • "00" (length 2): needs 2 // (2+1) = 0 flips
  • "1" (length 1): needs 1 // (2+1) = 0 flips
• Total flips needed: 1 ≤ 2 ✓
• Since we can achieve m=2, try smaller values

Iteration 3: Check if m = 1 is achievable

• This requires alternating pattern
• Pattern "010101": compare with "111001"
  • Positions needing flips: indices 0, 1, 3 → 3 flips
• Pattern "101010": compare with "111001"
  • Positions needing flips: indices 2, 4, 5 → 3 flips
• Minimum flips needed: 3 > 2 ✗
• Cannot achieve m=1

Result: The minimum achievable maximum consecutive length is 2.

To verify: With 1 flip at index 1, we get "101001" with maximum consecutive length of 2 ("00").

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 binary search, the check function is called:

• When m == 1: The function iterates through the string once to count mismatches with alternating patterns, 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.

Therefore, 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, k, and t (when m == 1)
• The string t = "01" when m == 1 is constant size
• The binary search itself only requires a few variables for bounds and mid-point calculations

The space used does not scale with the input size, resulting in O(1) space complexity.

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

Common Pitfalls

1. Incorrect Formula for Breaking Consecutive Groups

Pitfall: A common mistake is using the wrong formula to calculate the number of flips needed to break a consecutive group into segments of at most length m. Developers might incorrectly use:

• cnt += (k - 1) // m
• cnt += k // m
• cnt += (k + m - 1) // m

Why it's wrong: These formulas don't correctly calculate the minimum number of flips needed. For example, if we have a group of 7 consecutive '1's and want max length of 2:

• String: "1111111" (length 7)
• To achieve max length 2, we need to flip at positions to get: "1101101"
• We need 2 flips (positions 2 and 5)
• The correct formula 7 // (2 + 1) = 7 // 3 = 2 gives us 2 flips
• But (7 - 1) // 2 = 3 would give wrong answer

Solution: Use the correct formula k // (m + 1) where:

• k is the length of the consecutive group
• m is the target maximum length
• This formula works because we place flips every (m + 1) positions to break the group optimally

2. Off-by-One Error in Binary Search Range

Pitfall: Setting the binary search range incorrectly, such as:

# Wrong: searching from 0
bisect_left(range(string_length + 1), True, lo=0, key=can_achieve_max_length)

Why it's wrong:

• The minimum possible maximum consecutive length is 1, not 0
• A consecutive length of 0 doesn't make sense (empty substring)
• Starting from 0 would waste one iteration and could cause logical errors

Solution: Always start the search from lo=1:

bisect_left(range(string_length), True, lo=1, key=can_achieve_max_length)

3. Misunderstanding the Alternating Pattern Check

Pitfall: When checking for m = 1 (alternating pattern), incorrectly calculating the flips needed:

# Wrong approach 1: Only checking one pattern
cnt = sum(c != str(i % 2) for i, c in enumerate(s))

# Wrong approach 2: Not considering both starting patterns
cnt = sum(c == t[i % 2] for i, c in enumerate(s))  # Without taking min

Why it's wrong:

• For alternating patterns, we have two choices: start with '0' or start with '1'
• Pattern "010101..." and pattern "101010..." might require different numbers of flips
• We should choose the pattern that requires fewer flips

Solution: Check both patterns and take the minimum:

pattern = "01"
cnt = sum(c == pattern[i & 1] for i, c in enumerate(s))
cnt = min(cnt, n - cnt)  # n - cnt gives flips for opposite pattern

4. Not Resetting the Consecutive Counter

Pitfall: Forgetting to reset the consecutive counter after processing each group:

# Wrong: counter not reset
k = 0
for i, c in enumerate(s):
    k += 1
    if i == len(s) - 1 or c != s[i + 1]:
        cnt += k // (m + 1)
        # Missing: k = 0

Why it's wrong: Without resetting k, the counter continues to accumulate across different groups, leading to incorrect calculations for subsequent groups.

Solution: Always reset the counter after processing each group:

if i == len(s) - 1 or c != s[i + 1]:
    cnt += k // (m + 1)
    k = 0  # Reset for next group

5. Edge Case: Single Character Groups

Pitfall: Not handling strings with no consecutive identical characters properly (e.g., "0101010"):

• When all groups have length 1, the formula 1 // (m + 1) always gives 0 for m ≥ 1
• This might lead to confusion about whether the algorithm is working correctly

Solution: The algorithm handles this correctly by design. When groups are already small enough (length ≤ m), no flips are needed, which is the expected behavior. Ensure test cases cover such scenarios to validate correctness.