Solution Approach

The solution uses the boundary-finding binary search template to find the minimum possible maximum length of consecutive identical characters.

Binary Search Template Structure: We search for the minimum m where feasible(m) returns True. The search space is [1, n].

left, right = 1, n
first_true_index = -1

while left <= right:
    mid = (left + right) // 2
    if feasible(mid):
        first_true_index = mid
        right = mid - 1  # Look for smaller m
    else:
        left = mid + 1

return first_true_index

The Feasible Function: feasible(m) determines if we can achieve a maximum consecutive length of m using at most numOps flips.

Case 1: When m = 1

• We need the string to alternate completely
• Count operations for pattern "010101..." using sum(c == pattern[i & 1])
• Take minimum of this count and n - count (for pattern "101010...")

Case 2: When m > 1

• For each group of k consecutive identical characters:
  • Calculate flips needed: k // (m + 1)
  • This formula optimally breaks a segment into parts of at most m

Why This Template Works: The feasible function creates a monotonic pattern: False, False, ..., True, True, True. If we can achieve max length m, we can certainly achieve any length greater than m. We find the first True (minimum achievable m).

Time Complexity: O(n log n) where binary search runs O(log n) iterations, each check takes O(n).

Space Complexity: O(1) - only using a few variables.

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").

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. Using Wrong Binary Search Template Variant

A critical mistake is using the while left < right with right = mid pattern instead of the correct template.

Incorrect approach:

left, right = 1, n
while left < right:
    mid = (left + right) // 2
    if feasible(mid):
        right = mid
    else:
        left = mid + 1
return left

Correct approach (template-compliant):

left, right = 1, n
first_true_index = -1

while left <= right:
    mid = (left + right) // 2
    if feasible(mid):
        first_true_index = mid
        right = mid - 1
    else:
        left = mid + 1

return first_true_index

The template uses while left <= right with right = mid - 1 and tracks the answer in first_true_index.

2. Incorrect Formula for Breaking Consecutive Groups

Using the wrong formula to calculate flips needed:

• k // m - doesn't account for spacing correctly
• (k - 1) // m - underestimates

Correct formula: k // (m + 1) - places flips every (m + 1) positions optimally.

3. Misunderstanding the Alternating Pattern Check

When checking for m = 1, forgetting to consider both patterns ("010101..." and "101010..."):

# Wrong: Only checking one pattern
cnt = sum(c == pattern[i & 1] for i, c in enumerate(s))

# Correct: Take minimum of both patterns
cnt = min(cnt, n - cnt)

4. Not Resetting the Consecutive Counter

Forgetting to reset k = 0 after processing each group leads to incorrect calculations for subsequent groups.