Problem Description

You have a test answer key consisting of n true/false questions, where each answer is represented as either 'T' (true) or 'F' (false). The goal is to maximize the length of consecutive questions that have the same answer.

Given:

• A string answerKey where answerKey[i] represents the original answer to question i
• An integer k representing the maximum number of answer changes allowed

You can perform the following operation at most k times:

• Change any answer in the answer key from 'T' to 'F' or from 'F' to 'T'

The task is to find the maximum possible length of consecutive 'T's or consecutive 'F's that can be achieved after performing at most k changes.

For example, if answerKey = "TTFTTFTT" and k = 1, you could change one 'F' to 'T' to create a sequence of 5 consecutive 'T's: "TTTTTFTT" or "TTFTTTTT". The maximum consecutive length would be 5.

The problem essentially asks you to find the longest substring where all characters can be made the same (either all 'T' or all 'F') by changing at most k characters.

Intuition

The key insight is that we want to find the longest substring where we can make all characters the same by changing at most k characters. Since we only have two possible characters ('T' and 'F'), we need to consider both possibilities: making all characters 'T' or making all characters 'F'.

Think of it this way: if we're trying to make a substring all 'T's, then we can tolerate at most k 'F's in that substring. Similarly, if we're trying to make it all 'F's, we can tolerate at most k 'T's.

This naturally leads to a sliding window approach. We can maintain a window and expand it as long as the number of characters we need to change (the minority character in the window) doesn't exceed k. When it does exceed k, we need to shrink the window from the left.

The clever part of the solution is that instead of explicitly tracking both the window size and positions, we can use a simplified sliding window that never actually shrinks in size - it only slides forward. When we encounter too many characters to change (more than k), we move both the left and right boundaries forward by one position, maintaining the window size. This way, the window naturally grows to its maximum possible size and stays there.

We run this sliding window logic twice: once to find the longest substring that can be made all 'T's (by tolerating up to k 'F's), and once to find the longest substring that can be made all 'F's (by tolerating up to k 'T's). The answer is the maximum of these two values.

This approach works because we're essentially asking: "What's the longest substring where the minority character appears at most k times?" And since we don't know which character should be the minority, we try both possibilities.

Learn more about Binary Search, Prefix Sum and Sliding Window patterns.

Solution Approach

The solution implements a sliding window technique with a helper function f(c) that finds the maximum length of consecutive characters when we can change at most k characters to character c.

Here's how the algorithm works:

Helper Function f(c):

• Initialize two variables: cnt (count of character c in current window) and l (left pointer of the window, starting at 0)
• Iterate through each character in answerKey using an implicit right pointer
• For each character:
  • If it matches c, increment cnt
  • If cnt > k (meaning we need to change more than k characters), we need to adjust the window:
    • Check if the character at the left boundary (answerKey[l]) is c
    • If it is, decrement cnt (removing it from our count)
    • Move the left pointer right by incrementing l
• Return len(answerKey) - l, which gives us the final window size

Key Insight of the Window Logic:

The window never actually shrinks - when we exceed k changes needed, we slide the window forward by moving both boundaries. This maintains the maximum window size found so far. At the end, len(answerKey) - l gives us this maximum size.

Main Function:

• Call f("T") to find the maximum consecutive length when changing characters to 'T'
• Call f("F") to find the maximum consecutive length when changing characters to 'F'
• Return the maximum of these two values

Example Walkthrough:

For answerKey = "TTFTTFTT" and k = 1:

• f("T"): We can tolerate 1 'F'. The window grows until we hit the second 'F', then slides forward maintaining size 5
• f("F"): We can tolerate 1 'T'. The best we can do is a window of size 2 (the two 'F's)
• Result: max(5, 2) = 5

Time Complexity: O(n) where n is the length of answerKey, as we traverse the string twice (once for each character type)

Space Complexity: O(1) as we only use a constant amount of extra space for variables

Example Walkthrough

Let's walk through the solution with answerKey = "TFFT" and k = 1:

Goal: Find the longest consecutive sequence of same characters after at most 1 change.

Step 1: Try making all characters 'T' using f("T")

We can tolerate at most 1 'F' in our window.

• Position 0: Character is 'T'

  • Window: [T], cnt = 1, l = 0
  • 'T' matches our target, cnt becomes 1

• Position 1: Character is 'F'

  • Window: [T,F], cnt = 1, l = 0
  • 'F' doesn't match, cnt stays 1
  • We have 1 'F' in window (≤ k=1), so it's valid

• Position 2: Character is 'F'

  • Window: [T,F,F], cnt = 1, l = 0
  • 'F' doesn't match, cnt stays 1
  • We have 2 'F's in window (> k=1), need to slide!
  • Check answerKey[l=0] = 'T', it matches 'T', so cnt becomes 0
  • Move l to 1
  • New window: [F,F], cnt = 0, l = 1

• Position 3: Character is 'T'

  • Window: [F,F,T], cnt = 1, l = 1
  • 'T' matches our target, cnt becomes 1

Final: len(answerKey) - l = 4 - 1 = 3

Step 2: Try making all characters 'F' using f("F")

We can tolerate at most 1 'T' in our window.

• Position 0: Character is 'T'

  • Window: [T], cnt = 0, l = 0
  • 'T' doesn't match 'F', cnt stays 0

• Position 1: Character is 'F'

  • Window: [T,F], cnt = 1, l = 0
  • 'F' matches our target, cnt becomes 1
  • We have 1 'T' in window (≤ k=1), valid

• Position 2: Character is 'F'

  • Window: [T,F,F], cnt = 2, l = 0
  • 'F' matches our target, cnt becomes 2
  • We have 1 'T' in window (≤ k=1), valid

• Position 3: Character is 'T'

  • Window: [T,F,F,T], cnt = 2, l = 0
  • 'T' doesn't match 'F', cnt stays 2
  • We have 2 'T's in window (> k=1), need to slide!
  • Check answerKey[l=0] = 'T', doesn't match 'F', cnt stays 2
  • Move l to 1
  • New window: [F,F,T], cnt = 2, l = 1

Final: len(answerKey) - l = 4 - 1 = 3

Result: max(3, 3) = 3

We can achieve a consecutive sequence of length 3 by either:

• Changing the middle 'F' to 'T': "TFFT" → "TTTT" (positions 0-2)
• Changing the first 'T' to 'F': "TFFT" → "FFFT" (positions 0-2)

Solution Implementation

Time and Space Complexity

Time Complexity: O(n) where n is the length of the answerKey string.

The algorithm calls the helper function f twice - once with "T" and once with "F". Each call to f performs a single pass through the answerKey string using a sliding window approach. In each iteration of the loop, we perform constant time operations (comparison, addition/subtraction, and index access). Even though there's a conditional increment of the left pointer l, each character is visited at most twice (once by the right pointer implicitly through the loop, and potentially once when the left pointer moves past it). This gives us O(2n) which simplifies to O(n) for each call to f. Since we call f twice, the total time complexity is O(2n) = O(n).

Space Complexity: O(1)

The algorithm uses only a constant amount of extra space. The helper function f uses two integer variables (cnt and l) regardless of the input size. The parameter c is a single character. No additional data structures that grow with the input size are used. Therefore, the space complexity is constant O(1).

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

Common Pitfalls

1. Incorrect Character Counting Logic

The Pitfall: The most common mistake is counting the wrong characters. The helper function should count characters that are already the target character (don't need changing), not the characters that need to be flipped. Many developers mistakenly count the characters that need to be changed.

Incorrect Implementation:

def find_max_consecutive(target_char: str) -> int:
    flip_count = 0  # Counting flips needed
    left = 0
  
    for right, char in enumerate(answerKey):
        # WRONG: Counting characters that need flipping
        if char != target_char:
            flip_count += 1
      
        # This logic becomes inverted
        if flip_count > k:
            if answerKey[left] != target_char:
                flip_count -= 1
            left += 1
  
    return len(answerKey) - left

Correct Implementation:

def find_max_consecutive(target_char: str) -> int:
    same_count = 0  # Count characters already matching target
    left = 0
  
    for right, char in enumerate(answerKey):
        # Count characters that are already the target
        if char == target_char:
            same_count += 1
      
        # Window is valid when: (window_size - same_count) <= k
        # Rearranged: same_count >= window_size - k
        # Or: same_count > k when window_size = right - left + 1
        if same_count > k:
            if answerKey[left] == target_char:
                same_count -= 1
            left += 1
  
    return len(answerKey) - left

2. Window Size Calculation Error

The Pitfall: Returning right - left + 1 at the end instead of len(answerKey) - left. The algorithm maintains the maximum window size found, and the final position of left determines this maximum.

Incorrect:

max_length = 0
for right, char in enumerate(answerKey):
    # ... sliding window logic ...
    max_length = max(max_length, right - left + 1)
return max_length

Correct: The window never shrinks below the maximum size found. When we need to slide, we move both pointers, maintaining the size. The final len(answerKey) - left gives us the maximum window size achieved.

3. Misunderstanding the Problem Statement

The Pitfall: Thinking you need to find consecutive characters in the original string, rather than understanding that you can change up to k characters to create the consecutive sequence.

Example of Misunderstanding: For "TFFT" with k=1, someone might think the answer is 2 (the existing "FF"), but actually it's 3 because you can change one 'T' to get "FFFT" or "TFFF".

Solution: Always remember: the goal is to find the longest substring where after making at most k changes, all characters become the same. The sliding window maintains a valid substring where the number of characters that need changing doesn't exceed k.