Maximize the Confusion of an Exam solution and explanation
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The problem
A teacher is writing a test with n true/false questions, with 'T' denoting true and 'F' denoting false. He wants to confuse the students by maximizing the number of consecutive questions with the same answer (multiple trues or multiple falses in a row).
You are given a string answerKey, where answerKey[i] is the original answer to the ith question. In addition, you are given an integer k, the maximum number of times you may perform the following operation:
- Change the answer key for any question to
'T'or'F'(i.e., setanswerKey[i]to'T'or'F').
Return the maximum number of consecutive 'T's or 'F's in the answer key after performing the operation at most k times.
Example 1:
Input: answerKey = "TTFF", k = 2 Output: 4 Explanation: We can replace both the 'F's with 'T's to make answerKey = "TTTT". There are four consecutive 'T's.
Example 2:
Input: answerKey = "TFFT", k = 1 Output: 3 Explanation: We can replace the first 'T' with an 'F' to make answerKey = "FFFT". Alternatively, we can replace the second 'T' with an 'F' to make answerKey = "TFFF". In both cases, there are three consecutive 'F's.
Example 3:
Input: answerKey = "TTFTTFTT", k = 1 Output: 5 Explanation: We can replace the first 'F' to make answerKey = "TTTTTFTT" Alternatively, we can replace the second 'F' to make answerKey = "TTFTTTTT". In both cases, there are five consecutive 'T's.
Constraints:
n == answerKey.length1 <= n <= 5 * 104answerKey[i]is either'T'or'F'1 <= k <= n
Approach overview
Problem Overview: You receive a string representing exam answers where each character is 'T' or 'F'. You may change at most k answers. The goal is to maximize the length of a consecutive block of identical answers after performing those changes.
Approach 1: Sliding Window Technique (O(n) time, O(1) space)
The optimal strategy uses a sliding window. Treat the problem as finding the longest substring where you can flip at most k characters to make all values the same. Run the window twice: once assuming the final sequence should be all 'T', and once for 'F'. Expand the right pointer while counting mismatches. When mismatches exceed k, move the left pointer forward and shrink the window. The window size always represents the longest valid segment you can transform with ≤ k flips.
This works because the window always maintains a valid state where the number of required flips is within the allowed limit. Each character is visited at most twice (entering and leaving the window), giving linear complexity. This approach is the standard pattern used in many string optimization problems involving limited modifications.
Approach 2: Two Pointer Technique (O(n) time, O(1) space)
The two pointer method is conceptually the same as sliding window but framed more explicitly with left and right indices. Move the right pointer forward to expand the candidate segment and maintain counts of 'T' and 'F'. When the number of characters needing conversion exceeds k, advance the left pointer to restore validity. The current window length represents a segment that can be converted into all identical answers.
This formulation emphasizes pointer movement and state maintenance rather than a formal window abstraction. It is useful when practicing general pointer techniques used in binary search-style range problems and substring optimization tasks.
Recommended for interviews: The sliding window solution is what most interviewers expect. It demonstrates that you can transform the problem into a longest-valid-substring pattern and maintain constraints efficiently in O(n) time. A brute-force approach that checks all substrings would take O(n²) time and quickly becomes impractical. Showing the transition from brute force thinking to a sliding window optimization demonstrates strong problem-solving skills.
Approach 1: Sliding Window Technique
This approach uses the sliding window technique to maintain a window of the answerKey where the number of modifications does not exceed k. We attempt to maximize the length of this window by evaluating either 'T' or 'F' transformations separately and then taking the maximum length between both transformations.
def maxConsecutiveAnswers(answerKey, k):
def maxConsecutive(char):
left = maxLen = 0
count = 0
for right in range(len(answerKey)):
if answerKey[right] == char:
count += 1
while count > k:
if answerKey[left] == char:
count -= 1
left += 1
maxLen = max(maxLen, right - left + 1)
return maxLen
return max(maxConsecutive('T'), maxConsecutive('F'))
answerKey = "TTFTTFTT"
k = 1
print(maxConsecutiveAnswers(answerKey, k))public class MaxConsecutiveAnswers {
public static int maxConsecutiveAnswers(String answerKey, int k) {
int maxLen = 0, left = 0, countT = 0, countF = 0;
for (int right = 0; right < answerKey.length(); right++) {
char currentChar = answerKey.charAt(right);
if (currentChar == 'T') countT++;
else countF++;
while (Math.min(countT, countF) > k) {
if (answerKey.charAt(left) == 'T') countT--;
else countF--;
left++;
}
maxLen = Math.max(maxLen, right - left + 1);
}
return maxLen;
}
public static void main(String[] args) {
String answerKey = "TTFF";
int k = 2;
System.out.println(maxConsecutiveAnswers(answerKey, k));
}
}#include <iostream>
#include <string>
using namespace std;
int maxConsecutiveAnswers(string answerKey, int k) {
int left = 0, maxLen = 0, countT = 0, countF = 0;
for (int right = 0; right < answerKey.size(); ++right) {
answerKey[right] == 'T' ? countT++ : countF++;
while (min(countT, countF) > k) {
answerKey[left] == 'T' ? countT-- : countF--;
left++;
}
maxLen = max(maxLen, right - left + 1);
}
return maxLen;
}
int main() {
string answerKey = "TFFT";
int k = 1;
cout << maxConsecutiveAnswers(answerKey, k) << endl;
return 0;
}function maxConsecutiveAnswers(answerKey, k) {
const maxConsecutive = (char) => {
let left = 0, maxLen = 0, count = 0;
for (let right = 0; right < answerKey.length; right++) {
if (answerKey[right] === char) count++;
while (count > k) {
if (answerKey[left] === char) count--;
left++;
}
maxLen = Math.max(maxLen, right - left + 1);
}
return maxLen;
}
return Math.max(maxConsecutive('T'), maxConsecutive('F'));
}
console.log(maxConsecutiveAnswers('TTFTTFTT', 1));#include <stdio.h>
#include <string.h>
int maxConsecutiveAnswers(char *answerKey, int k) {
int maxLen = 0;
int left = 0, right = 0, maxT = 0, maxF = 0;
int len = strlen(answerKey);
for (; right < len; ++right) {
if (answerKey[right] == 'T')
maxT++;
else
maxF++;
while (maxT > k && maxF > k) {
if (answerKey[left] == 'T')
maxT--;
else
maxF--;
left++;
}
maxLen = maxLen > right - left + 1 ? maxLen : right - left + 1;
}
return maxLen;
}
int main() {
char answerKey[] = "TTFTTFTT";
int k = 1;
printf("%d\n", maxConsecutiveAnswers(answerKey, k));
return 0;
}using System;
class Program {
static int maxConsecutiveAnswers(string answerKey, int k) {
int maxLen = 0, left = 0, countT = 0, countF = 0;
for (int right = 0; right < answerKey.Length; right++) {
if (answerKey[right] == 'T')
countT++;
else
countF++;
while (Math.Min(countT, countF) > k) {
if (answerKey[left] == 'T')
countT--;
else
countF--;
left++;
}
maxLen = Math.Max(maxLen, right - left + 1);
}
return maxLen;
}
static void Main() {
string answerKey = "TFFT";
int k = 1;
Console.WriteLine(maxConsecutiveAnswers(answerKey, k));
}
}The Python solution respects the sliding window by evaluating both characters 'T' and 'F' distinctly through the nested maxConsecutive method. The method retains a count of the characters needing modification and retracts the left pointer when the count surpasses k.
Time Complexity: O(n), given the single looping through the string.
Space Complexity: O(1), due to the use of constant space structures, only managing indices and counts.
Approach 2: Two Pointer Technique
In this approach, we utilize two pointers to explore the string sequence, adjusting pointers as we evaluate possibilities involving updates utilizing k. The purpose is to ensure the longest alteration window without exceeding k modifications. It proceeds by specifically examining alternate possibilities and then concluding with the optimal maximum length.
def maxConsecutiveAnswers(answerKey, k):
n = len(answerKey)
def countMax(char):
left = count = 0
maxLen = 0
for right in range(n):
if answerKey[right] != char:
count += 1
while count > k:
if answerKey[left] != char:
count -= 1
left += 1
maxLen = max(maxLen, right - left + 1)
return maxLen
return max(countMax('T'), countMax('F'))
print(maxConsecutiveAnswers("TTFTTFTT", 1))public class MaxConsecutiveAnswers {
public static int maxConsecutiveAnswers(String answerKey, int k) {
int n = answerKey.length(), maxLen = 0;
char[] options = {'T', 'F'};
for (char x : options) {
int left = 0, count = 0;
for (int right = 0; right < n; right++) {
if (answerKey.charAt(right) != x) count++;
while (count > k) {
if (answerKey.charAt(left) != x) count--;
left++;
}
maxLen = Math.max(maxLen, right - left + 1);
}
}
return maxLen;
}
public static void main(String[] args) {
String answerKey = "TTFF";
int k = 2;
System.out.println(maxConsecutiveAnswers(answerKey, k));
}
}#include <iostream>
#include <string>
using namespace std;
int maxConsecutiveAnswers(string answerKey, int k) {
int n = answerKey.length(), result = 0;
char directions[] = {'T', 'F'};
for (char x : directions) {
int count = 0, left = 0;
for (int right = 0; right < n; ++right) {
if (answerKey[right] != x) count++;
while (count > k) {
if (answerKey[left] != x) count--;
left++;
}
result = max(result, right - left + 1);
}
}
return result;
}
int main() {
string answerKey = "TFFT";
int k = 1;
cout << maxConsecutiveAnswers(answerKey, k) << endl;
return 0;
}function maxConsecutiveAnswers(answerKey, k) {
const countMax = (char) => {
let left = 0;
let count = 0;
let maxLen = 0;
for (let right = 0; right < answerKey.length; ++right) {
if (answerKey[right] !== char) count++;
while (count > k) {
if (answerKey[left] !== char) count--;
left++;
}
maxLen = Math.max(maxLen, right - left + 1);
}
return maxLen;
};
return Math.max(countMax('T'), countMax('F'));
}
console.log(maxConsecutiveAnswers('TTFTTFTT', 1));#include <stdio.h>
#include <string.h>
int maxConsecutiveAnswers(char *answerKey, int k) {
int n = strlen(answerKey);
int result = 0;
for (char x : {'T', 'F'}) {
int count = 0, left = 0;
for (int right = 0; right < n; right++) {
if (answerKey[right] != x) count++;
while (count > k) {
if (answerKey[left] != x) count--;
left++;
}
result = result > right - left + 1 ? result : right - left + 1;
}
}
return result;
}
int main() {
char answerKey[] = "TTFTTFTT";
int k = 1;
printf("%d\n", maxConsecutiveAnswers(answerKey, k));
return 0;
}using System;
class Program {
static int maxConsecutiveAnswers(string answerKey, int k) {
int n = answerKey.Length;
int result = 0;
foreach (char x in new char[]{ 'T', 'F' }) {
int count = 0, left = 0;
for (int right = 0; right < n; right++) {
if (answerKey[right] != x) count++;
while (count > k) {
if (answerKey[left] != x) count--;
left++;
}
result = Math.Max(result, right - left + 1);
}
}
return result;
}
static void Main() {
string answerKey = "TFFT";
int k = 1;
Console.WriteLine(maxConsecutiveAnswers(answerKey, k));
}
}The Python variant distinctly analyzes answer conversions from separate character strains producing competitive responses at each tentative setting. Adjustments maintain plausible ranges to ensure altered strings sustain under k conversion restraint.
Time Complexity: O(n), which directly ties to the cumulative efficiency through incremental passes.
Space Complexity: O(1), relying strictly on essential count and index operation sufficiently therein.
Approach 3: Sliding Window
We design a function f(c), which represents the longest length of consecutive characters under the condition that at most k characters c can be replaced, where c can be 'T' or 'F'. The answer is max(f('T'), f('F')).
We iterate through the string answerKey, using a variable cnt to record the number of characters c within the current window. When cnt > k, we move the left pointer of the window one position to the right. After the iteration ends, the length of the window is the maximum length of consecutive characters.
Time complexity is O(n), where n is the length of the string. Space complexity is O(1).
Similar problems:
function maxConsecutiveAnswers(answerKey: string, k: number): number {
const n = answerKey.length;
const f = (c: string): number => {
let [l, cnt] = [0, 0];
for (const ch of answerKey) {
cnt += ch === c ? 1 : 0;
if (cnt > k) {
cnt -= answerKey[l++] === c ? 1 : 0;
}
}
return n - l;
};
return Math.max(f('T'), f('F'));
}func maxConsecutiveAnswers(answerKey string, k int) int {
f := func(c byte) int {
l, cnt := 0, 0
for _, ch := range answerKey {
if byte(ch) == c {
cnt++
}
if cnt > k {
if answerKey[l] == c {
cnt--
}
l++
}
}
return len(answerKey) - l
}
return max(f('T'), f('F'))
}impl Solution {
pub fn max_consecutive_answers(answer_key: String, k: i32) -> i32 {
let n = answer_key.len();
let k = k as usize;
let s: Vec<char> = answer_key.chars().collect();
let f = |c: char| -> usize {
let mut l = 0;
let mut cnt = 0;
for &ch in &s {
cnt += if ch == c { 1 } else { 0 };
if cnt > k {
cnt -= if s[l] == c { 1 } else { 0 };
l += 1;
}
}
n - l
};
std::cmp::max(f('T'), f('F')) as i32
}
}Complexity comparison
| Approach | Complexity |
|---|---|
| Sliding Window Technique | Time Complexity: O(n), given the single looping through the string. |
| Two Pointer Technique | Time Complexity: O(n), which directly ties to the cumulative efficiency through incremental passes. |
| Sliding Window | – |
Complexity analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Brute Force (Check All Substrings) | O(n²) | O(1) | Conceptual baseline to understand the problem constraints |
| Sliding Window | O(n) | O(1) | Best general solution for longest valid substring with limited modifications |
| Two Pointer Technique | O(n) | O(1) | When implementing substring expansion/shrinking logic using explicit pointer control |
Frequently asked questions
Is Maximize the Confusion of an Exam easy or hard?
The problem is classified as Medium difficulty. The core challenge is recognizing the sliding window pattern and maintaining the constraint of at most k modifications while scanning the string.
Maximize the Confusion of an Exam Python/Java solution
Implement the sliding window logic with two pointers and a mismatch counter. Expand the right pointer through the string, shrink the left pointer when flips exceed k, and track the maximum window length. The same algorithm works across Python, Java, C++, C#, and JavaScript with identical O(n) complexity.
How to solve Maximize the Confusion of an Exam in O(n)?
Use a sliding window that tracks how many characters inside the window differ from the target character (either 'T' or 'F'). Expand the right pointer while the number of required flips is ≤ k. When the flips exceed k, move the left pointer until the window becomes valid again. Track the maximum window size during the process.
What is the best approach for Maximize the Confusion of an Exam?
The sliding window approach is the most efficient solution. It maintains a dynamic substring and ensures the number of characters needing conversion does not exceed k. By expanding and shrinking the window with two pointers, you compute the longest valid segment in O(n) time and O(1) space.
Is Maximize the Confusion of an Exam asked at Google/Amazon/Meta?
Problems involving sliding window substring optimization frequently appear in interviews at companies like Amazon, Google, and Meta. This problem specifically tests the ability to maintain constraints while scanning a string in linear time.
What data structure is used in Maximize the Confusion of an Exam?
The solution mainly relies on a sliding window with two pointers and simple counters. No complex data structures are required since the input only contains two possible characters.
What is the time complexity of Maximize the Confusion of an Exam?
The optimal solution runs in O(n) time because each character is processed at most twice while expanding and shrinking the sliding window. Space complexity is O(1) since only a few counters and pointer variables are used.