Scale Engineer

Maximize the Confusion of an Exam

Med
#1843Time: O(N^2), where N is the length of `answerKey`. The nested loops iterate through all O(N^2) subarrays.Space: O(1), as we only use a few variables to store counts and pointers, regardless of the input size.1 company
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Sliding WindowPrefix Sum
Algorithms
Binary Search
Data structures
String
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Arcesium

Prompt

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., set answerKey[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.length
  • 1 <= n <= 5 * 104
  • answerKey[i] is either 'T' or 'F'
  • 1 <= k <= n

Approaches

3 approaches with complexity analysis and trade-offs.

This approach involves checking every possible contiguous subarray within the answerKey. For each subarray, we count the number of 'T's and 'F's. If the count of either character is less than or equal to k, it means we can make the entire subarray uniform by performing at most k changes. We keep track of the maximum length of such a valid subarray.

Algorithm

  • Initialize maxLength = 0.
  • Use a nested loop structure. The outer loop with index i determines the start of a potential subarray, and the inner loop with index j determines the end.
  • For each subarray answerKey[i...j], count the number of 'T's (countT) and 'F's (countF).
  • A subarray can be made uniform if the number of characters to change is at most k. This means either countT <= k (to make it all 'F's) or countF <= k (to make it all 'T's).
  • If the condition countT <= k || countF <= k is met, the current subarray of length j - i + 1 is a valid candidate for the maximum length.
  • Update maxLength = max(maxLength, j - i + 1).
  • After iterating through all possible subarrays, return maxLength.

Walkthrough

We use two nested loops to define the start (i) and end (j) of each subarray. The outer loop iterates from the first character to the last to set the starting point. The inner loop expands the subarray from the starting point to the end of the string. Inside the inner loop, we maintain counts of 'T's and 'F's for the current subarray [i, j]. For each subarray, we check if it can be made entirely of 'T's (by changing at most k 'F's) or entirely of 'F's (by changing at most k 'T's). If this is possible, we update our maxLength with the current subarray's length (j - i + 1). After checking all O(N^2) subarrays, maxLength will hold the result.

class Solution {    public int maxConsecutiveAnswers(String answerKey, int k) {        int n = answerKey.length();        int maxLength = 0;        if (k >= n / 2) {            return n;        }        for (int i = 0; i < n; i++) {            int countT = 0;            int countF = 0;            for (int j = i; j < n; j++) {                if (answerKey.charAt(j) == 'T') {                    countT++;                } else {                    countF++;                }                if (countT <= k || countF <= k) {                    maxLength = Math.max(maxLength, j - i + 1);                }            }        }        return maxLength;    }}

Complexity

Time

O(N^2), where N is the length of `answerKey`. The nested loops iterate through all O(N^2) subarrays.

Space

O(1), as we only use a few variables to store counts and pointers, regardless of the input size.

Trade-offs

Pros

  • Simple to understand and implement.

  • Correct for small input sizes.

Cons

  • Highly inefficient due to its O(N^2) time complexity.

  • Will result in a 'Time Limit Exceeded' (TLE) error on platforms like LeetCode for the given constraints.

Solutions

class Solution { public int maxConsecutiveAnswers ( String answerKey , int k ) { return Math . max ( get ( 'T' , k , answerKey ), get ( 'F' , k , answerKey )); } public int get ( char c , int k , String answerKey ) { int l = 0 , r = 0 ; while ( r < answerKey . length ()) { if ( answerKey . charAt ( r ++) == c ) { -- k ; } if ( k < 0 && answerKey . charAt ( l ++) == c ) { ++ k ; } } return r - l ; } }

Video walkthrough

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