LeetCode 2524: Maximum Frequency Score of a Subarray Solution
Master LeetCode problem 2524 (Maximum Frequency Score of a Subarray), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
2524. Maximum Frequency Score of a Subarray
Problem Explanation
Explanation:
Given an array of integers nums, the task is to find the maximum possible score of a subarray of nums where the score of a subarray is defined as the product of the frequency of the most frequent element in the subarray and the length of the subarray.
To solve this problem, we can use a sliding window approach. We will maintain two hash maps:
countMapto store the frequency of each element in the current window.freqMapto store the frequency of frequencies of elements in the current window.
We will iterate over the array using a sliding window of size k. At each step, we will update the frequency counts in the countMap and freqMap. We will track the maximum score we have seen so far.
Steps:
- Initialize variables
maxScoreto store the maximum score seen so far, andleftto 0. - Iterate over the array using a sliding window of size
k. - Update the frequency counts in
countMapandfreqMapfor each element in the current window. - Calculate the score for the current window as the product of the most frequent element's frequency and the window size.
- Update
maxScoreif the current score is greater. - Slide the window by incrementing
leftand updating the frequency counts accordingly. - Return the maximum score.
Time Complexity:
The time complexity of this approach is O(n), where n is the number of elements in the input array.
Space Complexity:
The space complexity of this approach is O(n) to store the frequency counts in the hash maps.
:
Solution Code
class Solution {
public int maxFrequencyScore(int[] nums, int k) {
int maxScore = 0;
Map<Integer, Integer> countMap = new HashMap<>();
Map<Integer, Integer> freqMap = new HashMap<>();
int left = 0;
for (int right = 0; right < nums.length; right++) {
countMap.put(nums[right], countMap.getOrDefault(nums[right], 0) + 1);
int maxFreq = countMap.get(nums[right]);
freqMap.put(maxFreq, freqMap.getOrDefault(maxFreq, 0) + 1);
while ((right - left + 1) - freqMap.get(maxFreq) > k) {
int leftFreq = countMap.get(nums[left]);
freqMap.put(leftFreq, freqMap.get(leftFreq) - 1);
if (freqMap.get(leftFreq) == 0) {
freqMap.remove(leftFreq);
}
countMap.put(nums[left], countMap.get(nums[left]) - 1);
left++;
}
maxScore = Math.max(maxScore, maxFreq * (right - left + 1));
}
return maxScore;
}
}Try It Yourself
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