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The sliding window technique is used to find the longest subarray where elements can be incremented to make them equal using at most k increments. We sort the array first to ensure it's easier to increment elements to become equal. We then use two pointers: one for the current end of the subarray and the other to represent the start. By checking the cost to transform the current subarray into an equal value using the difference between values and leveraging the sorted property, we can determine the maximum frequency we can achieve.
Time Complexity: O(n log n), due to sorting the array.
Space Complexity: O(1), as no extra space is used apart from variable allocations.
1function maxFrequency(nums, k) {
2 nums.sort((a, b) => a - b);
3 let sum = 0;
4
In JavaScript, the solution uses the Array.prototype.sort() method for sorting, and leverages a sliding window approach, adjusting the window size while tracking sums to achieve the maximum achievable frequency under k operations.
This approach makes use of binary search in combination with prefix sums to efficiently find the maximum frequency possible by transforming elements. We first sort the array. For a given target frequency, we check through calculating the required increment operations using a prefix sum array and binary searching over possible solutions. This allows us to leverage efficient range queries and reduce time complexity.
Time Complexity: O(n log n), because of sorting and binary search iterations.
Space Complexity: O(n), due to additional space for prefix sums.
The Java solution, like the C++ one, calculates a prefix sum array to help quickly check if a certain frequency is achievable under the constraints. Using binary search, we determine the largest frequency that can be obtained through valid operations.