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.
1using System;
2
3public class Solution {
4 public int MaxFrequency(int[] nums, int k) {
5 Array.Sort(nums);
6 long sum = 0;
7 int left = 0, res = 0;
8 for (int right = 0; right < nums.Length; ++right) {
9 sum += nums[right];
10 while ((long)nums[right] * (right - left + 1) > sum + k) {
11 sum -= nums[left++];
12 }
13 res = Math.Max(res, right - left + 1);
14 }
15 return res;
16 }
17
18 public static void Main() {
19 Solution sol = new Solution();
20 Console.WriteLine(sol.MaxFrequency(new int[] { 1, 2, 4 }, 5));
21 }
22}The C# solution utilizes the Array.Sort() method to sort the input array. It then applies the sliding window technique to effectively compute the maximum possible frequency by maintaining cumulative sums and adjusting the range via the left pointer.
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.
1import java.util.Arrays;
2
3public class Solution {
4 private boolean canAchieveFrequency(int[] nums, int freq, int k, long[] prefixSum) {
5 for (int end = freq - 1; end < nums.length; ++end) {
6 long totalNeeded = (long)nums[end] * freq;
7 long currentSum = prefixSum[end] - (end - freq >= 0 ? prefixSum[end - freq] : 0);
8 if (totalNeeded - currentSum <= k) return true;
9 }
10 return false;
11 }
12
13 public int maxFrequency(int[] nums, int k) {
14 Arrays.sort(nums);
15 long[] prefixSum = new long[nums.length];
16 for (int i = 0; i < nums.length; ++i) {
17 prefixSum[i] = (i > 0 ? prefixSum[i - 1] : 0) + nums[i];
18 }
19 int left = 1, right = nums.length, res = 1;
20 while (left <= right) {
21 int mid = left + (right - left) / 2;
22 if (canAchieveFrequency(nums, mid, k, prefixSum)) {
23 res = mid;
24 left = mid + 1;
25 } else {
26 right = mid - 1;
27 }
28 }
29 return res;
30 }
31
32 public static void main(String[] args) {
33 Solution sol = new Solution();
34 int[] nums = {1, 2, 4};
35 int k = 5;
36 System.out.println(sol.maxFrequency(nums, k));
37 }
38}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.