FrequencyOfTheMostFrequentElement.cpp

// Source : https://leetcode.com/problems/frequency-of-the-most-frequent-element/
// Author : Hao Chen
// Date   : 2021-04-25

/***************************************************************************************************** 
 *
 * The frequency of an element is the number of times it occurs in an array.
 * 
 * You are given an integer array nums and an integer k. In one operation, you can choose an index of 
 * nums and increment the element at that index by 1.
 * 
 * Return the maximum possible frequency of an element after performing at most k operations.
 * 
 * Example 1:
 * 
 * Input: nums = [1,2,4], k = 5
 * Output: 3
 * Explanation: Increment the first element three times and the second element two times to make nums 
 * = [4,4,4].
 * 4 has a frequency of 3.
 * 
 * Example 2:
 * 
 * Input: nums = [1,4,8,13], k = 5
 * Output: 2
 * Explanation: There are multiple optimal solutions:
 * - Increment the first element three times to make nums = [4,4,8,13]. 4 has a frequency of 2.
 * - Increment the second element four times to make nums = [1,8,8,13]. 8 has a frequency of 2.
 * - Increment the third element five times to make nums = [1,4,13,13]. 13 has a frequency of 2.
 * 
 * Example 3:
 * 
 * Input: nums = [3,9,6], k = 2
 * Output: 1
 * 
 * Constraints:
 * 
 * 	1 <= nums.length <= 10^5
 * 	1 <= nums[i] <= 10^5
 * 	1 <= k <= 10^5
 ******************************************************************************************************/

class Solution {
public:
    int maxFrequency(vector<int>& nums, int k) {
        sort(nums.begin(), nums.end());
        int m = 1;
        int start = 0;
        int i = 1;
        for(; i<nums.size(); i++){
            long delta = nums[i] - nums[i-1];
            k -= delta * (i - start);;
            if (k < 0 )  {
                // remove the first one
                k += (nums[i] - nums[start]) ;
                start++;
            }
            m = max(m, i - start +1);
            
        }
        return m;
    }
};