Frequency of the Most Frequent Element

Description

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 <= 105
• 1 <= nums[i] <= 105
• 1 <= k <= 105

Code

Time Complexity: $O(n\log n)$, Space Complexity: $O(1)$

要將鄰近的 element 都變成和自己具有相同的值，其實就是 sliding window 的概念（鄰近的 elements 組成一個 window）。

class Solution {
public:
    int maxFrequency(vector<int>& nums, int k) {
        sort(nums.begin(), nums.end());
        int n = nums.size();
        
        int j = 0;
        long count = 0;
        int res = 1;
        for(int i = 1; i < n; i++) {
	            count += (long) (i - j) * (nums[i] - nums[i - 1]);
            while(count > k) {
                count -=(nums[i] - nums[j]);
                j++;
            }
            res = max(res, i - j + 1);
        }
 
        return res;
    }
};

Source

• Frequency of the Most Frequent Element - LeetCode