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New Problem Solution - "1838. Frequency of the Most Frequent Element"
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algorithms/cpp/frequencyOfTheMostFrequentElement/FrequencyOfTheMostFrequentElement.cpp
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// Source : https://leetcode.com/problems/frequency-of-the-most-frequent-element/ | ||
// Author : Hao Chen | ||
// Date : 2021-04-25 | ||
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/***************************************************************************************************** | ||
* | ||
* 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 | ||
******************************************************************************************************/ | ||
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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); | ||
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} | ||
return m; | ||
} | ||
}; |