64 lines
2.0 KiB
C++
64 lines
2.0 KiB
C++
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// Source : https://leetcode.com/problems/frequency-of-the-most-frequent-element/
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// Author : Hao Chen
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// Date : 2021-04-25
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/*****************************************************************************************************
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*
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* The frequency of an element is the number of times it occurs in an array.
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*
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* You are given an integer array nums and an integer k. In one operation, you can choose an index of
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* nums and increment the element at that index by 1.
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*
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* Return the maximum possible frequency of an element after performing at most k operations.
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*
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* Example 1:
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*
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* Input: nums = [1,2,4], k = 5
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* Output: 3
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* Explanation: Increment the first element three times and the second element two times to make nums
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* = [4,4,4].
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* 4 has a frequency of 3.
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*
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* Example 2:
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*
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* Input: nums = [1,4,8,13], k = 5
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* Output: 2
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* Explanation: There are multiple optimal solutions:
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* - Increment the first element three times to make nums = [4,4,8,13]. 4 has a frequency of 2.
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* - Increment the second element four times to make nums = [1,8,8,13]. 8 has a frequency of 2.
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* - Increment the third element five times to make nums = [1,4,13,13]. 13 has a frequency of 2.
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*
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* Example 3:
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*
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* Input: nums = [3,9,6], k = 2
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* Output: 1
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*
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* Constraints:
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*
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* 1 <= nums.length <= 10^5
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* 1 <= nums[i] <= 10^5
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* 1 <= k <= 10^5
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******************************************************************************************************/
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class Solution {
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public:
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int maxFrequency(vector<int>& nums, int k) {
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sort(nums.begin(), nums.end());
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int m = 1;
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int start = 0;
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int i = 1;
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for(; i<nums.size(); i++){
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long delta = nums[i] - nums[i-1];
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k -= delta * (i - start);;
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if (k < 0 ) {
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// remove the first one
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k += (nums[i] - nums[start]) ;
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start++;
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}
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m = max(m, i - start +1);
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}
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return m;
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}
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};
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