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
Solution: Sliding Window
Sort the elements, maintain a window such that it takes at most k ops to make the all the elements equal to nums[i].
Time complexity: O(nlogn)
Space complexity: O(1)
C++
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class Solution { public: int maxFrequency(vector<int>& nums, int k) { sort(begin(nums), end(nums)); int l = 0; long sum = 0; int ans = 0; for (int r = 0; r < nums.size(); ++r) { sum += nums[r]; while (l < r && sum + k < static_cast<long>(nums[r]) * (r - l + 1)) sum -= nums[l++]; ans = max(ans, r - l + 1); } return ans; } }; |
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