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1838. 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 <= 10^{5}
1 <= nums[i] <= 10^{5}
1 <= k <= 10^{5}
Solutions
Method 1: Sorting + Sliding Window
We can first sort the array $nums$, then enumerate each number as the most frequent element, and use a sliding window to maintain the number of operations to increase all numbers from index $l$ to $r$ to $nums[r]$. If the number of operations is greater than $k$, the left end of the window moves to the right until the number of operations is less than or equal to $k$. In this way, we can find out the maximum frequency for each number as the most frequent element.
The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Where $n$ is the length of the array $nums$.
Method 2: Sorting + Prefix Sum + Binary Search
We observe that if a range length $cnt$ meets the condition, then the range length less than $cnt$ must also meet the condition. Therefore, we can use the method of binary search to find the maximum range length that meets the condition.
Before binary search, we need to sort the array $nums[r]$, then calculate the prefix sum array $s$ of the array $nums[r]$, where $s[i]$ represents the sum of the first $i$ numbers in the array $nums[r]$. In this way, we can find the sum of the range $[i, j]$ is $s[j + 1]  s[i]$ in $O(1)$ time.
Next, we define the left boundary of the binary search as $left=1$, $right=n$. Then we binary search the range length $mid$, if the current range length $mid$ meets the condition, then we update the left boundary of the binary search to $mid$, otherwise update the right boundary of the binary search to $mid1$. Finally, we return the left boundary of the binary search.
The problem is transformed into how to judge whether the range with length $cnt$ meets the condition. We enumerate the left endpoint $i$ in the range $[0,..ncnt]$, then the right endpoint of the range at this time is $j = i + cnt  1$. The number of operations required to increase all numbers in the range to $nums[j]$ is $nums[j] \times cnt  (s[j + 1]  s[i])$. If this number of operations is less than or equal to $k$, it means that the range with length $cnt$ meets the condition, return true
. Otherwise, the enumeration ends, return false
.
The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Where $n$ is the length of the array $nums$.

class Solution { public int maxFrequency(int[] nums, int k) { Arrays.sort(nums); int n = nums.length; int ans = 1, window = 0; for (int l = 0, r = 1; r < n; ++r) { window += (nums[r]  nums[r  1]) * (r  l); while (window > k) { window = (nums[r]  nums[l++]); } ans = Math.max(ans, r  l + 1); } return ans; } }

class Solution { public: int maxFrequency(vector<int>& nums, int k) { sort(nums.begin(), nums.end()); int n = nums.size(); int ans = 1; long long window = 0; for (int l = 0, r = 1; r < n; ++r) { window += 1LL * (nums[r]  nums[r  1]) * (r  l); while (window > k) { window = (nums[r]  nums[l++]); } ans = max(ans, r  l + 1); } return ans; } };

class Solution: def maxFrequency(self, nums: List[int], k: int) > int: nums.sort() l, r, n = 0, 1, len(nums) ans, window = 1, 0 while r < n: window += (nums[r]  nums[r  1]) * (r  l) while window > k: window = nums[r]  nums[l] l += 1 r += 1 ans = max(ans, r  l) return ans

func maxFrequency(nums []int, k int) int { sort.Ints(nums) ans, window := 1, 0 for l, r := 0, 1; r < len(nums); r++ { window += (nums[r]  nums[r1]) * (r  l) for window > k { window = nums[r]  nums[l] l++ } ans = max(ans, rl+1) } return ans }

function maxFrequency(nums: number[], k: number): number { nums.sort((a, b) => a  b); let ans = 1; let window = 0; const n = nums.length; for (let l = 0, r = 1; r < n; ++r) { window += (nums[r]  nums[r  1]) * (r  l); while (window > k) { window = nums[r]  nums[l++]; } ans = Math.max(ans, r  l + 1); } return ans; }

/** * @param {number[]} nums * @param {number} k * @return {number} */ var maxFrequency = function (nums, k) { nums.sort((a, b) => a  b); let ans = 1; let window = 0; const n = nums.length; for (let l = 0, r = 1; r < n; ++r) { window += (nums[r]  nums[r  1]) * (r  l); while (window > k) { window = nums[r]  nums[l++]; } ans = Math.max(ans, r  l + 1); } return ans; };