14. Frequency of the Most Frequent Element
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
Solution:(Greedy + Sliding Window)
Approach: Sorting the array Finding Sliding window condition: (arr[i] * (i-start) - sum)<=k
class Solution
{
public:
int maxFrequency(vector<int> &nums, int k)
{
sort(nums.begin(), nums.end());
long long int s = nums[0];
int maxLen = 1;
int start = 0;
for (int i = 1; i < nums.size(); i++)
{
s = s + nums[i];
while ((((long long int)nums[i] * ((i - start) + 1)) - s) > k && start < i)
{
s -= nums[start];
start++;
}
int len = (i - start) + 1;
maxLen = max(maxLen, len);
}
return maxLen;
}
};
Time Complexity: O(n log n)
Previous13. Shortest Subarray with Sum at Least K (Negative Numbers)Next15. Count Number of Nice Subarrays
Last updated
Was this helpful?