Problem
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 <= 10^51 <= nums[i] <= 10^51 <= k <= 10^5
Solution (Java)
class Solution {
public int maxFrequency(int[] nums, int k) {
countingSort(nums);
int start = 0;
int preSum = 0;
int total = 1;
for (int i = 0; i < nums.length; i++) {
int length = i - start + 1;
int product = nums[i] * length;
preSum += nums[i];
while (product - preSum > k) {
preSum -= nums[start++];
length--;
product = nums[i] * length;
}
total = Math.max(total, length);
}
return total;
}
private void countingSort(int[] nums) {
int max = Integer.MIN_VALUE;
for (int num : nums) {
max = Math.max(max, num);
}
int[] map = new int[max + 1];
for (int num : nums) {
map[num]++;
}
int i = 0;
int j = 0;
while (i <= max) {
if (map[i]-- > 0) {
nums[j++] = i;
} else {
i++;
}
}
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).