LeetCode 3347: Maximum Frequency of an Element After Performing Operations II Solution
Master LeetCode problem 3347 (Maximum Frequency of an Element After Performing Operations II), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
3347. Maximum Frequency of an Element After Performing Operations II
Problem Explanation
Explanation:
To maximize the frequency of an element in the array after performing operations, we can try to make the elements in the array as close to each other as possible. To achieve this, we can iterate through the array and for each element, calculate the range of values it can take after all operations. Then, we can sort the array and choose the mid-point element as the target value for all elements. This will maximize the frequency of the most common element in the array.
Algorithm:
- Iterate through the array and calculate the range of values each element can take after all operations.
- Sort the array.
- Choose the mid-point element as the target value for all elements.
- Calculate the frequency of the target value in the array.
Time Complexity:
The time complexity of this approach is O(n log n) where n is the length of the input array nums due to the sorting operation.
Space Complexity:
The space complexity is O(1) as we are not using any extra space apart from a few variables.
:
Solution Code
class Solution {
public int maxFrequency(int[] nums, int k, int numOperations) {
Arrays.sort(nums);
long sum = 0;
int i = 0;
int result = 1;
for (int j = 0; j < nums.length; j++) {
sum += nums[j];
while ((long)nums[j] * (j - i + 1) - sum > k) {
sum -= nums[i];
i++;
}
result = Math.max(result, j - i + 1);
}
return result;
}
}Try It Yourself
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