Introduction

In this blog post, we will explore how to solve the LeetCode problem "3346" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given an integer array nums and two integers k and numOperations. You must perform an operation numOperations times on nums, where in each operation you: Select an index i that was not selected in any previous operations. Add an integer in the range [-k, k] to nums[i]. Return the maximum possible frequency of any element in nums after performing the operations. Example 1: Input: nums = [1,4,5], k = 1, numOperations = 2 Output: 2 Explanation: We can achieve a maximum frequency of two by: Adding 0 to nums[1]. nums becomes [1, 4, 5]. Adding -1 to nums[2]. nums becomes [1, 4, 4]. Example 2: Input: nums = [5,11,20,20], k = 5, numOperations = 1 Output: 2 Explanation: We can achieve a maximum frequency of two by: Adding 0 to nums[1]. Constraints: 1 <= nums.length <= 105 1 <= nums[i] <= 105 0 <= k <= 105 0 <= numOperations <= nums.length

Explanation

Here's a solution approach, complexity analysis, and the code:

• Sort and Group: Sort the array to group similar numbers together. The core idea is to try to make elements equal to a chosen target value. Sorting enables an efficient sliding window approach.
• Sliding Window: Use a sliding window to check the feasibility of achieving a certain frequency. For each window, calculate the number of operations needed to make all elements within the window equal to the rightmost element of the window.

• Maximize Frequency: Iterate through possible frequencies (window sizes) and adjust the window to maximize frequency while respecting the operation limit.

• Runtime Complexity: O(n log n) due to sorting.

• Storage Complexity: O(1) (excluding space for sorting, which might be O(log n) or O(n) depending on the sorting algorithm).

Code

    def maxFrequency(nums, k, numOperations):
    nums.sort()
    n = len(nums)
    left = 0
    operations_used = 0
    max_freq = 0

    for right in range(n):
        operations_used += (nums[right] - nums[right - 1]) * (right - left) if right > 0 else 0

        while operations_used > numOperations:
            operations_used -= (nums[right] - nums[left])
            left += 1

        max_freq = max(max_freq, right - left + 1)

    return max_freq