Introduction

In this blog post, we will explore how to solve the LeetCode problem "3347" 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], after which nums becomes [1, 4, 5]. Adding -1 to nums[2], after which 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] <= 109 0 <= k <= 109 0 <= numOperations <= nums.length

Explanation

Here's an efficient solution to determine the maximum possible frequency of any element in the array after performing the operations.

• Core Idea: Sort the array. For each element nums[i], determine the maximum window size (frequency) centered around it that can be achieved using the given numOperations and k.
• Sliding Window: Use a sliding window approach. Expand the window to the left (smaller elements) and determine the operations needed to make all elements within the window equal to the current element nums[i]. If the operations exceed numOperations, shrink the window from the left.

• Binary Search (Optimization): For each element, instead of linearly expanding the sliding window, employ binary search to find the optimal window size. This speeds up the process of finding the longest valid window.

• Complexity: Time: O(n log n), Space: O(1) - where n is the length of the input array nums. Sorting takes O(n log n) time. The binary search within the loop for each element contributes O(n log n) in the worst case (binary search of O(log n) performed for each of the 'n' elements).

Code

    def maxFrequency(nums, k, numOperations):
    nums.sort()
    n = len(nums)
    max_freq = 1

    for i in range(n):
        low = 1
        high = i + 1
        ans = 1

        while low <= high:
            mid = (low + high) // 2
            cost = 0
            for j in range(i, i - mid, -1):
                if i-mid < 0:
                   cost = float('inf')
                   break
                cost += (nums[i] - nums[j])

            if cost <= numOperations * k:
                ans = mid
                low = mid + 1
            else:
                high = mid - 1

        max_freq = max(max_freq, ans)

    return max_freq