Introduction

In this blog post, we will explore how to solve the LeetCode problem "3434" 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 array nums of length n. You are also given an integer k. You perform the following operation on nums once: Select a subarray nums[i..j] where 0 <= i <= j <= n - 1. Select an integer x and add x to all the elements in nums[i..j]. Find the maximum frequency of the value k after the operation. Example 1: Input: nums = [1,2,3,4,5,6], k = 1 Output: 2 Explanation: After adding -5 to nums[2..5], 1 has a frequency of 2 in [1, 2, -2, -1, 0, 1]. Example 2: Input: nums = [10,2,3,4,5,5,4,3,2,2], k = 10 Output: 4 Explanation: After adding 8 to nums[1..9], 10 has a frequency of 4 in [10, 10, 11, 12, 13, 13, 12, 11, 10, 10]. Constraints: 1 <= n == nums.length <= 105 1 <= nums[i] <= 50 1 <= k <= 50

Explanation

Here's a breakdown of the approach and the Python code:

• Key Idea: The core idea is to iterate through all possible subarrays and, for each subarray, determine the value x that, when added to the subarray, maximizes the frequency of k in the modified array.
• Optimization: Instead of exhaustively trying all possible values of x, we can directly compute the optimal x for a given subarray. Specifically, x = k - nums[i] for each element nums[i] in the subarray.

• Frequency Calculation: After applying the optimal x for a specific element in a subarray, count the frequency of k in the modified array. Track the maximum frequency encountered across all subarrays and elements within the subarrays.

• Complexity:

  • Runtime Complexity: O(n³) due to the three nested loops (iterating through subarrays and elements within the subarrays).
  • Storage Complexity: O(n) for the copy of the input array.

Code

    def maxFrequency(nums, k):
    n = len(nums)
    max_freq = 0

    for i in range(n):
        for j in range(i, n):
            for l in range(i, j + 1):
                temp_nums = nums[:]
                x = k - nums[l]
                for m in range(i, j + 1):
                    temp_nums[m] += x

                freq = temp_nums.count(k)
                max_freq = max(max_freq, freq)

    return max_freq