# Solving Leetcode Interviews in Seconds with AI: Maximum Frequency After Subarray Operation


	# 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](https://www.chatmagic.app), 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(n<sup>3</sup>) 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
	```python
	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
	```