Introduction

In this blog post, we will explore how to solve the LeetCode problem "3505" 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, x and k. You can perform the following operation any number of times (including zero): Increase or decrease any element of nums by 1. Return the minimum number of operations needed to have at least k non-overlapping subarrays of size exactly x in nums, where all elements within each subarray are equal. Example 1: Input: nums = [5,-2,1,3,7,3,6,4,-1], x = 3, k = 2 Output: 8 Explanation: Use 3 operations to add 3 to nums[1] and use 2 operations to subtract 2 from nums[3]. The resulting array is [5, 1, 1, 1, 7, 3, 6, 4, -1]. Use 1 operation to add 1 to nums[5] and use 2 operations to subtract 2 from nums[6]. The resulting array is [5, 1, 1, 1, 7, 4, 4, 4, -1]. Now, all elements within each subarray [1, 1, 1] (from indices 1 to 3) and [4, 4, 4] (from indices 5 to 7) are equal. Since 8 total operations were used, 8 is the output. Example 2: Input: nums = [9,-2,-2,-2,1,5], x = 2, k = 2 Output: 3 Explanation: Use 3 operations to subtract 3 from nums[4]. The resulting array is [9, -2, -2, -2, -2, 5]. Now, all elements within each subarray [-2, -2] (from indices 1 to 2) and [-2, -2] (from indices 3 to 4) are equal. Since 3 operations were used, 3 is the output. Constraints: 2 <= nums.length <= 105 -106 <= nums[i] <= 106 2 <= x <= nums.length 1 <= k <= 15 2 <= k * x <= nums.length

Explanation

Here's the solution to the problem, addressing the efficiency and optimality requirements:

Key High-Level Approach:

• Dynamic Programming: The core idea is to use dynamic programming to explore different combinations of subarrays. The DP state represents the number of subarrays formed so far and the ending index.
• Cost Calculation: For each potential subarray, the median value is chosen as the target value to minimize the total operations. The cost to make the subarray elements equal to the median is then calculated.

• Optimization: The DP transitions consider both forming a new subarray at the current index and skipping the current index.

• Runtime & Storage Complexity: O(n k log(n)) time, O(n * k) space. The log(n) factor comes from sorting to find the median within each potential subarray.

Code

    import sys

def min_operations(nums, x, k):
    n = len(nums)
    dp = {}  # dp[(index, num_subarrays)] = min_cost

    def solve(index, num_subarrays):
        if num_subarrays == k:
            return 0
        if index > n - x or num_subarrays > k:
            return float('inf')

        if (index, num_subarrays) in dp:
            return dp[(index, num_subarrays)]

        # Option 1: Don't start a new subarray here
        cost = solve(index + 1, num_subarrays)

        # Option 2: Start a new subarray here
        subarray = nums[index:index + x]
        subarray_sorted = sorted(subarray)
        median = subarray_sorted[x // 2]

        subarray_cost = 0
        for val in subarray:
            subarray_cost += abs(val - median)

        cost = min(cost, subarray_cost + solve(index + x, num_subarrays + 1))

        dp[(index, num_subarrays)] = cost
        return cost

    result = solve(0, 0)
    return result if result != float('inf') else -1