Introduction

In this blog post, we will explore how to solve the LeetCode problem "1000" 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

There are n piles of stones arranged in a row. The ith pile has stones[i] stones. A move consists of merging exactly k consecutive piles into one pile, and the cost of this move is equal to the total number of stones in these k piles. Return the minimum cost to merge all piles of stones into one pile. If it is impossible, return -1. Example 1: Input: stones = [3,2,4,1], k = 2 Output: 20 Explanation: We start with [3, 2, 4, 1]. We merge [3, 2] for a cost of 5, and we are left with [5, 4, 1]. We merge [4, 1] for a cost of 5, and we are left with [5, 5]. We merge [5, 5] for a cost of 10, and we are left with [10]. The total cost was 20, and this is the minimum possible. Example 2: Input: stones = [3,2,4,1], k = 3 Output: -1 Explanation: After any merge operation, there are 2 piles left, and we can't merge anymore. So the task is impossible. Example 3: Input: stones = [3,5,1,2,6], k = 3 Output: 25 Explanation: We start with [3, 5, 1, 2, 6]. We merge [5, 1, 2] for a cost of 8, and we are left with [3, 8, 6]. We merge [3, 8, 6] for a cost of 17, and we are left with [17]. The total cost was 25, and this is the minimum possible. Constraints: n == stones.length 1 <= n <= 30 1 <= stones[i] <= 100 2 <= k <= 30

Explanation

Here's the breakdown of the solution:

• Dynamic Programming: Use dynamic programming to store the minimum cost to merge subarrays of stones into a single pile. dp[i][j][m] represents the minimum cost to merge stones from index i to j into m piles.

• Bottom-Up Approach: Iterate through subarrays of increasing length. For each subarray, try all possible merge points such that k consecutive piles are merged at a time.

• Prefix Sum: Utilize a prefix sum array to efficiently calculate the cost of merging k consecutive piles.

• Time & Space Complexity:

  • Time Complexity: O(n^3 * k)
  • Space Complexity: O(n^2 * n) - can be reduced to O(n^2)

Code

    def merge_stones(stones, k):
    n = len(stones)
    if (n - 1) % (k - 1) != 0:
        return -1

    prefix_sum = [0] * (n + 1)
    for i in range(1, n + 1):
        prefix_sum[i] = prefix_sum[i - 1] + stones[i - 1]

    dp = {}  # dp[i, j, m] = min cost to merge stones[i...j] into m piles

    def solve(i, j, m):
        if (i, j, m) in dp:
            return dp[i, j, m]

        if i == j:
            return 0 if m == 1 else float('inf')

        if m == 1:
            cost = solve(i, j, k)
            if cost == float('inf'):
                dp[i, j, m] = float('inf')
                return float('inf')
            else:
                dp[i, j, m] = cost + prefix_sum[j + 1] - prefix_sum[i]
                return dp[i, j, m]

        min_cost = float('inf')
        for l in range(i, j, k - 1):
            cost1 = solve(i, l, 1)
            cost2 = solve(l + 1, j, m - 1)
            min_cost = min(min_cost, cost1 + cost2)

        dp[i, j, m] = min_cost
        return min_cost

    result = solve(0, n - 1, 1)
    return result