LeetCode 1130: Minimum Cost Tree From Leaf Values Solution

Master LeetCode problem 1130 (Minimum Cost Tree From Leaf Values), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1130. Minimum Cost Tree From Leaf Values

Medium

Array Dynamic Programming Stack Greedy Monotonic Stack

Problem Explanation

Explanation:

To solve this problem, we can use dynamic programming. We can build a 2D DP array where dp[i][j] represents the minimum cost to form a tree using the subarray from index i to index j. We can then iterate over different subarrays and calculate the minimum cost for each subarray.

Here are the steps:

Initialize a 2D DP array dp of size n x n where n is the length of the input array.
Fill the diagonal of the DP array with zeros as the cost to form a tree with only one leaf node is zero.
Iterate over different subarray lengths starting from 2 up to n: a. For each subarray starting index i and length len, calculate the cost for forming a tree using this subarray. b. Divide the subarray into two parts and find the minimum cost for combining these two parts.
The minimum cost to form a tree using the entire array will be stored at dp[0][n-1].

Solution Code

class Solution {
    public int mctFromLeafValues(int[] arr) {
        int n = arr.length;
        int[][] dp = new int[n][n];

        for (int len = 2; len <= n; len++) {
            for (int i = 0; i <= n - len; i++) {
                int j = i + len - 1;
                dp[i][j] = Integer.MAX_VALUE;
                for (int k = i; k < j; k++) {
                    for (int l = i; l < k; l++) {
                        for (int r = k + 1; r <= j; r++) {
                            dp[i][j] = Math.min(dp[i][j], dp[l][k] + dp[k + 1][r] + getMax(arr, l, k) * getMax(arr, k + 1, r));
                        }
                    }
                }
            }
        }

        return dp[0][n - 1];
    }

    private int getMax(int[] arr, int start, int end) {
        int max = Integer.MIN_VALUE;
        for (int i = start; i <= end; i++) {
            max = Math.max(max, arr[i]);
        }
        return max;
    }
}

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