LeetCode 1547: Minimum Cost to Cut a Stick Solution

Master LeetCode problem 1547 (Minimum Cost to Cut a Stick), a hard 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.

1547. Minimum Cost to Cut a Stick

Hard

Array Dynamic Programming Sorting

Problem Explanation

Explanation

To solve this problem, we can use dynamic programming. We will sort the cuts array and then create a new array that includes 0 and n as the start and end points. We will then iterate through the cuts array and calculate the cost for cutting at each position. We will store these costs in a 2D array, where dp[i][j] represents the minimum cost to cut a stick from position i to position j.

We will then find the minimum cost by recursively calculating the cost for each possible subinterval. The base case is when the interval consists of adjacent positions.

The time complexity of this solution is O(n^3) and the space complexity is O(n^2), where n is the length of the cuts array.

Solution Code

class Solution {
    public int minCost(int n, int[] cuts) {
        Arrays.sort(cuts);
        int[] newCuts = new int[cuts.length + 2];
        newCuts[0] = 0;
        newCuts[newCuts.length - 1] = n;
        System.arraycopy(cuts, 0, newCuts, 1, cuts.length);
        
        int[][] dp = new int[newCuts.length][newCuts.length];
        
        for (int l = 2; l < dp.length; l++) {
            for (int i = 0; i < dp.length - l; i++) {
                int j = i + l;
                dp[i][j] = Integer.MAX_VALUE;
                for (int k = i + 1; k < j; k++) {
                    dp[i][j] = Math.min(dp[i][j], dp[i][k] + dp[k][j] + newCuts[j] - newCuts[i]);
                }
            }
        }
        
        return dp[0][dp[0].length - 1];
    }
}

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