LeetCode 3010: Divide an Array Into Subarrays With Minimum Cost I Solution

Master LeetCode problem 3010 (Divide an Array Into Subarrays With Minimum Cost I), a easy 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.

3010. Divide an Array Into Subarrays With Minimum Cost I

Easy

Array Sorting Enumeration

Problem Explanation

Explanation

To solve this problem, we can use dynamic programming. We will iterate over all possible positions to split the array into 3 subarrays. For each split position, we calculate the cost of the left, middle, and right subarrays. We keep track of the minimum cost obtained so far.

At each iteration, we update the minimum cost by considering the cost of the current split position compared to the previous minimum cost. Finally, we return the minimum cost found.

Algorithm:

Initialize a 2D array dp[][] of size n x n x n where n is the length of the input array nums.
Fill the dp[][][] array using dynamic programming to calculate the minimum cost of splitting the array into 3 subarrays at each possible split position.
Return the minimum cost obtained from dp[0][0][n-1], where [0][0][n-1] represents splitting the array from index 0 to n-1.

Time Complexity:

The time complexity of this approach is O(n^3), where n is the length of the input array.

Space Complexity:

The space complexity is O(n^3) due to the 3D array dp[][][].

Solution Code

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

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                for (int k = 0; k < n; k++) {
                    dp[i][j][k] = Integer.MAX_VALUE;
                }
            }
        }

        for (int i = 0; i < n; i++) {
            for (int j = i; j < n; j++) {
                int sum = 0;
                for (int k = i; k <= j; k++) {
                    sum += nums[k];
                }
                dp[i][j][j] = sum;
            }
        }

        for (int len = 1; len < n; len++) {
            for (int i = 0; i + len < n; i++) {
                int j = i + len;
                for (int k = i; k < j; k++) {
                    dp[i][j][2] = Math.min(dp[i][j][2], dp[i][k][0] + dp[k + 1][j][1]);
                }
            }
        }

        return dp[0][n - 1][2];
    }
}

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