LeetCode 3013: Divide an Array Into Subarrays With Minimum Cost II Solution

Master LeetCode problem 3013 (Divide an Array Into Subarrays With Minimum Cost II), 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.

3013. Divide an Array Into Subarrays With Minimum Cost II

Hard

Array Hash Table Sliding Window Heap (Priority Queue)

Problem Explanation

Explanation

To solve this problem, we can use dynamic programming. We will iterate over all possible positions to divide the array into k subarrays and calculate the minimum cost for each division. We will maintain a 2D DP array where DP[i][j] represents the minimum cost to divide the subarray nums[0..i] into j subarrays with the given constraints.

At each position i, we will consider all valid positions to split the array and update the minimum cost based on the previous calculations. The final result will be the minimum cost to divide the entire array into k subarrays.

The time complexity of this approach is O(n * k * dist) and the space complexity is O(n * k).

Solution Code

class Solution {
    public int minCost(int[] nums, int k, int dist) {
        int n = nums.length;
        int[][] dp = new int[n][k];
        
        for (int i = 0; i < n; i++) {
            int minCost = nums[i];
            for (int j = 0; j < k; j++) {
                dp[i][j] = minCost;
                if (j == k - 1) continue;
                for (int x = i - 1; x >= 0 && i - x <= dist; x--) {
                    minCost = Math.min(minCost, dp[x][j] + nums[i]);
                    dp[i][j] = Math.min(dp[i][j], minCost);
                }
            }
        }
        
        return dp[n - 1][k - 1];
    }
}

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