LeetCode 3500: Minimum Cost to Divide Array Into Subarrays Solution

Master LeetCode problem 3500 (Minimum Cost to Divide Array Into Subarrays), 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.

3500. Minimum Cost to Divide Array Into Subarrays

Hard

Array Dynamic Programming Prefix Sum

Problem Explanation

Explanation:

To solve this problem, we can use dynamic programming. We can define a dp array where dp[i] represents the minimum cost to divide the subarray nums[0...i] into subarrays according to the given constraints. We can iterate over all possible indices to divide the array and calculate the cost at each division point. The final answer will be dp[n-1], where n is the size of the input arrays.

Here are the steps:

Initialize a dp array of size n where n is the size of the input arrays.
Iterate over all possible indices to divide the array and calculate the cost at each division point.
Update dp[i] with the minimum cost obtained from all possible divisions before index i.
Return dp[n-1] as the final answer.

Time Complexity: O(n^2) where n is the size of the input arrays.

Space Complexity: O(n) where n is the size of the input arrays.

Solution Code

class Solution {
    public int minCost(int[] nums, int[] cost, int k) {
        int n = nums.length;
        int[] dp = new int[n];
        
        for (int i = 1; i < n; i++) {
            dp[i] = Integer.MAX_VALUE;
            int sum = nums[i];
            int totalCost = cost[i];
            
            for (int j = i; j >= 0 && j >= i - k + 1; j--) {
                dp[i] = Math.min(dp[i], dp[j - 1] + (sum + k * (i - j + 1)) * totalCost);
                
                if (j > 0) {
                    sum += nums[j - 1];
                    totalCost += cost[j - 1];
                }
            }
        }
        
        return dp[n - 1];
    }
}

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