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Leetcode 3010. Divide an Array Into Subarrays With Minimum Cost I

Problem Statement

You are given an array nums and an integer k. The array consists of distinct integers into subarrays, such that the result will be reorganized into subarrays, and the cost of dividing the array into those subarrays is minimized. The cost of having a number in a subarray of length x is x * k. Write a function minCost that will compute the minimum cost of the resulting subarrays.

Function Signature

public int minCost(int[] nums, int k)

Clarifying Questions

  1. What are the constraints on the size of the input array nums?
  2. Can k be zero or negative, or is it always a positive integer?
  3. What does the “cost of having a number in a subarray of length x is x * k” mean exactly? Is it the same cost for each number in the subarray?
  4. Should the elements in each subarray be distinct, or can duplicates exist within a subarray?

Strategy

  1. Understand Subarray Formation: The main aim is to divide the array into such subarrays where the cost is minimized. Each subarray has a cost related to its length.
  2. Dynamic Programming Insight: We can use dynamic programming to keep track of the minimum cost to partition the array up to each index.
  3. Cost Calculation: Define a dp[i] which represents the minimum cost to partition the subarray nums[0...i]. The transition would be to add a new subarray ending at i.
  4. Recursive Relationship:
    • Iterate through each possible partition ending at i and update the dp[i] as the minimum cost of forming valid subarrays.
  5. Minimize Cost: Ensure that we trace all possible subarrays and compute the dp value correctly.

Code

Here is the Java code for solving the problem:

import java.util.Arrays;

public class DivideArray {
    public int minCost(int[] nums, int k) {
        int n = nums.length;
        int[] dp = new int[n + 1];
        
        // Initialize the dp array with infinity
        Arrays.fill(dp, Integer.MAX_VALUE);
        dp[0] = 0; // Base case: cost to partition the first 0 elements is 0

        for (int end = 1; end <= n; end++) {
            for (int start = 0; start < end; start++) {
                int len = end - start;
                dp[end] = Math.min(dp[end], dp[start] + len * k);
            }
        }
        return dp[n];
    }
    
    public static void main(String[] args) {
        DivideArray solution = new DivideArray();
        int[] nums = {1, 3, 2, 4};
        int k = 10;
        System.out.println(solution.minCost(nums, k)); // Example output
    }
}

Time Complexity

The overall time complexity of this solution is O(n^2), where n is the length of the nums array. This is because for each index end, we iterate over all possible start indices to calculate the minimum cost for partition ending there. Given the constraints, this should be efficient enough unless the array size is extremely large.

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