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

Problem Statement

Given an integer array nums and an integer k, you need to divide the array into exactly k non-empty subarrays such that the cost is minimized. The cost of a subarray is the square of its length. The total cost is the sum of the costs of the individual subarrays.

Your task is to implement a function that finds the minimum possible total cost.

Function Signature

int minCost(vector<int>& nums, int k);

Example

Input: nums = [1, 2, 1, 2, 1, 2], k = 3
Output: 1
Explanation: We can divide the array as [1,2], [1,2], [1,2]. The cost is 2^2 + 2^2 + 2^2 = 4 + 4 + 4 = 12

Clarifying Questions

  1. Array Size: What constraints are there on the size of the array nums and the value of k?
    • Response: The size of the array nums ranges from 1 to 1000, and values for k range from 1 to nums.size().
  2. Array Contents: Are there any constraints on the values within the array nums?
    • Response: No, the values are just integer values and there are no specific constraints on them.
  3. Solution Permissibility: Is using dynamic programming permissible in solving this problem?
    • Response: Yes, you can use dynamic programming to solve it.

Strategy

We will use a dynamic programming approach to solve this problem.

Steps:

  1. Define DP State: Define dp[i][j] as the minimum cost to divide the first i elements into j subarrays.

  2. Base Case: dp[0][0] = 0, because the cost of dividing zero elements into zero subarrays is zero.

  3. Transition: For each i (from 1 to n) and j (from 1 to k):
    • We will consider dividing the array from every possible previous index m (from 0 to i-1), and calculate the cost of making a subarray from m+1 to i.
    • The cost of each subarray [m+1, i] will be (i - m) * (i - m).
    • Then update dp[i][j] as the minimum of its current value and dp[m][j-1] + cost of the subarray.
  4. Final Answer: The value dp[n][k] will hold the minimum cost to divide the array into exactly k subarrays.

Code

#include <vector>
#include <algorithm>
#include <limits.h>

using namespace std;

int minCost(vector<int>& nums, int k) {
    int n = nums.size();
    vector<vector<int>> dp(n + 1, vector<int>(k + 1, INT_MAX));
    
    // Base case
    dp[0][0] = 0;
    
    // DP array to keep result
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= k; ++j) {
            for (int m = 0; m < i; ++m) {
                int length_squared = (i - m) * (i - m);
                if (dp[m][j-1] != INT_MAX) {
                    dp[i][j] = min(dp[i][j], dp[m][j-1] + length_squared);
                }
            }
        }
    }
    
    return dp[n][k];
}

Time Complexity

  1. Space Complexity: The space complexity is O(n * k) due to the use of the dp table with dimensions (n+1) x (k+1).

  2. Time Complexity:

    • The dp table filling takes O(n^2 * k) since for each i and j, we’re iterating over all possible previous indices m.
    • Therefore, the overall time complexity is O(n^2 * k).

This solution should be efficient enough for the input constraints provided (considering n and k up to 1000).

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