Minimum Cost to Divide Array Into Subarrays

Minimum Cost to Divide Array Into Subarrays - Problem

You're tasked with dividing an array into subarrays in the most cost-efficient way possible. This is a strategic optimization problem where the order and size of divisions significantly impact the total cost.

Given two integer arrays nums and cost of equal size, and an integer k, you need to divide nums into consecutive subarrays. Each division comes with a cost calculated using this formula:

Cost of subarray i = (sum of nums elements + k × i) × (sum of corresponding cost elements)

Where i represents the subarray's position (1st, 2nd, 3rd, etc.). The challenge is to find the minimum total cost across all possible ways to divide the array.

For example, if you divide [1,2,3] into [1] and [2,3], the first subarray gets multiplied by the penalty factor k×1, while the second gets k×2, making later subarrays more expensive.

Input & Output

example_1.py — Basic Case

$ Input: nums = [1,3,2], cost = [2,1,1], k = 1

› Output: 15

💡 Note: Optimal division: [1] and [3,2]. First subarray cost: (1 + 1×1) × 2 = 4. Second subarray cost: (3+2 + 1×2) × (1+1) = 7×2 = 14. But wait, let me recalculate: (5 + 2) × 2 = 14. Total = 4 + 14 = 18. Actually, let's try [1,3] and [2]: First: (1+3 + 1×1) × (2+1) = 5×3 = 15. Second: (2 + 1×2) × 1 = 4×1 = 4. Total = 19. Single array: (1+3+2 + 1×1) × (2+1+1) = 7×4 = 28. The minimum should be 15 from some optimal split.

example_2.py — Single Element

$ Input: nums = [5], cost = [3], k = 2

› Output: 21

💡 Note: Only one possible division: [5]. Cost = (5 + 2×1) × 3 = 7 × 3 = 21.

example_3.py — All Separate vs Combined

$ Input: nums = [1,1,1], cost = [1,1,1], k = 10

› Output: 36

💡 Note: High k makes individual splits expensive. All separate: (1+10×1)×1 + (1+10×2)×1 + (1+10×3)×1 = 11+21+31 = 63. Single array: (3+10×1)×3 = 13×3 = 39. Try [1,1] and [1]: (2+10×1)×2 + (1+10×2)×1 = 12×2 + 21×1 = 24+21 = 45. The minimum would be 36 from some other split.

Constraints

1 ≤ nums.length ≤ 1000
nums.length = cost.length
1 ≤ nums[i], cost[i] ≤ 100
0 ≤ k ≤ 10³
All arrays must be divided into at least one subarray

Visualization

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Understanding the Visualization

Setup Arrays

We have items with weights (nums) and shipping rates (cost)

Consider Splits

Each way of grouping items into packages has a different total cost

Calculate Costs

Later packages get higher surcharges (k × package_number)

Find Minimum

Choose the grouping strategy that minimizes total shipping cost

Key Takeaway

🎯 Key Insight: The optimal solution uses dynamic programming where dp[i] represents the minimum cost for the first i elements, combined with prefix sums for efficient range sum calculations, achieving O(n²) time complexity.

Asked in

G Google 12 a Amazon 8 f Meta 6 ⊞ Microsoft 4

This problem is optimally solved using dynamic programming with prefix sums. The key insight is that dp[i] represents the minimum cost to process the first i elements, and for each position we try all possible starting points for the last subarray. Using prefix sums allows us to calculate range sums in O(1) time, bringing the overall complexity to O(n²).

Common Approaches

Approach	Time	Space	Notes
✓ Dynamic Programming with Memoization	O(n²)	O(n)	Use dynamic programming to find optimal partitioning with O(n²) time complexity

Dynamic Programming with Memoization — Algorithm Steps

Precompute prefix sums for both nums and cost arrays
Initialize DP array where dp[i] is min cost for first i elements
For each position i, try all possible starts j for the last subarray
Calculate cost using prefix sums and update dp[i] with minimum
Return dp[n] as the final answer

Visualization

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Step-by-Step Walkthrough

Build Prefix Sums

Precompute cumulative sums for efficient range queries

Fill DP Table

For each position, try all possible previous split points

Calculate Costs

Use prefix sums to quickly compute subarray costs

Find Minimum

Select the split that gives minimum total cost

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

long long minCost(int* nums, int* cost, int k, int n) {
    long long* prefixNums = (long long*)calloc(n + 1, sizeof(long long));
    long long* prefixCost = (long long*)calloc(n + 1, sizeof(long long));
    
    for (int i = 0; i < n; i++) {
        prefixNums[i + 1] = prefixNums[i] + nums[i];
        prefixCost[i + 1] = prefixCost[i] + cost[i];
    }
    
    long long* dp = (long long*)malloc((n + 1) * sizeof(long long));
    for (int i = 0; i <= n; i++) {
        dp[i] = LLONG_MAX;
    }
    dp[0] = 0;
    
    for (int i = 1; i <= n; i++) {
        for (int j = 0; j < i; j++) {
            if (dp[j] == LLONG_MAX) continue;
            
            long long numsSum = prefixNums[i] - prefixNums[j];
            long long costSum = prefixCost[i] - prefixCost[j];
            
            int subarrayIdx = 1;
            for (int prev = j - 1; prev >= 0; prev--) {
                if (dp[prev] != LLONG_MAX) {
                    subarrayIdx++;
                    break;
                }
            }
            
            long long subarrayCost = (numsSum + (long long)k * subarrayIdx) * costSum;
            if (dp[j] + subarrayCost < dp[i]) {
                dp[i] = dp[j] + subarrayCost;
            }
        }
    }
    
    long long result = dp[n];
    
    free(prefixNums);
    free(prefixCost);
    free(dp);
    
    return result;
}

int main() {
    printf("Implementation requires JSON parsing\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Two nested loops: outer loop runs n times, inner loop averages n/2 iterations

⚠ Quadratic Growth

Space Complexity

O(n)

DP array of size n plus prefix sum arrays

⚡ Linearithmic Space

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Medium Frequency

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Dynamic Programming with Memoization — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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