Divide an Array Into Subarrays With Minimum Cost I

Divide an Array Into Subarrays With Minimum Cost I - Problem

You are given an array of integers nums of length n.

The cost of an array is the value of its first element. For example, the cost of [1,2,3] is 1 while the cost of [3,4,1] is 3.

You need to divide nums into 3 disjoint contiguous subarrays.

Return the minimum possible sum of the cost of these subarrays.

Input & Output

Example 1 — Basic Case

$ Input: nums = [1,2,3,4]

› Output: 6

💡 Note: Divide into [1], [2], [3,4]. Costs are 1, 2, 3. Total = 1+2+3 = 6. This is minimum among all possible divisions.

Example 2 — Different Values

$ Input: nums = [6,2,7,4,1]

› Output: 9

💡 Note: Divide into [6], [2,7], [4,1]. Costs are 6, 2, 4. Total = 6+2+4 = 12. Alternative: [6,2], [7], [4,1] gives 6+7+4=17. Best is [6], [2], [7,4,1] = 6+2+7=15. Actually [6], [2,7,4], [1] = 6+2+1=9.

Example 3 — Minimum Size

$ Input: nums = [20,6,2]

› Output: 28

💡 Note: Only one way to divide into 3 subarrays: [20], [6], [2]. Costs are 20, 6, 2. Total = 20+6+2 = 28.

Constraints

3 ≤ nums.length ≤ 50
1 ≤ nums[i] ≤ 50

Visualization

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Understanding the Visualization

Input Array

Array [1,2,3,4] needs to be split into 3 parts

Make 2 Cuts

Try all valid positions to make exactly 2 cuts

Calculate Cost

Sum the first element of each resulting subarray

Key Takeaway

🎯 Key Insight: Cost equals sum of first elements, so try all valid cut positions to find minimum

Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is to enumerate all possible ways to make 2 cuts in the array, creating 3 contiguous subarrays, and calculate the cost as the sum of the first elements. Best approach is brute force enumeration with nested loops. Time: O(n²), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ Single Pass with Early Optimization	O(n²)	O(1)	Optimize the enumeration by avoiding redundant calculations
Brute Force Enumeration	O(n²)	O(1)	Try all possible ways to divide the array into 3 subarrays

Single Pass with Early Optimization — Algorithm Steps

Step 1: Extract nums[0] as constant part of cost
Step 2: Find minimum of nums[i] + nums[j] for valid i,j pairs
Step 3: Return nums[0] + minimum sum

Visualization

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Step-by-Step Walkthrough

Fixed Cost

nums[0] is always the first subarray's cost

Variable Part

Find minimum nums[i] + nums[j]

Final Result

Add fixed + minimum variable

Code -

solution.c — C

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

int solution(int* nums, int n) {
    int firstCost = nums[0];  // Cost of first subarray is always nums[0]
    int minRemaining = INT_MAX;
    
    // Find minimum nums[i] + nums[j] for valid cuts
    for (int i = 1; i < n-1; i++) {  // Start of second subarray
        for (int j = i+1; j < n; j++) {  // Start of third subarray
            int remainingCost = nums[i] + nums[j];
            if (remainingCost < minRemaining) {
                minRemaining = remainingCost;
            }
        }
    }
    
    return firstCost + minRemaining;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse JSON array manually
    int nums[1000];
    int n = 0;
    char* token = strtok(line, "[],\n");
    while (token != NULL) {
        if (strlen(token) > 0 && token[0] != ' ') {
            nums[n++] = atoi(token);
        }
        token = strtok(NULL, "[],\n");
    }
    
    int result = solution(nums, n);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Still need nested loops to try all cut position combinations

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra variables

✓ Linear Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Single Pass with Early Optimization — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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