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, 1. Total = 6+2+1 = 9. This is minimum among all possible divisions.

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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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

✓ Brute Force Enumeration

⏱️ Time: O(n²) Space: O(1)

For each possible first cut position and second cut position, calculate the cost (first element of each subarray) and track the minimum sum. Since we need exactly 3 subarrays, we need 2 cuts.

Single Pass with Early Optimization

⏱️ Time: O(n²) Space: O(1)

Still enumerate all possible cuts, but optimize by recognizing that nums[0] is always part of the cost, so we only need to minimize nums[i] + nums[j]. We can also break early in some cases.

Brute Force Enumeration — Algorithm Steps

Step 1: Try all positions for first cut (i from 1 to n-2)
Step 2: Try all positions for second cut (j from i+1 to n-1)
Step 3: Calculate cost = nums[0] + nums[i] + nums[j] and track minimum

Visualization

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Step-by-Step Walkthrough

Input Array

Original array with indices

Try All Cuts

Enumerate all valid (i,j) cut positions

Calculate Costs

Sum first elements of each subarray

Code -

solution.c — C

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

int solution(int* nums, int n) {
    int minCost = INT_MAX;
    
    // Try all possible ways to make 2 cuts
    for (int i = 1; i < n-1; i++) {  // First cut after position i-1
        for (int j = i+1; j < n; j++) {  // Second cut after position j-1
            // Three subarrays: [0:i], [i:j], [j:n]
            // Cost is first element of each subarray
            int cost = nums[0] + nums[i] + nums[j];
            if (cost < minCost) {
                minCost = cost;
            }
        }
    }
    
    return minCost;
}

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)

Nested loops: outer loop runs n-3 times, inner loop runs up to n-3 times

⚠ Quadratic Growth

Space Complexity

O(1)

Only using a few variables to track minimum cost

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force Enumeration — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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