Astronaut Food Rationing — Problem Statement & Solution Guide

Two PointersMediumMixed

TimeO(n)

SpaceO(1)

Problem Description

You are tasked with optimizing the storage configuration for a long-duration space mission. The food supply is represented by a sorted array of positive integers, where each element denotes the mass (in kilograms) of a specific ration unit. The array is non-decreasing, meaning lighter rations appear before heavier ones.

Your objective is to partition this inventory into two contiguous segments: a prefix consisting of the first k elements and a suffix consisting of the remaining n-k elements. The partition point k must be chosen such that the absolute difference between the sum of the prefix and the sum of the suffix is minimized. This ensures a balanced load distribution between the two storage modules.

Given the array of ration masses, determine the minimum possible absolute difference between the sum of the two segments. If multiple partition points yield the same minimum difference, return that minimum difference value. Note that k can range from 0 to n, where k=0 implies an empty prefix and k=n implies an empty suffix, though typically in such balancing problems, we consider non-empty segments or allow empty segments as per standard partition definitions. For this problem, assume k can be any integer from 0 to n.

DSA Pattern Breakdown

"Astronaut Food Rationing"

medium

WHY DOES IT MATTER?

The two‑pointer pattern is essential because it transforms a potentially quadratic problem into a linear one by leveraging the sorted property of the data. It eliminates redundant computations and ensures that each element is processed only once, which is critical for large datasets typical in production systems.

OPTIMIZATION CHALLENGE

The key insight is that the sum of a segment can be updated in O(1) when a pointer moves, because you only add or subtract a single element. This eliminates the need for recomputing sums from scratch at each step, which would otherwise lead to O(n^2) complexity.

REAL-WORLD CONNECTION

Think of a warehouse with shelves sorted by product weight. A forklift operator needs to load two trucks: one with lighter items and one with heavier items. By moving a single pointer from each end of the shelf and adjusting the load based on weight thresholds, the operator can efficiently decide where to split the shipment without scanning the entire shelf repeatedly.

When explaining this to an interviewer, emphasize the monotonicity of the sums and how moving a pointer can never invalidate a previous decision. Also, be ready to discuss edge cases like equal sums or when the optimal split is at the extremes of the array.

COMPLEXITY AT A GLANCE

⏱ Time:O(n)

💾 Space:O(1)

Core Theory — Why This Approach?

The core of this problem lies in recognizing that the array is sorted in non‑decreasing order, which guarantees that any contiguous segment starting from the left will contain lighter rations and any segment starting from the right will contain heavier rations. A naive approach would iterate over every possible split point, compute the sums of the two resulting subarrays, and compare them, leading to an O(n^2) time complexity due to repeated summations. By contrast, the optimal two‑pointer strategy maintains two running sums—one for the left segment and one for the right segment—and moves the pointers inward while adjusting the sums in constant time. This reduces the problem to a single linear scan, achieving O(n) time and O(1) extra space.

The two‑pointer paradigm is powerful when the input has an inherent order that can be exploited. In this case, the sorted nature of the array allows us to decide which side to shrink based on the relative sizes of the current sums: if the left sum is too large, we move the right pointer left to include heavier rations in the right segment; if the right sum is too large, we move the left pointer right to include lighter rations in the left segment. This greedy adjustment guarantees that we converge on the optimal partition without backtracking.

Moreover, the algorithm’s correctness hinges on the fact that moving a pointer only increases the sum of the segment it belongs to, never decreases it. Thus, once a pointer has moved past a certain index, that index can never be part of the optimal partition again, ensuring linear progress. This monotonicity is what makes the two‑pointer technique both efficient and elegant for this class of partition problems.

Interview Questions on This Problem

Q1How would you modify the two‑pointer solution if the array were not sorted?

If the array is unsorted, the two‑pointer technique no longer guarantees that moving a pointer will monotonically increase the sum of its segment. In that case, you would need to compute prefix sums first, then use a binary search or two‑pass approach to find the split point, resulting in O(n) time but requiring O(n) extra space for the prefix sums.

Q2Can you explain a real‑world scenario where this partitioning logic would be applied in a fintech product?

In a fintech app that balances user portfolios, you might need to split a list of assets into a low‑risk and high‑risk bucket based on their volatility scores. Since assets are sorted by volatility, a two‑pointer approach can quickly find the threshold where the cumulative risk of the low‑risk bucket stays below a target, ensuring a balanced portfolio.

Q3What would be the impact on time complexity if you were asked to return all possible partition indices that satisfy the condition instead of just one?

If you need to return every valid split point, you would still traverse the array once, but you would record each index where the condition holds. The time complexity remains O(n) because you only perform constant work per element; however, the output size could be O(n) in the worst case, so the overall complexity becomes O(n) time and O(n) space for the output.

Examples

Example 1

Input

rations = [1, 2, 3, 4, 5]

Output

Explanation: Total sum is 15. We test partition points: - k=0: Prefix sum=0, Suffix sum=15, Diff=15 - k=1: Prefix sum=1, Suffix sum=14, Diff=13 - k=2: Prefix sum=3, Suffix sum=12, Diff=9 - k=3: Prefix sum=6, Suffix sum=9, Diff=3 - k=4: Prefix sum=10, Suffix sum=5, Diff=5 - k=5: Prefix sum=15, Suffix sum=0, Diff=15 The minimum difference is 3, but wait, let's re-evaluate. Actually, for [1,2,3,4,5], total=15. Let's check k=3: sum(1,2,3)=6, sum(4,5)=9, diff=3. Let's check if there is a better one. Wait, the example output in the prompt draft was 0? No, 15 is odd, so diff cannot be 0. Let's pick a better example for output 0. Let's use [1, 2, 3, 4]. Total=10. k=2: sum(1,2)=3, sum(3,4)=7, diff=4. k=1: sum(1)=1, sum(2,3,4)=9, diff=8. k=3: sum(1,2,3)=6, sum(4)=4, diff=2. Min is 2. Let's create a new example for output 0. Input: [1, 1, 1, 1]. Total=4. k=2: sum(1,1)=2, sum(1,1)=2, diff=0. Let's stick to the generated examples in the final JSON. Example 1: [1, 2, 3, 4, 5] Total = 15. Prefix sums: 0, 1, 3, 6, 10, 15. Suffix sums: 15, 14, 12, 9, 5, 0. Diffs: 15, 13, 9, 3, 5, 15. Min diff is 3. Example 2: [1, 1, 1, 1] Total = 4. Prefix sums: 0, 1, 2, 3, 4. Suffix sums: 4, 3, 2, 1, 0. Diffs: 4, 2, 0, 2, 4. Min diff is 0. Example 3: [2, 2, 2, 2, 2] Total = 10. Prefix sums: 0, 2, 4, 6, 8, 10. Suffix sums: 10, 8, 6, 4, 2, 0. Diffs: 10, 6, 2, 2, 6, 10. Min diff is 2. Example 4: [1, 2, 3, 4, 5, 6] Total = 21. Prefix sums: 0, 1, 3, 6, 10, 15, 21. Suffix sums: 21, 20, 18, 15, 11, 6, 0. Diffs: 21, 19, 15, 9, 1, 9, 21. Min diff is 1.

Example 2

Input

rations = [1, 1, 1, 1]

Output

Explanation: Total sum is 4. Partition at k=2 gives prefix [1, 1] with sum 2 and suffix [1, 1] with sum 2. The absolute difference is |2 - 2| = 0. This is the minimum possible difference.

Example 3

Input

rations = [2, 2, 2, 2, 2]

Output

Explanation: Total sum is 10. Possible partitions: - k=2: Prefix sum = 4, Suffix sum = 6, Diff = 2. - k=3: Prefix sum = 6, Suffix sum = 4, Diff = 2. All other partitions yield larger differences. The minimum difference is 2.

Example 4

Input

rations = [1, 2, 3, 4, 5, 6]

Output

Explanation: Total sum is 21. Partition at k=4 gives prefix [1, 2, 3, 4] with sum 10 and suffix [5, 6] with sum 11. The absolute difference is |10 - 11| = 1. This is the minimum possible difference.

Constraints

1 <= rations.length <= 10^5
1 <= rations[i] <= 10^4
rations is sorted in non-decreasing order

Optimal Approach & Strategy

Maintain two running sums with two pointers moving inward; adjust the sums in constant time based on which side is heavier, achieving O(n) time and O(1) space.

Brute Force Approach

Check every possible split point, compute the sum of the left and right segments for each, and compare them. This requires O(n^2) time because you recompute sums for each split.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n)

function minRationDifference(rations) {
    const n = rations.length;
    if (n === 0) return 0;
    if (n === 1) return rations[0];
    
    let totalSum = 0;
    for (let i = 0; i < n; i++) {
        totalSum += rations[i];
    }
    
    let prefixSum = 0;
    let minDiff = Infinity;
    
    for (let k = 1; k < n; k++) {
        prefixSum += rations[k - 1];
        const suffixSum = totalSum - prefixSum;
        const diff = Math.abs(prefixSum - suffixSum);
        if (diff < minDiff) {
            minDiff = diff;
        }
    }
    
    return minDiff;
}

// Driver code
const readline = require('readline');
const rl = readline.createInterface({
    input: process.stdin,
    terminal: false
});

let lines = [];
rl.on('line', line => lines.push(line));
rl.on('close', () => {
    const n = parseInt(lines[0]);
    const rations = lines[1].split(' ').map(Number);
    console.log(minRationDifference(rations));
});

#include <iostream>
#include <vector>
#include <cmath>
#include <algorithm>

using namespace std;

int minRationDifference(vector<int>& rations) {
    int n = rations.size();
    if (n == 0) return 0;
    if (n == 1) return rations[0];
    
    long long totalSum = 0;
    for (int i = 0; i < n; i++) {
        totalSum += rations[i];
    }
    
    long long prefixSum = 0;
    long long minDiff = LLONG_MAX;
    
    for (int k = 1; k < n; k++) {
        prefixSum += rations[k - 1];
        long long suffixSum = totalSum - prefixSum;
        long long diff = abs(prefixSum - suffixSum);
        if (diff < minDiff) {
            minDiff = diff;
        }
    }
    
    return (int)minDiff;
}

int main() {
    int n;
    cin >> n;
    vector<int> rations(n);
    for (int i = 0; i < n; i++) {
        cin >> rations[i];
    }
    
    cout << minRationDifference(rations) << endl;
    
    return 0;
}

import java.util.*;

public class Main {
    public static int minRationDifference(int[] rations) {
        int n = rations.length;
        if (n == 0) return 0;
        if (n == 1) return rations[0];
        
        long totalSum = 0;
        for (int i = 0; i < n; i++) {
            totalSum += rations[i];
        }
        
        long prefixSum = 0;
        long minDiff = Long.MAX_VALUE;
        
        for (int k = 1; k < n; k++) {
            prefixSum += rations[k - 1];
            long suffixSum = totalSum - prefixSum;
            long diff = Math.abs(prefixSum - suffixSum);
            if (diff < minDiff) {
                minDiff = diff;
            }
        }
        
        return (int)minDiff;
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int[] rations = new int[n];
        for (int i = 0; i < n; i++) {
            rations[i] = scanner.nextInt();
        }
        
        System.out.println(minRationDifference(rations));
        scanner.close();
    }
}

def min_ration_difference(rations):
    n = len(rations)
    if n == 0:
        return 0
    if n == 1:
        return rations[0]
    
    total_sum = sum(rations)
    prefix_sum = 0
    min_diff = float('inf')
    
    for k in range(1, n):
        prefix_sum += rations[k - 1]
        suffix_sum = total_sum - prefix_sum
        diff = abs(prefix_sum - suffix_sum)
        if diff < min_diff:
            min_diff = diff
    
    return min_diff

if __name__ == "__main__":
    import sys
    input = sys.stdin.read
    data = input().split()
    n = int(data[0])
    rations = list(map(int, data[1:n+1]))
    print(min_ration_difference(rations))

function minRationDifference(rations) {
    const n = rations.length;
    if (n === 0) return 0;
    if (n === 1) return rations[0];
    
    let totalSum = 0;
    for (let i = 0; i < n; i++) {
        totalSum += rations[i];
    }
    
    let prefixSum = 0;
    let minDiff = Infinity;
    
    for (let k = 1; k < n; k++) {
        prefixSum += rations[k - 1];
        const suffixSum = totalSum - prefixSum;
        const diff = Math.abs(prefixSum - suffixSum);
        if (diff < minDiff) {
            minDiff = diff;
        }
    }
    
    return minDiff;
}

// Driver code
const readline = require('readline');
const rl = readline.createInterface({
    input: process.stdin,
    terminal: false
});

let lines = [];
rl.on('line', line => lines.push(line));
rl.on('close', () => {
    const n = parseInt(lines[0]);
    const rations = lines[1].split(' ').map(Number);
    console.log(minRationDifference(rations));
});

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Astronaut Food Rationing Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Input

Output

Constraints

More Two Pointers

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Astronaut Food Rationing — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor