Space Station Supply Chain Optimization — Problem Statement & Solution Guide

ArraysMediumAlternating sum subarray

TimeO(n)

SpaceO(1)

Problem Description

Given an integer array nums, determine the greatest possible alternating sum obtainable from any non‑empty contiguous subarray. For a chosen subarray nums[l..r] the alternating sum is defined as nums[l]-nums[l+1]+nums[l+2]-nums[l+3]+… (the first element is added, signs then alternate). The program receives the array and must output the maximum alternating sum over all such subarrays. The solution must run in O(n) time and O(1) extra memory.

DSA Pattern Breakdown

"Space Station Supply Chain Optimization"

medium

WHY DOES IT MATTER?

The pattern exemplifies how to adapt classic maximum‑subarray techniques to problems where the contribution of each element depends on its relative position, a common twist in financial and signal‑processing domains.

OPTIMIZATION CHALLENGE

Recognizing that only two DP states are needed – one for each possible sign of the last element – collapses the naïve O(n²) search to O(n) time and O(1) space, the key insight being the sign‑flip recurrence.

REAL-WORLD CONNECTION

Think of a satellite’s power budget where charging (+) and consumption (‑) alternate each orbit; optimizing the net energy over a contiguous sequence of orbits mirrors the alternating‑sum subarray problem.

During an interview, compute the recurrence on a small example first; it often reveals that bestNeg is simply bestPos from the previous step minus the current value, letting you write the update in a single line.

COMPLEXITY AT A GLANCE

⏱ Time:O(n)

💾 Space:O(1)

Core Theory — Why This Approach?

The alternating sum of a subarray can be expressed as a linear combination of the original elements with signs that depend only on the parity of the index relative to the subarray start. By pre‑multiplying the original array with a sign pattern (+,‑, +,‑, …) that flips at every position, the problem reduces to finding a maximum difference between two prefix sums where the parity of the start index is taken into account. A naïve solution would enumerate every O(n²) subarray and compute its alternating sum, which quickly exceeds time limits for n up to 10⁵ or more. The optimal paradigm treats the task as a variant of Kadane’s algorithm: we maintain two DP states – the best alternating sum ending at the current position with a ‘+’ sign (odd length) and with a ‘‑’ sign (even length). Each state updates in O(1) using the previous opposite‑sign state, yielding a linear‑time solution with constant extra space.

Interview Questions on This Problem

Q1How would you modify Kadane’s algorithm to handle alternating signs in a subarray sum problem?

Maintain two DP variables: bestPos for subarrays ending at i with a positive sign on nums[i], and bestNeg for those ending with a negative sign. Update bestPos = max(nums[i], bestNeg + nums[i]) and bestNeg = bestPosPrev - nums[i]; the answer is the maximum bestPos seen.

Q2Why does a simple prefix‑sum approach fail for this alternating‑sum problem, and how can you fix it?

A plain prefix sum ignores the sign flip caused by the subarray’s start parity, so differences of two prefixes do not represent the alternating sum. The fix is to keep two prefix‑sum arrays – one assuming the global start is even, the other odd – and compute the maximum difference between a current prefix and the smallest earlier prefix of the same parity.

Q3In a real‑time streaming scenario where numbers arrive one by one, how can you maintain the maximum alternating subarray sum efficiently?

Use the same DP recurrence in an online fashion: keep bestPos and bestNeg as state variables and update them with each incoming element. The global maximum is updated simultaneously, giving O(1) amortized time per element and O(1) memory.

Examples

Example 1

Input

5\n5 -3 2 7 -1

Output

Explanation: The subarray [5,-3,2] yields 5-(-3)+2=10, which is larger than any other contiguous segment.

Example 2

Input

3\n-4 -2 -7

Output

Explanation: The subarray [-2,-7] gives -2-(-7)=5, the highest achievable alternating sum.

Example 3

Input

4\n1 2 3 4

Output

Explanation: A single element 4 yields an alternating sum of 4, exceeding all longer subarrays.

Constraints

1 <= nums.length <= 200000
-10^9 <= nums[i] <= 10^9
Time complexity O(n)
Auxiliary space O(1)

Optimal Approach & Strategy

Use two DP variables representing the best alternating sum ending with a positive or negative sign and update them in a single pass, achieving O(n) time and O(1) space.

Brute Force Approach

Enumerate every possible subarray, compute its alternating sum in O(length) and keep the maximum, resulting in O(n³) or O(n²) with prefix tricks but still too slow for large n.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n)

const fs = require('fs');
const data = fs.readFileSync(0, 'utf8').trim().split(/\s+/).map(Number);
let p = 0;
const n = data[p++];
const nums = data.slice(p, p + n);

function maxAlternatingSum(nums) {
    const NEG_INF = Number.NEGATIVE_INFINITY;
    let dpPos = NEG_INF, dpNeg = NEG_INF;
    let best = NEG_INF;
    for (const x of nums) {
        const newPos = Math.max(x, dpNeg + x);
        const newNeg = Math.max(-x, dpPos - x);
        dpPos = newPos;
        dpNeg = newNeg;
        if (dpPos > best) best = dpPos;
    }
    return best;
}

console.log(maxAlternatingSum(nums).toString());

#include <bits/stdc++.h>
using namespace std;

// Kadane‑like DP for alternating sum.
// dpPos – maximum alternating sum of a subarray ending at i where nums[i] is added (+).
// dpNeg – maximum alternating sum of a subarray ending at i where nums[i] is subtracted (-).
long long maxAlternatingSum(const vector<int>& nums) {
    const long long INF_NEG = LLONG_MIN / 4; // safe negative infinity
    long long dpPos = INF_NEG, dpNeg = INF_NEG;
    long long best = INF_NEG;
    for (int x : nums) {
        long long newPos = max<long long>(x, dpNeg + x);
        long long newNeg = max<long long>(-x, dpPos - x);
        dpPos = newPos;
        dpNeg = newNeg;
        best = max(best, dpPos);
    }
    return best;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    if (!(cin >> n)) return 0;
    vector<int> nums(n);
    for (int i = 0; i < n; ++i) cin >> nums[i];
    cout << maxAlternatingSum(nums) << "\n";
    return 0;
}

import java.io.*;
import java.util.*;

public class Main {
    // Kadane‑like DP. Returns the maximum alternating sum of any non‑empty subarray.
    public static long maxAlternatingSum(int[] nums) {
        final long NEG_INF = Long.MIN_VALUE / 4;
        long dpPos = NEG_INF; // subarray ending here with '+' sign
        long dpNeg = NEG_INF; // subarray ending here with '-' sign
        long best = NEG_INF;
        for (int x : nums) {
            long newPos = Math.max((long) x, dpNeg + x);
            long newNeg = Math.max(-(long) x, dpPos - x);
            dpPos = newPos;
            dpNeg = newNeg;
            if (dpPos > best) best = dpPos;
        }
        return best;
    }

    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());
        int n = Integer.parseInt(st.nextToken());
        int[] nums = new int[n];
        int filled = 0;
        while (filled < n) {
            if (!st.hasMoreTokens()) {
                st = new StringTokenizer(br.readLine());
                continue;
            }
            nums[filled++] = Integer.parseInt(st.nextToken());
        }
        System.out.println(maxAlternatingSum(nums));
    }
}

import sys

def max_alternating_sum(nums):
    """Kadane‑like DP returning the maximum alternating sum of any subarray."""
    NEG_INF = -10**30
    dp_pos = NEG_INF  # ends with +
    dp_neg = NEG_INF  # ends with -
    best = NEG_INF
    for x in nums:
        new_pos = max(x, dp_neg + x)
        new_neg = max(-x, dp_pos - x)
        dp_pos, dp_neg = new_pos, new_neg
        if dp_pos > best:
            best = dp_pos
    return best

def main():
    data = sys.stdin.read().strip().split()
    if not data:
        return
    n = int(data[0])
    nums = list(map(int, data[1:1 + n]))
    print(max_alternating_sum(nums))

if __name__ == "__main__":
    main()

const fs = require('fs');
const data = fs.readFileSync(0, 'utf8').trim().split(/\s+/).map(Number);
let p = 0;
const n = data[p++];
const nums = data.slice(p, p + n);

function maxAlternatingSum(nums) {
    const NEG_INF = Number.NEGATIVE_INFINITY;
    let dpPos = NEG_INF, dpNeg = NEG_INF;
    let best = NEG_INF;
    for (const x of nums) {
        const newPos = Math.max(x, dpNeg + x);
        const newNeg = Math.max(-x, dpPos - x);
        dpPos = newPos;
        dpNeg = newNeg;
        if (dpPos > best) best = dpPos;
    }
    return best;
}

console.log(maxAlternatingSum(nums).toString());

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Space Station Supply Chain Optimization Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Constraints

More Arrays

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Space Station Supply Chain Optimization — 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