Optimizing Galactic Trade Routes — Problem Statement & Solution Guide

ArraysMediummax-subarray-sum

TimeO(n)

SpaceO(1)

Problem Description

Given an integer array nums, each element represents an asteroid: a positive value denotes a valuable resource, a negative value denotes a hazardous asteroid. Find the maximum possible sum of a contiguous subarray that satisfies two conditions: (1) the subarray starts with a positive element, and (2) the signs of consecutive elements strictly alternate (positive, negative, positive, …). A subarray of length 1 consisting of a single positive element is valid. If no such subarray exists, return 0. The algorithm must run in linear time relative to the array length.

DSA Pattern Breakdown

"Optimizing Galactic Trade Routes"

medium

WHY DOES IT MATTER?

Alternating‑sign subarrays appear in financial risk modeling, signal processing and load‑balancing where positive and negative impacts must offset each other; mastering this pattern teaches you to embed extra constraints into classic DP frameworks.

OPTIMIZATION CHALLENGE

The key insight is that only the immediately previous sign matters, so you can collapse the DP to two scalar variables instead of an O(n) table, cutting both time and space to linear and constant respectively.

REAL-WORLD CONNECTION

Think of a power grid that alternates between generation (positive) and consumption (negative) phases; maximizing net surplus while respecting the alternating schedule mirrors the algorithmic requirement.

During an interview, compute dpPos and dpNeg on the fly, update the global answer only when dpPos is valid, and reset dpNeg to a sentinel when the sign pattern breaks – this avoids hidden bugs and keeps the code tidy.

COMPLEXITY AT A GLANCE

⏱ Time:O(n)

💾 Space:O(1)

Core Theory — Why This Approach?

The problem is a constrained variant of the classic maximum subarray (Kadane) where the subarray must start with a positive element and the signs must strictly alternate. A naïve solution enumerates every possible start‑end pair, checks the alternating condition and computes the sum, leading to O(n^2) time which explodes for large n. The optimal paradigm treats the alternating sign requirement as a state machine and uses dynamic programming: for each index we keep two values – the best sum of a valid subarray ending at that index with a positive last element (dpPos) and with a negative last element (dpNeg). Transition is simple: if nums[i] is positive we can either start a new subarray (value = nums[i]) or extend a subarray that previously ended with a negative element (dpNeg+nums[i]); similarly for a negative nums[i] we can only extend dpPos. Scanning the array once while updating these two states yields the global maximum in linear time.

Interview Questions on This Problem

Q1How would you adapt Kadane's algorithm to enforce an alternating sign constraint and a positive start?

Maintain two DP variables dpPos and dpNeg. When the current element is positive, dpPos = max(nums[i], dpNeg+nums[i]); when negative, dpNeg = dpPos+nums[i] (or reset to negative infinity). Track the overall maximum of dpPos only because the subarray must end on a positive element.

Q2What edge case must you handle when the array contains only negative numbers?

Since the subarray must start with a positive element, the answer is 0 (or "no valid subarray") because no valid subarray can be formed; the algorithm should initialize the answer to 0 and never update it from a negative dpPos.

Q3Can the same DP idea be extended to a pattern that repeats every k signs (e.g., +, -, +, +, - …)?

Yes. Generalize the state array to k slots, each representing the best sum ending with the i‑th position of the pattern. Transition uses the previous slot in the pattern, yielding O(k·n) time and O(k) space.

Examples

Example 1

Input

[4,-1,3,-2,5,-3,-4]

Output

Explanation: The longest alternating segment beginning with a positive number is [4,-1,3,-2,5]; its sum is 4-1+3-2+5=9, which is larger than any other valid segment.

Example 2

Input

[-5,2,-1,4,-2,6]

Output

Explanation: The subarray starting at index 1, [2,-1,4,-2,6], alternates signs and yields 2-1+4-2+6=9, which is the maximum achievable sum.

Example 3

Input

[1,2,3,-1,-2,4]

Output

Explanation: Only single‑element positive subarrays satisfy the alternating rule because any longer segment would contain two consecutive positives. The largest positive element is 4, so the answer is 4.

Constraints

1 <= nums.length <= 100000
-1000000000 <= nums[i] <= 1000000000
The solution must use O(1) additional memory beyond the input array
Time complexity must be O(n)

Optimal Approach & Strategy

Use two DP variables (dpPos, dpNeg) updated in a single left‑to‑right pass, yielding O(n) time and O(1) extra space.

Brute Force Approach

Check every possible start index, expand right while maintaining alternating signs, compute sums, and keep the best – O(n^2) time.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n)

function maxAlternatingSum(nums){
    let maxSum = 0;
    let curSum = 0;
    let expect = 0; // 1 positive, -1 negative, 0 not started
    for(const v of nums){
        if(curSum===0){
            if(v>0){curSum=v; expect=-1; if(curSum>maxSum)maxSum=curSum;}
        }else{
            const ok = (expect===1 && v>0) || (expect===-1 && v<0);
            if(ok){
                curSum+=v; expect=-expect; if(curSum>maxSum)maxSum=curSum;
            }else{
                if(v>0){curSum=v; expect=-1; if(curSum>maxSum)maxSum=curSum;}
                else{curSum=0; expect=0;}
            }
        }
    }
    return maxSum;
}
const fs=require('fs');
const data=fs.readFileSync(0,'utf8').trim().split(/\s+/).map(Number);
if(data.length===0)process.exit(0);
let n=data[0];
let arr=data.slice(1,1+n);
console.log(maxAlternatingSum(arr).toString());

#include <bits/stdc++.h>
using namespace std;
long long maxAlternatingSum(const vector<int>& nums){
    long long maxSum = 0; // if no valid subarray, answer is 0
    long long curSum = 0;
    int expect = 0; // 1 for positive, -1 for negative, 0 means not started
    for(int v: nums){
        if(curSum==0){
            if(v>0){
                curSum=v; expect=-1; maxSum=max(maxSum,curSum);
            }
        }else{
            bool ok = (expect==1 && v>0) || (expect==-1 && v<0);
            if(ok){
                curSum+=v; expect=-expect; maxSum=max(maxSum,curSum);
            }else{
                if(v>0){
                    curSum=v; expect=-1; maxSum=max(maxSum,curSum);
                }else{
                    curSum=0; expect=0;
                }
            }
        }
    }
    return maxSum;
}
int main(){ios::sync_with_stdio(false);cin.tie(nullptr);int n; if(!(cin>>n)) return 0;vector<int> a(n);for(int i=0;i<n;++i)cin>>a[i];cout<<maxAlternatingSum(a);}

import java.io.*;
import java.util.*;
public class Main {
    public static long maxAlternatingSum(int[] nums){
        long maxSum = 0;
        long curSum = 0;
        int expect = 0; // 1 positive, -1 negative, 0 not started
        for(int v: nums){
            if(curSum==0){
                if(v>0){
                    curSum=v; expect=-1; if(curSum>maxSum)maxSum=curSum;
                }
            }else{
                boolean ok = (expect==1 && v>0) || (expect==-1 && v<0);
                if(ok){
                    curSum+=v; expect=-expect; if(curSum>maxSum)maxSum=curSum;
                }else{
                    if(v>0){
                        curSum=v; expect=-1; if(curSum>maxSum)maxSum=curSum;
                    }else{
                        curSum=0; expect=0;
                    }
                }
            }
        }
        return maxSum;
    }
    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String line = br.readLine();
        if(line==null||line.isEmpty()) return;
        int n = Integer.parseInt(line.trim());
        int[] nums = new int[n];
        StringTokenizer st = new StringTokenizer(br.readLine());
        for(int i=0;i<n;i++) nums[i]=Integer.parseInt(st.nextToken());
        System.out.println(maxAlternatingSum(nums));
    }
}

def max_alternating_sum(nums):
    max_sum = 0
    cur_sum = 0
    expect = 0  # 1 -> positive, -1 -> negative, 0 -> not started
    for v in nums:
        if cur_sum == 0:
            if v > 0:
                cur_sum = v
                expect = -1
                max_sum = max(max_sum, cur_sum)
        else:
            ok = (expect == 1 and v > 0) or (expect == -1 and v < 0)
            if ok:
                cur_sum += v
                expect = -expect
                max_sum = max(max_sum, cur_sum)
            else:
                if v > 0:
                    cur_sum = v
                    expect = -1
                    max_sum = max(max_sum, cur_sum)
                else:
                    cur_sum = 0
                    expect = 0
    return max_sum
if __name__ == "__main__":
    import sys
    data = sys.stdin.read().strip().split()
    if not data:
        sys.exit(0)
    n = int(data[0])
    nums = list(map(int, data[1:1+n]))
    print(max_alternating_sum(nums))

function maxAlternatingSum(nums){
    let maxSum = 0;
    let curSum = 0;
    let expect = 0; // 1 positive, -1 negative, 0 not started
    for(const v of nums){
        if(curSum===0){
            if(v>0){curSum=v; expect=-1; if(curSum>maxSum)maxSum=curSum;}
        }else{
            const ok = (expect===1 && v>0) || (expect===-1 && v<0);
            if(ok){
                curSum+=v; expect=-expect; if(curSum>maxSum)maxSum=curSum;
            }else{
                if(v>0){curSum=v; expect=-1; if(curSum>maxSum)maxSum=curSum;}
                else{curSum=0; expect=0;}
            }
        }
    }
    return maxSum;
}
const fs=require('fs');
const data=fs.readFileSync(0,'utf8').trim().split(/\s+/).map(Number);
if(data.length===0)process.exit(0);
let n=data[0];
let arr=data.slice(1,1+n);
console.log(maxAlternatingSum(arr).toString());

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Optimizing Galactic Trade Routes Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Constraints

More Arrays

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Optimizing Galactic Trade Routes — 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