Galactic Trade Route Optimization 4 — Problem Statement & Solution Guide

ArraysMediumMixed

TimeO(n)

SpaceO(1)

Problem Description

Given an integer array nums, select a non‑empty contiguous subarray whose elements strictly alternate in sign (positive, negative, positive, … or negative, positive, negative, …). The subarray may start with either a positive or a negative number, but no two adjacent elements may share the same sign. Compute the maximum possible sum of such a subarray and output that sum.

DSA Pattern Breakdown

"Galactic Trade Route Optimization 4"

medium

WHY DOES IT MATTER?

This pattern is essential for problems where constraints on adjacent elements affect the validity of subarrays. It teaches how to extend dynamic programming techniques to handle additional constraints, which is a common requirement in real-world optimization problems.

OPTIMIZATION CHALLENGE

The key insight is to track two states (max sum ending with positive and negative) instead of a single state. This allows you to handle the sign alternation constraint without needing to backtrack or use a more complex data structure.

REAL-WORLD CONNECTION

This is analogous to scheduling tasks in a distributed system where certain tasks must alternate between two types (e.g., CPU-intensive and I/O-bound) to optimize resource utilization. The goal is to find the longest sequence of tasks that adheres to this alternation while maximizing throughput.

In interviews, clearly articulate the two states and how they transition based on the current element's sign. This demonstrates your ability to model complex constraints with simple state machines, which is a highly valued skill.

COMPLEXITY AT A GLANCE

⏱ Time:O(n)

💾 Space:O(1)

Core Theory — Why This Approach?

The problem requires finding the maximum sum of a contiguous subarray where adjacent elements have strictly opposite signs. This is a variation of the classic Kadane's algorithm, which finds the maximum subarray sum without sign constraints. The key difference here is the additional constraint on the sign pattern, which means we cannot simply extend a subarray if the next element violates the alternating sign rule. Instead, we must track two states: the maximum sum ending at the current index if the subarray starts with a positive sign, and the maximum sum ending at the current index if it starts with a negative sign. This dual-state tracking allows us to handle the sign alternation constraint efficiently.

Interview Questions on This Problem

Q1How would you modify your solution if the array could contain zeros?

Zeros can be treated as neutral elements that do not break the alternating sign pattern. You can include zeros in the subarray without affecting the sign alternation, but they also do not contribute to the sum. The state tracking remains the same, but you need to ensure that zeros do not reset the subarray unless they are part of a valid alternating sequence.

Q2What if the array is very large, and you need to process it in a streaming fashion?

In a streaming scenario, you can maintain the two states (max sum ending with positive and negative) as you process each element. Since the states only depend on the previous element, you can update them in constant time per element, making the solution suitable for streaming data.

Q3How would you handle the case where no valid subarray exists?

If no valid subarray exists (e.g., all elements have the same sign), you should return a specific value, such as the maximum single element or a sentinel value like -infinity, depending on the problem requirements. Ensure your initialization and state transitions handle this edge case correctly.

Examples

Example 1

Input

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

Output

Explanation: The whole array 4,-1,2,-3,5 alternates signs and its sum is 4+(-1)+2+(-3)+5=7, which is larger than any shorter alternating segment.

Example 2

Input

[-5,6,-2,8,-1]

Output

Explanation: The subarray 6,-2,8 alternates signs and yields 6+(-2)+8=12, which exceeds all other alternating subarrays such as the full array (-5+6-2+8-1=6) or 6,-2,8,-1 (sum 11).

Example 3

Input

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

Output

Explanation: Only single‑element subarrays satisfy the alternating condition because any longer contiguous segment contains two consecutive positives or two consecutive negatives. The largest element is 5, so the maximum sum is 5.

Constraints

1 <= nums.length <= 200000
-10^9 <= nums[i] <= 10^9
The answer fits in a 64‑bit signed integer

Optimal Approach & Strategy

Use a modified Kadane's algorithm that tracks two states: the maximum sum ending at the current index for subarrays starting with a positive sign and those starting with a negative sign. Update these states based on the current element's sign and keep track of the overall maximum sum.

Brute Force Approach

Check all possible contiguous subarrays and verify if they satisfy the alternating sign constraint. Compute the sum for each valid subarray and keep track of the maximum sum found.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n)

/**
 * @param {number[]} nums - The input array of integers.
 * @return {number} - The maximum sum of a contiguous subarray with strictly alternating signs.
 */
function maxAlternatingSum(nums) {
    const n = nums.length;
    if (n === 0) return 0;
    
    let maxSum = -Infinity;
    let currentSum = 0;
    
    for (let i = 0; i < n; i++) {
        if (i === 0) {
            currentSum = nums[i];
        } else {
            // Check if signs alternate
            if ((nums[i] > 0 && nums[i-1] < 0) || (nums[i] < 0 && nums[i-1] > 0)) {
                currentSum += nums[i];
            } else {
                // Reset the subarray starting from current element
                currentSum = nums[i];
            }
        }
        
        maxSum = Math.max(maxSum, currentSum);
    }
    
    return maxSum;
}

// Driver code
const fs = require('fs');
const input = fs.readFileSync('/dev/stdin', 'utf8').trim().split('\n');
const n = parseInt(input[0]);
const nums = input[1].split(' ').map(Number);

console.log(maxAlternatingSum(nums));

#include <iostream>
#include <vector>
#include <algorithm>
#include <climits>

using namespace std;

int maxAlternatingSum(vector<int>& nums) {
    int n = nums.size();
    if (n == 0) return 0;
    
    int maxSum = INT_MIN;
    int currentSum = 0;
    
    for (int i = 0; i < n; ++i) {
        if (i == 0) {
            currentSum = nums[i];
        } else {
            // Check if signs alternate
            if ((nums[i] > 0 && nums[i-1] < 0) || (nums[i] < 0 && nums[i-1] > 0)) {
                currentSum += nums[i];
            } else {
                // Reset the subarray starting from current element
                currentSum = nums[i];
            }
        }
        
        maxSum = max(maxSum, currentSum);
    }
    
    return maxSum;
}

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

import java.util.*;

public class Main {
    /**
     * @param nums The input array of integers.
     * @return The maximum sum of a contiguous subarray with strictly alternating signs.
     */
    public static int maxAlternatingSum(int[] nums) {
        int n = nums.length;
        if (n == 0) return 0;
        
        int maxSum = Integer.MIN_VALUE;
        int currentSum = 0;
        
        for (int i = 0; i < n; i++) {
            if (i == 0) {
                currentSum = nums[i];
            } else {
                // Check if signs alternate
                if ((nums[i] > 0 && nums[i-1] < 0) || (nums[i] < 0 && nums[i-1] > 0)) {
                    currentSum += nums[i];
                } else {
                    // Reset the subarray starting from current element
                    currentSum = nums[i];
                }
            }
            
            maxSum = Math.max(maxSum, currentSum);
        }
        
        return maxSum;
    }

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

def maxAlternatingSum(nums):
    """
    Calculate the maximum sum of a contiguous subarray with strictly alternating signs.
    
    Args:
        nums (list[int]): The input array of integers.
        
    Returns:
        int: The maximum sum of such a subarray.
    """
    n = len(nums)
    if n == 0:
        return 0
    
    max_sum = float('-inf')
    current_sum = 0
    
    for i in range(n):
        if i == 0:
            current_sum = nums[i]
        else:
            # Check if signs alternate
            if (nums[i] > 0 and nums[i-1] < 0) or (nums[i] < 0 and nums[i-1] > 0):
                current_sum += nums[i]
            else:
                # Reset the subarray starting from current element
                current_sum = nums[i]
        
        max_sum = max(max_sum, current_sum)
    
    return max_sum

if __name__ == "__main__":
    import sys
    input_data = sys.stdin.read().split()
    n = int(input_data[0])
    nums = list(map(int, input_data[1:n+1]))
    
    print(maxAlternatingSum(nums))

/**
 * @param {number[]} nums - The input array of integers.
 * @return {number} - The maximum sum of a contiguous subarray with strictly alternating signs.
 */
function maxAlternatingSum(nums) {
    const n = nums.length;
    if (n === 0) return 0;
    
    let maxSum = -Infinity;
    let currentSum = 0;
    
    for (let i = 0; i < n; i++) {
        if (i === 0) {
            currentSum = nums[i];
        } else {
            // Check if signs alternate
            if ((nums[i] > 0 && nums[i-1] < 0) || (nums[i] < 0 && nums[i-1] > 0)) {
                currentSum += nums[i];
            } else {
                // Reset the subarray starting from current element
                currentSum = nums[i];
            }
        }
        
        maxSum = Math.max(maxSum, currentSum);
    }
    
    return maxSum;
}

// Driver code
const fs = require('fs');
const input = fs.readFileSync('/dev/stdin', 'utf8').trim().split('\n');
const n = parseInt(input[0]);
const nums = input[1].split(' ').map(Number);

console.log(maxAlternatingSum(nums));

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Galactic Trade Route Optimization 4 Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Constraints

More Arrays

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Galactic Trade Route Optimization 4 — 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