Galactic Trade Route Optimization 2 — Problem Statement & Solution Guide

ArraysMediumprefix sum and sign tracking

TimeO(n)

SpaceO(1)

Problem Description

Given an integer array nums representing resource prices along a linear trade route, determine the maximum total price of a contiguous segment (subarray) whose consecutive price differences strictly alternate in sign (positive, negative, positive, … or negative, positive, …). The segment must contain at least two elements; a difference of zero breaks the alternation. If no such segment exists, output 0. The function should run in linear time and use O(1) extra space.

DSA Pattern Breakdown

"Galactic Trade Route Optimization 2"

medium

WHY DOES IT MATTER?

Wiggle‑pattern subarrays appear in financial time‑series, signal processing, and any domain where volatility direction matters; mastering this pattern teaches you to encode relational constraints efficiently.

OPTIMIZATION CHALLENGE

The key insight is that the alternation property depends only on the sign of the most recent difference, allowing a constant‑time state transition rather than re‑examining the whole prefix.

REAL-WORLD CONNECTION

Think of a convoy of ships adjusting speed: each ship must speed up then slow down alternately to avoid collisions, and the total fuel consumption (sum) of a valid convoy segment is what we want to maximize.

During an interview, compute the sign of nums[i]-nums[i-1] on the fly and update two running sums; never store the whole DP table—just two variables and a global max.

COMPLEXITY AT A GLANCE

⏱ Time:O(n)

💾 Space:O(1)

Core Theory — Why This Approach?

The alternating‑sign subarray condition is a classic example of a "wiggle" constraint applied to the differences between adjacent elements. A naive scan that checks every possible subarray would be O(n²) and quickly becomes infeasible for n up to 10⁵ because each candidate requires recomputing both the sum and the sign pattern. The optimal paradigm treats the problem as a dynamic‑programming walk over the array, maintaining two states: the best sum ending at the current index with the last difference positive, and the best sum ending with the last difference negative. By updating these states in O(1) per element, we propagate the wiggle property forward while simultaneously accumulating the maximum total price, achieving linear time.

Interview Questions on This Problem

Q1How would you modify the solution if the requirement changed from maximizing the sum to maximizing the length of the alternating‑sign subarray?

Keep the same two DP states but store lengths instead of sums; when the sign alternates, extend the previous length, otherwise reset to 1 (or 2 for a new pair). The answer is the maximum length recorded, still O(n) time.

Q2Can the algorithm be extended to handle circular trade routes where the subarray may wrap around the end of the array?

Yes. Duplicate the array (concatenate it to itself) and run the linear DP on the 2n‑length array while limiting window size to n, or use a sliding‑window variant that respects the wrap‑around constraint, preserving O(n) complexity.

Q3Why does a zero difference break the alternation, and how do you handle it in the DP formulation?

A zero difference has no sign, so it cannot satisfy the strict positive/negative alternation. In DP we treat a zero as a reset: both positive‑last and negative‑last states are re‑initialized to the value of the current element, effectively starting a new segment after the zero.

Examples

Example 1

Input

[4,2,5,1,6]

Output

Explanation: Differences: 4‑2 = -2 (negative), 2‑5 = +3 (positive), 5‑1 = -4 (negative), 1‑6 = +5 (positive). Signs alternate for the whole array, so the sum 4+2+5+1+6 = 18 is valid and maximal.

Example 2

Input

[1,3,2,4,3]

Output

Explanation: Differences: 1‑3 = -2 (negative), 3‑2 = +1 (positive), 2‑4 = -2 (negative), 4‑3 = +1 (positive). The entire array alternates, giving sum 1+3+2+4+3 = 13, which is the largest possible.

Example 3

Input

[5,5,5]

Output

Explanation: All consecutive differences are 0, which does not satisfy the strict sign‑alternation rule. No valid segment of length ≥2 exists, so the answer is 0.

Constraints

1 <= nums.length <= 100000
-1000000000 <= nums[i] <= 1000000000
At least two elements are required for a valid segment
Differences of zero break alternation

Optimal Approach & Strategy

Maintain two DP variables representing the best sum ending with a positive or negative last difference and update them in a single pass while tracking the global maximum.

Brute Force Approach

Enumerate every possible subarray, check if its consecutive differences strictly alternate, and compute its sum; keep the maximum among valid ones.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n)

const fs = require('fs');
const data = fs.readFileSync(0, 'utf8').trim().split(/\s+/).map(Number);
function maxAlternatingSubarraySum(nums) {
    const n = nums.length;
    if (n < 2) return 0;
    let best = -Infinity;
    let curSum = 0;
    let lastSign = 0; // 1, -1, or 0 (undefined)
    let startIdx = 0;
    for (let i = 1; i < n; ++i) {
        const diff = nums[i] - nums[i-1];
        if (diff === 0) {
            lastSign = 0;
            curSum = 0;
            continue;
        }
        const sign = diff > 0 ? 1 : -1;
        if (lastSign === 0) {
            curSum = nums[i-1] + nums[i];
            startIdx = i-1;
        } else if (sign !== lastSign) {
            curSum += nums[i];
        } else {
            curSum = nums[i-1] + nums[i];
            startIdx = i-1;
        }
        lastSign = sign;
        if (i - startIdx + 1 >= 2) best = Math.max(best, curSum);
    }
    return best === -Infinity ? 0 : best;
}
const result = maxAlternatingSubarraySum(data);
process.stdout.write(String(result));

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

int maxAlternatingSubarraySum(const vector<int>& nums) {
    int n = nums.size();
    if (n < 2) return 0;
    long long best = LLONG_MIN;
    long long curSum = 0;
    int lastSign = 0; // 1 for positive, -1 for negative, 0 for undefined
    int startIdx = 0;
    for (int i = 1; i < n; ++i) {
        long long diff = (long long)nums[i] - nums[i-1];
        if (diff == 0) {
            // zero breaks any alternating segment
            lastSign = 0;
            curSum = 0;
            continue;
        }
        int sign = diff > 0 ? 1 : -1;
        if (lastSign == 0) {
            // start a new segment of length 2
            curSum = (long long)nums[i-1] + nums[i];
            startIdx = i-1;
        } else if (sign != lastSign) {
            // extend current segment
            curSum += nums[i];
        } else {
            // same sign as previous diff -> start new segment at i-1
            curSum = (long long)nums[i-1] + nums[i];
            startIdx = i-1;
        }
        lastSign = sign;
        if (i - startIdx + 1 >= 2) {
            best = max(best, curSum);
        }
    }
    return best == LLONG_MIN ? 0 : (int)best;
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    vector<int> nums;
    int x; while (cin>>x) nums.push_back(x);
    cout<<maxAlternatingSubarraySum(nums);
    return 0;
}

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

public class Main {
    public static long maxAlternatingSubarraySum(int[] nums) {
        int n = nums.length;
        if (n < 2) return 0;
        long best = Long.MIN_VALUE;
        long curSum = 0;
        int lastSign = 0; // 1, -1, or 0 (undefined)
        int startIdx = 0;
        for (int i = 1; i < n; ++i) {
            long diff = (long)nums[i] - nums[i-1];
            if (diff == 0) {
                lastSign = 0;
                curSum = 0;
                continue;
            }
            int sign = diff > 0 ? 1 : -1;
            if (lastSign == 0) {
                curSum = (long)nums[i-1] + nums[i];
                startIdx = i-1;
            } else if (sign != lastSign) {
                curSum += nums[i];
            } else {
                curSum = (long)nums[i-1] + nums[i];
                startIdx = i-1;
            }
            lastSign = sign;
            if (i - startIdx + 1 >= 2) {
                best = Math.max(best, curSum);
            }
        }
        return best == Long.MIN_VALUE ? 0 : best;
    }
    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        List<Integer> list = new ArrayList<>();
        String line;
        while ((line = br.readLine()) != null) {
            line = line.trim();
            if (line.isEmpty()) continue;
            for (String s : line.split("\\s+")) {
                list.add(Integer.parseInt(s));
            }
        }
        int[] nums = list.stream().mapToInt(i -> i).toArray();
        System.out.print(maxAlternatingSubarraySum(nums));
    }
}

import sys

def max_alternating_subarray_sum(nums):
    n = len(nums)
    if n < 2:
        return 0
    best = None
    cur_sum = 0
    last_sign = 0  # 1, -1, or 0 (undefined)
    start_idx = 0
    for i in range(1, n):
        diff = nums[i] - nums[i-1]
        if diff == 0:
            last_sign = 0
            cur_sum = 0
            continue
        sign = 1 if diff > 0 else -1
        if last_sign == 0:
            cur_sum = nums[i-1] + nums[i]
            start_idx = i-1
        elif sign != last_sign:
            cur_sum += nums[i]
        else:
            cur_sum = nums[i-1] + nums[i]
            start_idx = i-1
        last_sign = sign
        if i - start_idx + 1 >= 2:
            if best is None or cur_sum > best:
                best = cur_sum
    return 0 if best is None else best

data = sys.stdin.read().strip().split()
nums = list(map(int, data))
print(max_alternating_subarray_sum(nums))

const fs = require('fs');
const data = fs.readFileSync(0, 'utf8').trim().split(/\s+/).map(Number);
function maxAlternatingSubarraySum(nums) {
    const n = nums.length;
    if (n < 2) return 0;
    let best = -Infinity;
    let curSum = 0;
    let lastSign = 0; // 1, -1, or 0 (undefined)
    let startIdx = 0;
    for (let i = 1; i < n; ++i) {
        const diff = nums[i] - nums[i-1];
        if (diff === 0) {
            lastSign = 0;
            curSum = 0;
            continue;
        }
        const sign = diff > 0 ? 1 : -1;
        if (lastSign === 0) {
            curSum = nums[i-1] + nums[i];
            startIdx = i-1;
        } else if (sign !== lastSign) {
            curSum += nums[i];
        } else {
            curSum = nums[i-1] + nums[i];
            startIdx = i-1;
        }
        lastSign = sign;
        if (i - startIdx + 1 >= 2) best = Math.max(best, curSum);
    }
    return best === -Infinity ? 0 : best;
}
const result = maxAlternatingSubarraySum(data);
process.stdout.write(String(result));

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Galactic Trade Route Optimization 2 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 2 — 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