Alternating Subarray Sum — Problem Statement & Solution Guide
Quick Answer & Algorithm Key Takeaway
Master coding challenges related to Arrays and solve the Alternating Subarray Sum problem optimally.
O(n)O(1)Problem Description
Given an integer array values, determine the largest possible sum of any contiguous subarray that strictly alternates between positive and negative numbers, beginning with a positive element. The subarray must contain at least one element. If the array does not contain a positive element or no alternating subarray can be formed, return 0.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Alternating Subarray Sum"
WHY DOES IT MATTER?
Alternating‑sign patterns appear in financial signal processing, error‑correction codes, and load‑balancing where opposite‑signed metrics indicate state flips; efficiently extracting the best contiguous alternating segment is crucial for anomaly detection and profit maximization.
OPTIMIZATION CHALLENGE
The key insight is that a valid alternating subarray can only be extended by the opposite‑sign sum from the previous index, allowing us to collapse the DP state to two scalars (posSum, negSum) instead of O(n) tables, thus achieving O(1) space.
REAL-WORLD CONNECTION
Think of a server cluster where request latency spikes (positive) must be followed by cooldown periods (negative) to avoid overload. Finding the longest profitable burst of alternating high‑load and recovery phases mirrors the alternating subarray sum problem.
During the interview, write the recurrence first, then immediately convert it to two rolling variables; this shows you understand optimal substructure and can translate DP into a greedy‑style implementation.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The alternating‑sign subarray problem is a variant of the classic maximum subarray (Kadane) where the sign of consecutive elements must flip, and the first element must be positive. A naive scan that simply adds numbers fails because a negative element can invalidate the alternating property, forcing a restart of the subarray; likewise, a positive element that follows another positive breaks the pattern. To solve it efficiently we maintain two running sums: one for subarrays ending with a positive element (posSum) and one for those ending with a negative element (negSum). When we encounter a positive value we can extend a previously valid negative‑ending subarray (negSum) or start a new subarray; similarly a negative value can extend a positive‑ending subarray (posSum). The maximum of all posSum values encountered is the answer. This DP‑like recurrence runs in linear time and constant extra space, overcoming the O(n²) brute‑force that enumerates all O(n²) subarrays.
The optimal paradigm blends greedy selection with dynamic programming. At each index we make a locally optimal decision—whether to extend the existing alternating chain or to reset—based solely on the sign of the current element and the best sum achievable with the opposite sign at the previous index. Because the decision depends only on the immediate predecessor, the global optimum is guaranteed by optimal substructure, a hallmark of DP solutions like Kadane’s algorithm. This yields an O(n) time, O(1) space solution suitable for large inputs where O(n²) would time out.
Interview Questions on This Problem
Q1How would you modify Kadane’s algorithm to handle the alternating‑sign constraint and ensure the subarray starts with a positive number?
Maintain two variables, posSum and negSum. For a positive a[i], set posSum = a[i] + max(0, negSum) and reset negSum to 0; for a negative a[i], set negSum = a[i] + max(0, posSum) and reset posSum to 0. Track the maximum posSum seen; if no positive element exists, return 0.
Q2What is the time and space complexity of the optimal solution, and why can’t we achieve better than O(n) time?
The optimal solution runs in O(n) time and O(1) extra space because each element is processed once with constant‑time updates. Any algorithm must inspect each element at least once to know whether it can start or extend an alternating subarray, so O(n) is optimal.
Q3In a streaming setting where numbers arrive one‑by‑one, how would you compute the answer without storing the entire array?
Keep only the current posSum, negSum, and global maxPos. Update them as each new value arrives using the same recurrence; this uses O(1) memory and works for an infinite stream, outputting the max alternating sum seen so far.
Examples
Input
[5,-2,3,-1,2,-4,6]
Output
9
Explanation: Starting at index 0 yields the longest alternating sequence 5,-2,3,-1,2,-4,6. Its sum is 5-2+3-1+2-4+6=9, which is larger than any other valid subarray.
Input
[-3,2,-5,4,-1]
Output
4
Explanation: Valid alternating subarrays that start with a positive number are: [2] (sum 2), [2,-5] (‑3), [2,-5,4] (1), [2,-5,4,-1] (0), [4] (4), [4,-1] (3). The maximum sum is 4.
Input
[-1,-2,-3]
Output
0
Explanation: The array contains no positive element, therefore no subarray can satisfy the required pattern; the answer is 0.
Constraints
- 1 <= values.length <= 100000
- -1000000000 <= values[i] <= 1000000000
- The algorithm should run in O(n) time and O(1) additional space
Optimal Approach & Strategy
Traverse once, maintaining posSum and negSum using sign‑aware transitions; update a global maxPos whenever posSum improves.
Brute Force Approach
Enumerate every possible subarray, check if it alternates starting with a positive, and compute its sum, keeping the maximum.
Code Solutions
function maxAlternatingSubarraySum(values){
let maxSum = 0;
let curSum = 0;
let expectNeg = false; // after a positive we expect a negative
for(const v of values){
if(v>0){
if(!expectNeg){
curSum += v;
expectNeg = true;
}else{ // two positives in a row -> restart at this positive
curSum = v;
expectNeg = true;
}
}else if(v<0){
if(expectNeg){
curSum += v;
expectNeg = false;
}else{ // negative where positive expected -> reset
curSum = 0;
expectNeg = false;
}
}else{ // zero breaks the pattern
curSum = 0;
expectNeg = false;
}
if(curSum>maxSum) maxSum = curSum;
}
return maxSum;
}#include <bits/stdc++.h>
using namespace std;
long long maxAlternatingSubarraySum(const vector<int>& values){
long long maxSum = 0, curSum = 0;
bool expectNeg = false; // after a positive we expect negative
for(int v: values){
if(v>0){
if(!expectNeg){ // either start or continuation of positive after negative
curSum += v;
expectNeg = true;
}else{ // two positives in a row -> break, start new at this positive
curSum = v;
expectNeg = true;
}
}else if(v<0){
if(expectNeg){ // correct sign
curSum += v;
expectNeg = false;
}else{ // negative where positive expected -> cannot start here
curSum = 0;
expectNeg = false;
}
}else{ // zero breaks the alternating pattern
curSum = 0;
expectNeg = false;
}
maxSum = max(maxSum, curSum);
}
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 &x:a) cin>>x;
cout<<maxAlternatingSubarraySum(a);
return 0;
}
import java.io.*;
import java.util.*;
public class Main {
public static long maxAlternatingSubarraySum(int[] values) {
long maxSum = 0;
long curSum = 0;
boolean expectNeg = false; // after a positive we expect a negative
for (int v : values) {
if (v > 0) {
if (!expectNeg) {
curSum += v;
expectNeg = true;
} else { // two positives in a row, restart at this positive
curSum = v;
expectNeg = true;
}
} else if (v < 0) {
if (expectNeg) {
curSum += v;
expectNeg = false;
} else { // negative where positive expected
curSum = 0;
expectNeg = false;
}
} else { // zero breaks the pattern
curSum = 0;
expectNeg = false;
}
if (curSum > maxSum) maxSum = curSum;
}
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[] values = new int[n];
StringTokenizer st = new StringTokenizer(br.readLine());
for (int i = 0; i < n; i++) {
values[i] = Integer.parseInt(st.nextToken());
}
System.out.println(maxAlternatingSubarraySum(values));
}
}
def max_alternating_subarray_sum(values):
max_sum = 0
cur_sum = 0
expect_neg = False # after a positive we expect a negative
for v in values:
if v > 0:
if not expect_neg:
cur_sum += v
expect_neg = True
else: # two positives in a row, restart at this positive
cur_sum = v
expect_neg = True
elif v < 0:
if expect_neg:
cur_sum += v
expect_neg = False
else: # negative where positive expected
cur_sum = 0
expect_neg = False
else: # zero breaks the alternating requirement
cur_sum = 0
expect_neg = False
if cur_sum > max_sum:
max_sum = cur_sum
return max_sum
function maxAlternatingSubarraySum(values){
let maxSum = 0;
let curSum = 0;
let expectNeg = false; // after a positive we expect a negative
for(const v of values){
if(v>0){
if(!expectNeg){
curSum += v;
expectNeg = true;
}else{ // two positives in a row -> restart at this positive
curSum = v;
expectNeg = true;
}
}else if(v<0){
if(expectNeg){
curSum += v;
expectNeg = false;
}else{ // negative where positive expected -> reset
curSum = 0;
expectNeg = false;
}
}else{ // zero breaks the pattern
curSum = 0;
expectNeg = false;
}
if(curSum>maxSum) maxSum = curSum;
}
return maxSum;
}
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