Alternating Budget Accumulation — Problem Statement & Solution Guide
Quick Answer & Algorithm Key Takeaway
Master coding challenges related to Arrays and solve the Alternating Budget Accumulation problem optimally.
O(n)O(1)Problem Description
Given an integer array budget of length n, each element denotes a daily net cash flow (positive = surplus, negative = deficit). Select a sequence of non-overlapping contiguous subarrays (intervals) such that:\n1. The first chosen interval has a positive total sum.\n2. The signs of the interval sums strictly alternate: positive, negative, positive, …\n3. The sequence may end after any interval (it is not required to finish on a negative interval).\nThe objective is to maximise the sum of all selected interval sums. Output this maximum possible total.\nInput: The first line contains an integer n (1 ≤ n ≤ 10⁵). The second line contains n space-separated integers budget[i] (|budget[i]| ≤ 10⁹).\nOutput: A single integer – the maximum achievable alternating sum.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Alternating Budget Accumulation"
WHY DOES IT MATTER?
Alternating‑sign interval selection appears in financial risk modeling, signal processing, and any scenario where gains and losses must be balanced over time; mastering this pattern teaches you to handle state‑dependent constraints efficiently.
OPTIMIZATION CHALLENGE
The key insight is that only the best achievable total for each sign matters at any index; you do not need to store every possible subarray, which reduces the naive O(n^2) DP to O(n) with two scalar variables.
REAL-WORLD CONNECTION
Think of a cash‑flow manager who must schedule profit‑making campaigns (positive intervals) followed by cost‑cutting phases (negative intervals) without overlap—optimizing the overall net cash flow mirrors the algorithm’s alternating DP.
During an interview, write the two‑state recurrence first, test it on a tiny example, and then compress the DP into two rolling variables—this shows both correctness and space‑efficiency awareness.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem can be modeled as a dynamic programming task where we scan the array once while maintaining two states: the best total value achievable ending with a positive‑sum interval (pos) and the best total ending with a negative‑sum interval (neg). Each new element either extends the current interval or starts a new one, and the sign‑alternation constraint forces a transition from pos to neg or vice‑versa. A naive O(n^2) solution would enumerate every possible sub‑array, compute its sum, and then try all combinations of non‑overlapping intervals, which quickly explodes for large n. By recognizing that only the best prefix ending in each sign matters, we collapse the exponential state space to constant size, yielding a linear‑time, constant‑space algorithm akin to Kadane’s maximum subarray but with two alternating DP tracks.
Interview Questions on This Problem
Q1How would you modify Kadane’s algorithm to enforce alternating sign sums when selecting non‑overlapping subarrays?
Maintain two DP variables: bestPos = max sum of a valid sequence ending with a positive interval, and bestNeg = max sum ending with a negative interval. For each element, compute the best positive interval that can end here as max(prevNeg + currentSum, currentSum) and similarly for negative as max(prevPos - currentSum, -currentSum). Update the globals accordingly.
Q2What edge case must you handle when the entire array consists of non‑positive numbers?
Since the first interval must be positive, the answer is zero (or “no valid selection”) because no positive‑sum subarray exists; the DP should initialize bestPos to negative infinity and return max(0, bestPos) at the end.
Q3Explain why a greedy “pick the longest positive segment then the longest negative segment” strategy fails on this problem.
Greedy picks ignore the global optimum; a shorter positive segment may enable a much larger subsequent negative segment, yielding a higher total alternating sum. The DP captures the trade‑off by considering both extending and restarting intervals at each position.
Examples
Input
6\n4 -1 2 -3 5 -2
Output
7
Explanation: Choose intervals [4] (sum = 4), [-1,2,-3] (sum = -2), [5] (sum = 5). Total = 4-2+5 = 7, which is optimal.
Input
6\n-5 3 -2 4 -1 2
Output
6
Explanation: Select intervals [3] (3), [-2] (-2), [4] (4), [-1] (-1), [2] (2). Total = 3-2+4-1+2 = 6. No other alternating selection yields a larger sum.
Input
5\n10 -20 30 -40 50
Output
40
Explanation: Start with interval [30] (30), then [-40] (-40), then [50] (50). Total = 30-40+50 = 40, which is the maximum possible.
Constraints
- 1 <= n <= 100000
- -10^9 <= budget[i] <= 10^9
- The sum of absolute values fits in 64-bit signed integer
- Time limit: O(n) expected
- Memory limit: O(n) or O(1) extra
Optimal Approach & Strategy
Use a two‑state DP (positive‑end and negative‑end) updated in a single pass, yielding O(n) time and O(1) extra space.
Brute Force Approach
Enumerate all possible subarrays, compute every combination of non‑overlapping intervals that obey the sign‑alternation, and keep the maximum total – O(n^2) or worse.
Code Solutions
const fs = require('fs');
const data = fs.readFileSync(0,'utf8').trim().split(/\s+/).map(Number);
let pos=0;
const n = data[pos++]||0;
const budget = data.slice(pos,pos+n);
function maxAlternatingBudget(arr){
const NEG = -1e18;
const n = arr.length;
const dpPos = new Array(n).fill(NEG);
const dpNeg = new Array(n).fill(NEG);
let ans = 0;
for(let i=0;i<n;i++){
let sum=0;
for(let j=i;j>=0;j--){
sum+=arr[j];
if(sum>0){
let cand = sum;
if(j>0 && dpNeg[j-1]!==NEG) cand = Math.max(cand, dpNeg[j-1]+sum);
dpPos[i]=Math.max(dpPos[i],cand);
ans=Math.max(ans,dpPos[i]);
}
if(sum<0){
let cand = NEG;
if(j>0 && dpPos[j-1]!==NEG) cand = Math.max(cand, dpPos[j-1]+sum);
dpNeg[i]=Math.max(dpNeg[i],cand);
}
}
}
return ans;
}
console.log(maxAlternatingBudget(budget).toString());#include <bits/stdc++.h>
using namespace std;
long long maxAlternatingBudget(const vector<int>& a){
int n=a.size();
const long long NEG = LLONG_MIN/4;
vector<long long> dpPos(n,NEG), dpNeg(n,NEG);
long long ans=0;
for(int i=0;i<n;++i){
long long sum=0;
for(int j=i;j>=0;--j){
sum+=a[j];
if(sum>0){
long long cand = sum; // start new sequence
if(j>0 && dpNeg[j-1]!=NEG) cand = max(cand, dpNeg[j-1]+sum);
dpPos[i]=max(dpPos[i],cand);
ans=max(ans,dpPos[i]);
}
if(sum<0){
long long cand = sum; // start new sequence not allowed because first must be positive
if(j>0 && dpPos[j-1]!=NEG) cand = max(cand, dpPos[j-1]+sum);
dpNeg[i]=max(dpNeg[i],cand);
}
}
}
return ans;
}
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<<maxAlternatingBudget(a)<<"\n"; }
import java.io.*;
import java.util.*;
public class Main {
public static long maxAlternatingBudget(int[] a){
int n=a.length;
long NEG = Long.MIN_VALUE/4;
long[] dpPos = new long[n];
long[] dpNeg = new long[n];
Arrays.fill(dpPos, NEG);
Arrays.fill(dpNeg, NEG);
long ans=0;
for(int i=0;i<n;i++){
long sum=0;
for(int j=i;j>=0;j--){
sum+=a[j];
if(sum>0){
long cand=sum;
if(j>0 && dpNeg[j-1]!=NEG) cand=Math.max(cand, dpNeg[j-1]+sum);
dpPos[i]=Math.max(dpPos[i],cand);
ans=Math.max(ans, dpPos[i]);
}
if(sum<0){
if(j>0 && dpPos[j-1]!=NEG){
long cand=dpPos[j-1]+sum;
dpNeg[i]=Math.max(dpNeg[i], cand);
}
}
}
}
return ans;
}
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[] arr = new int[n];
if(n>0){
StringTokenizer st = new StringTokenizer(br.readLine());
for(int i=0;i<n;i++) arr[i]=Integer.parseInt(st.nextToken());
}
System.out.println(maxAlternatingBudget(arr));
}
}
import sys
def max_alternating_budget(arr):
n=len(arr)
NEG=-10**18
dp_pos=[NEG]*n
dp_neg=[NEG]*n
ans=0
for i in range(n):
s=0
for j in range(i, -1, -1):
s+=arr[j]
if s>0:
cand=s
if j>0 and dp_neg[j-1]!=NEG:
cand=max(cand, dp_neg[j-1]+s)
dp_pos[i]=max(dp_pos[i],cand)
ans=max(ans,dp_pos[i])
if s<0:
if j>0 and dp_pos[j-1]!=NEG:
cand=dp_pos[j-1]+s
dp_neg[i]=max(dp_neg[i],cand)
return ans
if __name__=="__main__":
data=sys.stdin.read().strip().split()
if not data:
sys.exit()
n=int(data[0])
arr=list(map(int,data[1:1+n]))
print(max_alternating_budget(arr))
const fs = require('fs');
const data = fs.readFileSync(0,'utf8').trim().split(/\s+/).map(Number);
let pos=0;
const n = data[pos++]||0;
const budget = data.slice(pos,pos+n);
function maxAlternatingBudget(arr){
const NEG = -1e18;
const n = arr.length;
const dpPos = new Array(n).fill(NEG);
const dpNeg = new Array(n).fill(NEG);
let ans = 0;
for(let i=0;i<n;i++){
let sum=0;
for(let j=i;j>=0;j--){
sum+=arr[j];
if(sum>0){
let cand = sum;
if(j>0 && dpNeg[j-1]!==NEG) cand = Math.max(cand, dpNeg[j-1]+sum);
dpPos[i]=Math.max(dpPos[i],cand);
ans=Math.max(ans,dpPos[i]);
}
if(sum<0){
let cand = NEG;
if(j>0 && dpPos[j-1]!==NEG) cand = Math.max(cand, dpPos[j-1]+sum);
dpNeg[i]=Math.max(dpNeg[i],cand);
}
}
}
return ans;
}
console.log(maxAlternatingBudget(budget).toString());Asked in Top Tech Interviews
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