Maximum Alternate Chest Sum — Problem Statement & Solution Guide
Quick Answer & Algorithm Key Takeaway
Master coding challenges related to Arrays and solve the Maximum Alternate Chest Sum problem optimally.
O(N*M)O(1)Problem Description
Given a two‑dimensional integer array chests, each inner array represents a row of values. For a row, define its even‑index sum as the sum of elements whose positions are 0,2,4,… (0‑based). You may select any subset of rows and add their even‑index sums together. Rows whose even‑index sum is less than or equal to zero are never beneficial and can be omitted. Return the maximum total sum achievable.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Maximum Alternate Chest Sum"
WHY DOES IT MATTER?
This pattern exemplifies the "independent sub‑problem aggregation" paradigm where global optimum is the sum of locally optimal, non‑conflicting choices, a cornerstone in greedy algorithms and DP base cases.
OPTIMIZATION CHALLENGE
The key insight is recognizing that rows do not interact, allowing us to replace exponential subset enumeration with a single pass that filters positive contributions, collapsing both time and space to linear bounds.
REAL-WORLD CONNECTION
Think of each row as a microservice reporting profit from even‑indexed transactions; the system only activates services with positive profit, akin to scaling out only profitable instances in a distributed cloud deployment.
During an interview, first state the independence observation, then write a loop that computes the even‑index sum per row and accumulates only if >0 – this shows both analytical clarity and coding efficiency.
COMPLEXITY AT A GLANCE
O(N*M)O(1)Core Theory — Why This Approach?
The problem reduces to a simple aggregation over independent rows. Each row contributes a value equal to the sum of its elements at even positions (0‑based). Because rows are independent, the global optimum is obtained by selecting every row whose contribution is positive – a classic greedy selection on a set of non‑interacting items. A naïve solution might try all subsets of rows, leading to O(2^n) exponential time, which is infeasible for large n. Recognizing the separability lets us replace the exponential search with a linear scan, computing each row’s even‑index sum in O(m) where m is the row length, and then summing only the positive contributions, achieving O(N·M) overall where N is the number of rows.
Interview Questions on This Problem
Q1How would you modify the solution if the definition of a "beneficial" row changed to "even‑index sum greater than a given threshold T"?
Compute each row’s even‑index sum as before, then add it to the answer only if sum > T. The algorithmic complexity remains O(N·M) because the threshold check is O(1) per row.
Q2In a streaming scenario where rows arrive one‑by‑one, how can you maintain the maximum total without storing all rows?
Maintain a running total of positive even‑index sums; for each incoming row compute its even‑index sum and, if positive, add it to the accumulator. No storage of past rows is needed, yielding O(1) extra space.
Q3If the array were three‑dimensional (layers × rows × columns) and you could pick any subset of layers, each layer’s contribution being the sum of its rows’ even‑index sums, does the greedy strategy still hold?
Yes, because layer contributions are independent; you still sum all layer totals that are positive. The algorithm extends by first aggregating each layer’s total (a linear pass) and then adding positive layer totals.
Examples
Input
[[5,-2,3,1],[-4,2,-1],[7,0,-3,4,2]]
Output
14
Explanation: Row0 even‑index sum =5+3=8 (positive). Row1 even‑index sum =-4+(-1)=-5 (negative, skip). Row2 even‑index sum =7+(-3)+2=6 (positive). Total =8+6=14.
Input
[[-1,-2],[0],[10,-10,5]]
Output
15
Explanation: Row0 sum=-1 (skip). Row1 sum=0 (skip). Row2 sum=10+5=15 (positive). Total=15.
Input
[[2,3,4],[1,1,1,1],[-5,-5,-5,-5]]
Output
8
Explanation: Row0 sum=2+4=6. Row1 sum=1+1=2. Row2 sum=-5+(-5)=-10 (skip). Total=6+2=8.
Constraints
- 1 <= chests.length <= 100000
- 1 <= chests[i].length <= 100000
- Sum of all chests[i].length across rows <= 200000
- -10^9 <= chests[i][j] <= 10^9
Optimal Approach & Strategy
Compute each row’s even‑index sum once, add it to the answer only if it’s positive – linear time.
Brute Force Approach
Enumerate every subset of rows, compute the total even‑index sum for each subset, and keep the maximum – exponential time.
Code Solutions
const fs = require('fs');
const data = fs.readFileSync(0,'utf8').trim().split(/\s+/).map(Number);
let idx = 0;
function maximumAlternateChestSum(chests){
let total = 0;
for(const row of chests){
let sum = 0;
for(let i=0;i<row.length;i+=2) sum += row[i];
if(sum>0) total += sum;
}
return total;
}
if(data.length===0){process.exit(0);}
const n = data[idx++];
let chests = [];
for(let i=0;i<n;i++){
const m = data[idx++];
const row = [];
for(let j=0;j<m;j++) row.push(data[idx++]);
chests.push(row);
}
console.log(maximumAlternateChestSum(chests));#include <bits/stdc++.h>
using namespace std;
long long maximumAlternateChestSum(const vector<vector<int>>& chests){
long long total = 0;
for(const auto& row: chests){
long long sum = 0;
for(size_t i=0;i<row.size();i+=2) sum += row[i];
if(sum>0) total += sum;
}
return total;
}
int main(){
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n; if(!(cin>>n)) return 0;
vector<vector<int>> chests(n);
for(int i=0;i<n;++i){
int m;cin>>m; chests[i].resize(m);
for(int j=0;j<m;++j) cin>>chests[i][j];
}
cout<<maximumAlternateChestSum(chests);
return 0;
}import java.io.*;
import java.util.*;
public class Main {
public static long maximumAlternateChestSum(List<List<Integer>> chests) {
long total = 0;
for(List<Integer> row : chests){
long sum = 0;
for(int i=0;i<row.size();i+=2){
sum += row.get(i);
}
if(sum>0) total += sum;
}
return total;
}
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());
List<List<Integer>> chests = new ArrayList<>();
for(int i=0;i<n;i++){
StringTokenizer st = new StringTokenizer(br.readLine());
int m = Integer.parseInt(st.nextToken());
List<Integer> row = new ArrayList<>();
for(int j=0;j<m;j++){
if(!st.hasMoreTokens()) st = new StringTokenizer(br.readLine());
row.add(Integer.parseInt(st.nextToken()));
}
chests.add(row);
}
System.out.print(maximumAlternateChestSum(chests));
}
}import sys
def maximum_alternate_chest_sum(chests):
total = 0
for row in chests:
s = sum(row[i] for i in range(0, len(row), 2))
if s > 0:
total += s
return total
def main():
data = list(map(int, sys.stdin.read().split()))
if not data:
return
it = iter(data)
n = next(it)
chests = []
for _ in range(n):
m = next(it)
row = [next(it) for __ in range(m)]
chests.append(row)
print(maximum_alternate_chest_sum(chests))
if __name__ == "__main__":
main()const fs = require('fs');
const data = fs.readFileSync(0,'utf8').trim().split(/\s+/).map(Number);
let idx = 0;
function maximumAlternateChestSum(chests){
let total = 0;
for(const row of chests){
let sum = 0;
for(let i=0;i<row.length;i+=2) sum += row[i];
if(sum>0) total += sum;
}
return total;
}
if(data.length===0){process.exit(0);}
const n = data[idx++];
let chests = [];
for(let i=0;i<n;i++){
const m = data[idx++];
const row = [];
for(let j=0;j<m;j++) row.push(data[idx++]);
chests.push(row);
}
console.log(maximumAlternateChestSum(chests));Asked in Top Tech Interviews
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