Galactic Resource Allocation 4 — Problem Statement & Solution Guide
Problem Description
Galactic Resource Allocation 4
You are given an integer n (n ≥ 2) and two integer arrays weight[0…n‑1] and volume[0…n‑1]. Each index i represents a crate with weight[i] units of mass and volume[i] units of space. Find an index p (0 ≤ p < n‑1) such that:
• sum_{i=0}^{p} weight[i] = sum_{i=p+1}^{n-1} weight[i]
• sum_{i=0}^{p} volume[i] = sum_{i=p+1}^{n-1} volume[i]
If multiple indices satisfy the condition, return the smallest one. If no such index exists, return -1.
Input format:
The first line contains a single integer n.
The second line contains n space‑separated integers representing weight[0] … weight[n‑1].
The third line contains n space‑separated integers representing volume[0] … volume[n‑1].
Output format:
Print a single integer – the required index p or -1 if it does not exist.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Galactic Resource Allocation 4"
WHY DOES IT MATTER?
The two‑pointer pattern transforms a quadratic search into a linear one by exploiting cumulative properties, making it essential for large‑scale data where performance is critical.
OPTIMIZATION CHALLENGE
The core optimization is recognizing that the suffix sum can be derived from the total minus the prefix, eliminating the need for a second pass or nested loops.
REAL-WORLD CONNECTION
Think of load balancing in a data center: you want to split servers into two clusters with equal CPU and memory usage. The two‑pointer method is like a single pass through the server list, updating running totals until both resources are balanced.
When explaining this in an interview, emphasize that you compute the total once, then use a single loop with two running sums—this showcases both algorithmic insight and code simplicity.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem asks for an index that splits two parallel arrays—weight and volume—into two halves with equal sums. A naive solution would compute the sum of each side for every possible split, leading to an O(n^2) time complexity, which becomes infeasible for large n. The optimal approach uses the two‑pointer (or prefix‑sum) technique: first compute the total sums of weight and volume. Then iterate once, maintaining running prefixes for both arrays. At each index, compare the prefix sums to the remaining suffix sums (total minus prefix). When both weight and volume prefixes equal their respective suffixes, the current index is the desired split. This reduces the problem to a single linear scan with constant additional space, achieving O(n) time and O(1) space.
The key insight is that the equality conditions for weight and volume can be checked independently but simultaneously during the same traversal. Because the arrays are independent, we only need to keep two running totals, not a full prefix array. This eliminates the need for nested loops or multiple passes, which is why the two‑pointer paradigm is the optimal solution for this class of partition problems.
Interview Questions on This Problem
Q1How would you modify the algorithm if the arrays could contain negative numbers?
The two‑pointer approach still works because we only rely on cumulative sums. However, we must ensure that the total sums are correctly computed and that the comparison uses the updated prefix sums. If the total sums are zero, any index where the prefix equals half the total works; otherwise, we still compare prefix to total minus prefix.
Q2In a distributed system, how could you parallelize the search for the split index across multiple machines?
You could partition the arrays into chunks, compute local prefix sums and local totals, then perform a prefix‑sum reduction across machines to obtain global totals. Each machine can then scan its chunk using the global totals to determine if a split exists locally. Finally, a coordinator aggregates the results to find the global split index.
Q3What would be the impact on time complexity if you were required to find all possible split indices instead of just one?
You would still perform a single linear scan, but you would record every index where the prefix equals the suffix for both arrays. The time complexity remains O(n) and space increases to O(k) where k is the number of valid splits, which is at most n.
Examples
Input
6 1 2 3 3 2 1 4 5 6 6 5 4
Output
2
Explanation: Prefix up to index 2: weight = 1+2+3 = 6, volume = 4+5+6 = 15. Suffix from index 3: weight = 3+2+1 = 6, volume = 6+5+4 = 15. Both sums match, so the smallest valid index is 2.
Input
4 0 0 0 0 1 2 3 6
Output
2
Explanation: For p = 2: weight prefix = 0+0+0 = 0, weight suffix = 0. Volume prefix = 1+2+3 = 6, volume suffix = 6. Both conditions hold, and no smaller index works, therefore the answer is 2.
Input
5 1 2 3 4 5 1 1 1 1 1
Output
-1
Explanation: Checking every possible p (0…3) shows that while some indices balance the weight sums, none balance the volume sums simultaneously. Hence no index satisfies both equations and the result is -1.
Constraints
- 2 <= n <= 200000
- 0 <= weight[i] <= 10^9
- 0 <= volume[i] <= 10^9
- All calculations fit into 64‑bit signed integers.
Optimal Approach & Strategy
Compute total sums once, then iterate once while maintaining running prefixes; compare prefixes to remaining totals to find the split in O(n) time and O(1) space.
Brute Force Approach
Check every possible split index by recomputing the sums of the left and right halves for both arrays, leading to O(n^2) time.
Verified Code Solutions
const fs = require('fs');
const data = fs.readFileSync(0,'utf8').trim().split(/\s+/).map(Number);
let pos = 0;
function findSplit(weight, volume) {
let totalW = 0n, totalV = 0n;
for (let i = 0; i < weight.length; ++i) {
totalW += BigInt(weight[i]);
totalV += BigInt(volume[i]);
}
let prefW = 0n, prefV = 0n;
for (let i = 0; i + 1 < weight.length; ++i) {
prefW += BigInt(weight[i]);
prefV += BigInt(volume[i]);
if (prefW * 2n === totalW && prefV * 2n === totalV) return i;
}
return -1;
}
if (data.length===0) process.exit(0);
const n = data[pos++];
const w = new Array(n), v = new Array(n);
for(let i=0;i<n;i++){
w[i]=data[pos++];
v[i]=data[pos++];
}
console.log(findSplit(w,v).toString());#include <bits/stdc++.h>
using namespace std;
int findSplit(const vector<long long>& weight, const vector<long long>& volume) {
long long totalW = 0, totalV = 0;
for (size_t i = 0; i < weight.size(); ++i) {
totalW += weight[i];
totalV += volume[i];
}
long long prefW = 0, prefV = 0;
for (size_t i = 0; i + 1 < weight.size(); ++i) {
prefW += weight[i];
prefV += volume[i];
if (prefW * 2 == totalW && prefV * 2 == totalV) return static_cast<int>(i);
}
return -1;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n;
if(!(cin>>n)) return 0;
vector<long long> w(n), v(n);
for(int i=0;i<n;++i) cin>>w[i]>>v[i];
cout<<findSplit(w,v);
return 0;
}import java.io.*;
import java.util.*;
public class Main {
private static int findSplit(long[] weight, long[] volume) {
long totalW = 0, totalV = 0;
for (int i = 0; i < weight.length; i++) {
totalW += weight[i];
totalV += volume[i];
}
long prefW = 0, prefV = 0;
for (int i = 0; i + 1 < weight.length; i++) {
prefW += weight[i];
prefV += volume[i];
if (prefW * 2 == totalW && prefV * 2 == totalV) {
return i;
}
}
return -1;
}
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());
long[] w = new long[n];
long[] v = new long[n];
for (int i = 0; i < n; i++) {
StringTokenizer st = new StringTokenizer(br.readLine());
w[i] = Long.parseLong(st.nextToken());
v[i] = Long.parseLong(st.nextToken());
}
System.out.print(findSplit(w, v));
}
}import sys
def find_split(weight, volume):
total_w = sum(weight)
total_v = sum(volume)
pref_w = pref_v = 0
for i in range(len(weight)-1):
pref_w += weight[i]
pref_v += volume[i]
if pref_w * 2 == total_w and pref_v * 2 == total_v:
return i
return -1
def main():
data = sys.stdin.read().strip().split()
if not data:
return
it = iter(data)
n = int(next(it))
w = []
v = []
for _ in range(n):
w.append(int(next(it)))
v.append(int(next(it)))
print(find_split(w, v))
if __name__ == "__main__":
main()const fs = require('fs');
const data = fs.readFileSync(0,'utf8').trim().split(/\s+/).map(Number);
let pos = 0;
function findSplit(weight, volume) {
let totalW = 0n, totalV = 0n;
for (let i = 0; i < weight.length; ++i) {
totalW += BigInt(weight[i]);
totalV += BigInt(volume[i]);
}
let prefW = 0n, prefV = 0n;
for (let i = 0; i + 1 < weight.length; ++i) {
prefW += BigInt(weight[i]);
prefV += BigInt(volume[i]);
if (prefW * 2n === totalW && prefV * 2n === totalV) return i;
}
return -1;
}
if (data.length===0) process.exit(0);
const n = data[pos++];
const w = new Array(n), v = new Array(n);
for(let i=0;i<n;i++){
w[i]=data[pos++];
v[i]=data[pos++];
}
console.log(findSplit(w,v).toString());Asked in Top Tech Interviews
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