Interstellar Cargo Distribution — Problem Statement & Solution Guide
Problem Description
Given an array of positive integers representing the weights of cargo packages, determine how many distinct ways the packages can be split between two spacecraft, Starblade and NovaSpur, so that the total weight carried by each spacecraft is exactly the same. A split is defined by a subset of packages assigned to Starblade; the remaining packages automatically belong to NovaSpur. Two splits are considered identical if one can be obtained from the other by swapping the spacecraft, i.e., the unordered partition of the set matters. Return the count of such unordered partitions. The solution must be derived using a recursive approach (with memoization or pruning as needed).
DSA Pattern Breakdown
DSA Pattern Breakdown
"Interstellar Cargo Distribution"
WHY DOES IT MATTER?
Subset‑sum counting is a fundamental DP pattern that appears in resource allocation, load balancing, and cryptographic knapsack problems; mastering it equips engineers to tackle many NP‑hard variants efficiently on bounded inputs.
OPTIMIZATION CHALLENGE
The key insight is to recognize that the total sum must be even and then to count subsets that hit sum/2, allowing us to prune half the search space and replace exponential recursion with memoized states (i,remaining).
REAL-WORLD CONNECTION
Think of distributing cargo across two parallel data‑center clusters where each cluster must handle equal load; the DP mirrors how a scheduler decides which jobs (packages) go to which cluster to achieve perfect balance.
During an interview, first compute total weight, early‑exit if odd, then implement a recursive function with a cache (e.g., unordered_map or vector) – this shows you understand both the mathematical reduction and practical memoization.
COMPLEXITY AT A GLANCE
O(n*target)O(n*target)Core Theory — Why This Approach?
The problem reduces to counting subsets whose sum equals half of the total weight. A naive recursive enumeration tries every inclusion/exclusion choice, leading to O(2^n) time which explodes even for moderate n. By recognizing the sub‑problem – “how many ways can we achieve a target sum using the first i items?” – we can apply recursion with memoization (top‑down DP) or iterative DP (bottom‑up) to reuse overlapping sub‑problems. This transforms the exponential search space into a pseudo‑polynomial one, O(n*target), where target = total/2, because each state (i,remaining) is solved at most once. The optimal paradigm is thus a classic subset‑sum counting DP, leveraging the principle of optimality and overlapping sub‑problems inherent to recursion with memoization.
Interview Questions on This Problem
Q1How would you modify the solution if the cargo weights could be negative?
Negative numbers break the simple DP table indexed by sum because the range can become unbounded. You would shift the possible sum range by adding an offset equal to the absolute sum of negative numbers, or use a hashmap‑based memoization that maps (index, currentSum) to count, thereby handling arbitrary integer sums.
Q2What is the time‑space trade‑off when using a 1‑dimensional DP array versus a 2‑dimensional DP table for this problem?
A 2‑D table dp[i][s] stores counts for each prefix i and sum s, using O(n*target) space. By iterating items and updating a 1‑D array from high to low sums, we collapse the dimension to O(target) space, but we lose the ability to reconstruct the exact subsets without extra bookkeeping.
Q3Can you extend the algorithm to return the actual subsets, not just the count, while keeping the same asymptotic complexity?
Returning all subsets inherently requires O(k) extra space where k is the number of valid subsets, which can be exponential. However, you can generate subsets on‑the‑fly using backtracking guided by the DP table, still O(n*target) time for enumeration, but the overall space becomes O(n+target) plus the output size.
Examples
Input
[1,2,3,4,6]
Output
2
Explanation: Total weight = 1+2+3+4+6 = 16, half = 8. Subsets that sum to 8 are {2,6} and {1,3,4}. Each subset defines a unique unordered partition, giving 2 ways.
Input
[5,5,5,5]
Output
6
Explanation: Total weight = 20, half = 10. Any choice of two 5‑weight packages forms a subset summing to 10. There are C(4,2)=6 such choices, and each leaves the remaining two 5‑weight packages for the other spacecraft. Hence 6 distinct partitions.
Input
[1,1,1,1,1]
Output
0
Explanation: Total weight = 5, which is odd, so it is impossible to divide the cargo into two equal‑weight groups. Therefore the answer is 0.
Constraints
- 1 <= nums.length <= 30
- 1 <= nums[i] <= 10^4
- All nums[i] are integers
Optimal Approach & Strategy
Use DP (recursion + memoization or iterative table) to count ways to reach the target sum, reducing complexity to O(n*target).
Brute Force Approach
Enumerate every possible subset (2^n) and count those whose sum equals half of the total weight.
Verified Code Solutions
function countWays(weights){
const total=weights.reduce((a,b)=>a+b,0);
if(total%2!==0) return 0;
const target=total/2;
const n=weights.length;
const memo=Array.from({length:n},()=>Array(target+1).fill(undefined));
function dfs(idx,rem){
if(rem===0) return 1;
if(idx===n||rem<0) return 0;
if(memo[idx][rem]!==undefined) return memo[idx][rem];
const take=dfs(idx+1,rem-weights[idx]);
const skip=dfs(idx+1,rem);
return memo[idx][rem]=take+skip;
}
return dfs(0,target);
}
function main(){
const fs=require('fs');
const data=fs.readFileSync(0,'utf8').trim().split(/\s+/).map(Number);
if(data.length===0) return;
const n=data[0];
const weights=data.slice(1,1+n);
console.log(countWays(weights));
}
main();#include <bits/stdc++.h>
using namespace std;
long long dfs(int idx,int target,const vector<int>& a,vector<vector<long long>>& memo){
if(target==0) return 1; // empty subset forms required sum
if(idx==a.size()||target<0) return 0;
if(memo[idx][target]!=-1) return memo[idx][target];
long long take=dfs(idx+1,target-a[idx],a,memo);
long long skip=dfs(idx+1,target,a,memo);
return memo[idx][target]=take+skip;
}
long long countWays(const vector<int>& weights){
long long total=0; for(int v:weights) total+=v;
if(total%2) return 0; // odd total cannot be split
int half=total/2;
int n=weights.size();
vector<vector<long long>> memo(n, vector<long long>(half+1,-1));
return dfs(0,half,weights,memo);
}
int main(){
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n; if(!(cin>>n)) return 0;
vector<int> w(n);
for(int i=0;i<n;++i)cin>>w[i];
cout<<countWays(w);
return 0;
}
import java.io.*;
import java.util.*;
public class Main {
static long dfs(int idx,int target,int[] a,Long[][] memo){
if(target==0) return 1L;
if(idx==a.length||target<0) return 0L;
if(memo[idx][target]!=null) return memo[idx][target];
long take=dfs(idx+1,target-a[idx],a,memo);
long skip=dfs(idx+1,target,a,memo);
return memo[idx][target]=take+skip;
}
static long countWays(int[] weights){
long total=0; for(int v:weights) total+=v;
if((total&1L)==1L) return 0L;
int target=(int)(total/2);
int n=weights.length;
Long[][] memo=new Long[n][target+1];
return dfs(0,target,weights,memo);
}
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[] w=new int[n];
StringTokenizer st=new StringTokenizer(br.readLine());
for(int i=0;i<n;i++) w[i]=Integer.parseInt(st.nextToken());
System.out.println(countWays(w));
}
}
def countWays(weights):
total=sum(weights)
if total%2:
return 0
target=total//2
n=len(weights)
memo=[[None]*(target+1) for _ in range(n)]
def dfs(idx,rem):
if rem==0:
return 1
if idx==n or rem<0:
return 0
if memo[idx][rem] is not None:
return memo[idx][rem]
take=dfs(idx+1,rem-weights[idx])
skip=dfs(idx+1,rem)
memo[idx][rem]=take+skip
return memo[idx][rem]
return dfs(0,target)
if __name__=="__main__":
import sys
data=list(map(int,sys.stdin.read().strip().split()))
if not data:
sys.exit()
n=data[0]
weights=data[1:1+n]
print(countWays(weights))
function countWays(weights){
const total=weights.reduce((a,b)=>a+b,0);
if(total%2!==0) return 0;
const target=total/2;
const n=weights.length;
const memo=Array.from({length:n},()=>Array(target+1).fill(undefined));
function dfs(idx,rem){
if(rem===0) return 1;
if(idx===n||rem<0) return 0;
if(memo[idx][rem]!==undefined) return memo[idx][rem];
const take=dfs(idx+1,rem-weights[idx]);
const skip=dfs(idx+1,rem);
return memo[idx][rem]=take+skip;
}
return dfs(0,target);
}
function main(){
const fs=require('fs');
const data=fs.readFileSync(0,'utf8').trim().split(/\s+/).map(Number);
if(data.length===0) return;
const n=data[0];
const weights=data.slice(1,1+n);
console.log(countWays(weights));
}
main();
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