Equal Payload Partition — Problem Statement & Solution Guide
Quick Answer & Algorithm Key Takeaway
Master coding challenges related to Recursion and solve the Equal Payload Partition problem optimally.
O(n * sum)O(sum)Problem Description
Given an integer array weights, determine how many distinct ways the elements can be divided into two disjoint groups such that the sum of the numbers in each group is identical. Every element must belong to exactly one group, and the internal ordering of a group is irrelevant. Two divisions are considered different if the set of indices placed in the first group differs from another division (the complementary set automatically forms the second group). Return the count of valid divisions. If no division yields equal sums, return 0.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Equal Payload Partition"
WHY DOES IT MATTER?
This pattern is essential for solving combinatorial optimization problems where the goal is to count the number of valid configurations under sum constraints. It is a fundamental building block for more complex problems involving resource allocation, scheduling, and load balancing.
OPTIMIZATION CHALLENGE
The key insight is to reduce the 2D DP table to a 1D array by iterating through the target sum in reverse order. This ensures that each element is only used once in the current iteration, preventing overcounting and reducing space complexity from O(n*sum) to O(sum).
REAL-WORLD CONNECTION
In distributed systems, this pattern is used to balance load across servers or data centers. For example, when deploying microservices, you might want to distribute instances such that the total resource usage (CPU, memory) is balanced across two clusters to ensure optimal performance and fault tolerance.
During the interview, clearly articulate the transformation from the partition problem to the subset sum counting problem. Emphasize the importance of checking for an odd total sum early to avoid unnecessary computation. Also, mention the space optimization as a sign of your ability to think about memory efficiency.
COMPLEXITY AT A GLANCE
O(n * sum)O(sum)Core Theory — Why This Approach?
The problem of partitioning an array into two subsets with equal sums is a classic variant of the Subset Sum problem, which is NP-complete in its general form. However, when the target sum is bounded by the total sum of the array, it can be solved efficiently using Dynamic Programming (DP). The core insight is that if the total sum of the array is odd, no such partition exists. If the total sum is even, we need to find the number of subsets that sum to exactly half of the total sum. This transforms the problem into a 0/1 Knapsack problem where the 'capacity' is the target sum, and the 'value' of each item is 1 (since we are counting the number of ways, not maximizing value).
Interview Questions on This Problem
Q1At a fintech company, you need to balance transaction loads across two servers such that the total transaction value on each server is identical. How would you model this as a DP problem, and what are the space optimization techniques you can apply?
Model it as a 0/1 Knapsack problem where the target capacity is half the total transaction value. Use a 1D DP array dp[j] representing the number of ways to achieve sum j. Iterate through each transaction and update the DP array from right to left to avoid using the same element multiple times. This reduces space complexity from O(n*sum) to O(sum).
Q2In a high-growth startup, you are designing a load balancer that distributes requests to two clusters. If the total load is 1000 and you need equal distribution, how do you handle the case where the total load is odd, and how do you count the distinct partitions?
First, check if the total load is odd; if so, return 0 immediately. If even, set the target to total/2. Use DP to count the number of subsets that sum to the target. Each valid subset corresponds to a unique partition because the complement subset is determined. Be careful with integer overflow if the number of ways is large, using long integers or modular arithmetic if required.
Q3At a global product company, you are optimizing a data replication strategy where data chunks must be split into two groups of equal size for redundancy. How does the 'counting' aspect of this problem differ from the 'existence' aspect, and how does it affect the DP state definition?
For existence, the DP state is boolean (true/false). For counting, the DP state is an integer representing the number of ways. The recurrence relation changes from dp[i][j] = dp[i-1][j] || dp[i-1][j-weights[i]] to dp[i][j] = dp[i-1][j] + dp[i-1][j-weights[i]]. This requires careful handling of base cases and ensuring that each element is considered only once in the transition.
Examples
Input
[1,2,3,4,6]
Output
2
Explanation: Total sum = 16, half = 8. Subsets that sum to 8 are {2,6} and {1,3,4}. Each subset defines a unique partition, giving 2 ways.
Input
[5,5,5,5]
Output
6
Explanation: Total sum = 20, half = 10. Any pair of the four 5‑s forms a subset of sum 10. There are C(4,2)=6 such pairs, so 6 distinct partitions.
Input
[1,1,1,1,1]
Output
0
Explanation: Total sum = 5, which is odd; an equal split is impossible, so the answer is 0.
Constraints
- 1 <= weights.length <= 20
- -10^4 <= weights[i] <= 10^4
- |sum(weights)| <= 2*10^5
Optimal Approach & Strategy
Use dynamic programming to count the number of subsets that sum to half the total sum. Use a 1D DP array to optimize space, iterating through each element and updating the DP array from right to left.
Brute Force Approach
Generate all possible subsets of the array and check if any subset sums to half the total sum. This takes O(2^n) time, which is infeasible for large arrays.
Code Solutions
const fs = require('fs');
const data = fs.readFileSync(0,'utf8').trim().split(/\s+/).map(Number);
if(data.length===0){process.exit(0);}
let pos=0;
const n=data[pos++];
const weights=data.slice(pos, pos+n);
function countPartitions(arr){
const total = arr.reduce((a,b)=>a+b,0);
if(total%2!==0) return 0;
const target = total/2;
const memo = new Map(); // key: idx|sum
function dfs(idx,sum){
if(idx===arr.length) return sum===target ? 1 : 0;
const key = idx+','+sum;
if(memo.has(key)) return memo.get(key);
let ways = dfs(idx+1,sum+arr[idx]) + dfs(idx+1,sum);
memo.set(key,ways);
return ways;
}
return dfs(0,0);
}
console.log(countPartitions(weights).toString());#include <bits/stdc++.h>
using namespace std;
using int64 = long long;
int64 dfs(int idx, int cur, int target, const vector<int>& w, vector<unordered_map<int,int64>>& memo){
if(idx==(int)w.size()) return cur==target ? 1 : 0;
auto it=memo[idx].find(cur);
if(it!=memo[idx].end()) return it->second;
int64 ways=0;
// take element
ways+=dfs(idx+1, cur+w[idx], target, w, memo);
// skip element
ways+=dfs(idx+1, cur, target, w, memo);
memo[idx][cur]=ways;
return ways;
}
int countPartitions(const vector<int>& weights){
long long total=0; for(int v:weights) total+=v;
if(total%2!=0) return 0;
int target=total/2;
vector<unordered_map<int,int64>> memo(weights.size());
return (int)dfs(0,0,target,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<<countPartitions(w);
return 0;
}
import java.io.*;
import java.util.*;
public class Main {
private static long dfs(int idx, int cur, int target, int[] w, Map<Long,Long> memo){
if(idx==w.length) return cur==target ? 1L : 0L;
long key = ((long)idx<<32) | (cur & 0xffffffffL);
Long cached = memo.get(key);
if(cached!=null) return cached;
long ways = dfs(idx+1, cur + w[idx], target, w, memo) + dfs(idx+1, cur, target, w, memo);
memo.put(key, ways);
return ways;
}
public static long countPartitions(int[] weights){
long total=0; for(int v:weights) total+=v;
if((total & 1L)==1L) return 0L;
int target = (int)(total/2);
Map<Long,Long> memo = new HashMap<>();
return dfs(0,0,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(countPartitions(w));
}
}
import sys
from functools import lru_cache
def count_partitions(weights):
total = sum(weights)
if total % 2:
return 0
target = total // 2
@lru_cache(maxsize=None)
def dfs(idx, cur):
if idx == len(weights):
return 1 if cur == target else 0
# include
take = dfs(idx+1, cur + weights[idx])
# exclude
skip = dfs(idx+1, cur)
return take + skip
return dfs(0, 0)
def main():
data = sys.stdin.read().strip().split()
if not data:
return
n = int(data[0])
weights = list(map(int, data[1:1+n]))
print(count_partitions(weights))
if __name__ == "__main__":
main()
const fs = require('fs');
const data = fs.readFileSync(0,'utf8').trim().split(/\s+/).map(Number);
if(data.length===0){process.exit(0);}
let pos=0;
const n=data[pos++];
const weights=data.slice(pos, pos+n);
function countPartitions(arr){
const total = arr.reduce((a,b)=>a+b,0);
if(total%2!==0) return 0;
const target = total/2;
const memo = new Map(); // key: idx|sum
function dfs(idx,sum){
if(idx===arr.length) return sum===target ? 1 : 0;
const key = idx+','+sum;
if(memo.has(key)) return memo.get(key);
let ways = dfs(idx+1,sum+arr[idx]) + dfs(idx+1,sum);
memo.set(key,ways);
return ways;
}
return dfs(0,0);
}
console.log(countPartitions(weights).toString());
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