How to divide a set into two subsets such that difference between the sum of numbers in two sets is minimal?

Question

Given a set of numbers, divide the numbers into two subsets such that difference between the sum of numbers in two subsets is minimal.

T

Answer 1

No, Your algorithm is wrong. Your algo follows a greedy approach. I implemented your approach and it failed over this test case: (You may try here)

A greedy algorithm:

#include
#define rep(i,_n) for(int i=0;i<_n;i++)
using namespace std;

#define MXN 55
int a[MXN];

int main() {
    //code
    int t,n,c;
    cin>>t;
    while(t--){
        cin>>n;
        rep(i,n) cin>>a[i];
        sort(a, a+n);
        reverse(a, a+n);
        ll sum1 = 0, sum2 = 0;
        rep(i,n){
            cout<

Test case:

1
8 
16 14 13 13 12 10 9 3

Wrong Ans: 6
16 13 10 9
14 13 12 3

Correct Ans: 0
16 13 13 3
14 12 10 9

The reason greedy algorithm fails is that it does not consider cases when taking a larger element in current larger sum set and later a much smaller in the larger sum set may result much better results. It always try to minimize current difference without exploring or knowing further possibilities, while in a correct solution you might include an element in a larger set and include a much smaller element later to compensate this difference, same as in above test case.

Correct Solution:

To understand the solution, you will need to understand all below problems in order:

• 0/1 Knapsack with Dynamic Programming
• Partition Equal Subset Sum with DP
• Solution

My Code (Same logic as this):

#include
#define rep(i,_n) for(int i=0;i<_n;i++)
using namespace std;

#define MXN 55
int arr[MXN];
int dp[MXN][MXN*MXN];

int main() {
    //code
    int t,N,c;
    cin>>t;
    while(t--){
        rep(i,MXN) fill(dp[i], dp[i]+MXN*MXN, 0);

        cin>>N;
        rep(i,N) cin>>arr[i];
        int sum = accumulate(arr, arr+N, 0);
        dp[0][0] = 1;
        for(int i=1; i<=N; i++)
            for(int j=sum; j>=0; j--)
                dp[i][j] |= (dp[i-1][j] | (j>=arr[i-1] ? dp[i-1][j-arr[i-1]] : 0));

        int res = sum;

        for(int i=0; i<=sum/2; i++)
            if(dp[N][i]) res = min(res, abs(i - (sum-i)));

        cout<