Minimum sum partition

Given an array, the task is to divide it into two sets S1 and S2 such that the absolute difference between their sums is minimum.

DynamicProgramming/Minimum sum partition

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+// Partition a set into two subsets such that the difference of subset sums is minimum
+
+import java.util.*;
+import java.lang.*;
+import java.io.*;
+class GFG
+ {
+	public static void main (String[] args)
+	 {
+	 Scanner sc=new Scanner(System.in);
+	 int t=sc.nextInt();
+	 while(t-->0)
+	 {
+	     int n=sc.nextInt();
+	     int arr[]=new int[n];
+	     int sum=0;
+	     for(int i=0;i<n;i++)
+	     {
+	         arr[i]=sc.nextInt();
+	         sum+=arr[i];
+	     }
+	     int ans[]=new int[sum];
+	     ans=subset(arr,sum);
+	     int min=Integer.MAX_VALUE;
+	     for(int i=0;i<ans.length;i++)
+	     min=Math.min(min,(sum-2*ans[i]));
+	     
+	     System.out.println(min);
+	     
+	     
+	 }
+	 }
+	 static int[] subset(int arr[],int sum)
+	 {
+	     int n=arr.length;
+	     boolean dp[][]=new boolean[n+1][sum+1];
+	     for(int i=0;i<=n;i++)
+	     dp[i][0]=true;
+	     for(int i=1;i<=sum;i++)
+	     dp[0][i]=false;
+	     
+	     // subset sum concept
+	    for(int i=1;i<=n;i++)
+	    {
+	        for(int j=1;j<=sum;j++)
+	        {
+	            if(arr[i-1]<=j)
+	            dp[i][j]=dp[i-1][j-arr[i-1]] || dp[i-1][j];
+	            else
+	            dp[i][j]=dp[i-1][j];
+	        }
+	    }
+	    
+	    //storing last dp column whose value is true till sum/2
+	    int index[]=new int[sum];
+	    int p=0;
+	    for(int i=0;i<=sum/2;i++)
+	    {
+	        if(dp[n][i]==true)
+	            index[p++]=i;
+	    }
+	    return index;
+	 }
+}