minimum_sum_partition.c

/*

=============================Problem Statement=============================
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.

If there is a set S with n elements, then if we assume Subset1 has m elements, Subset2 must have n-m elements and the value of abs(sum(Subset1) – sum(Subset2)) should be minimum.

Example:

Input:  arr[] = {1, 6, 11, 5}
Output: 1
Explanation:
Subset1 = {1, 5, 6}, sum of Subset1 = 12
Subset2 = {11}, sum of Subset2 = 11

============================LOGIC============================
The problem can be solved using dynamic programming when the sum of the elements is not too big. 
We can create a 2D array dp[n+1][(sum/2)+1] where n is number of elements in given set and sum is sum of all elements. 
We can construct the solution in bottom up manner maximizing the sum of a partion having sum less than or equal to (sum/2).

then minimum difference between partion is (sum-2*dp[n][sum/2])


*/



#include <stdio.h>
#include <string.h>

int minDiff(int a[], int n){
    int sum1 = 0;                                               //     sum1 stores the sum of all elements of array a
    for(int i=0; i<n; i++){
        sum1 += a[i];
    }


    const int t1 = n+1;
    const int t2 = (sum1/2)+1;
    int dp[t1][t2];                                             //      dp[i][j] stores the maximum sum of subset of a[0...i-1] such that sum of subset is less than or equal to j
    
    memset( dp, 0, t1*t2*sizeof(int) );                         //      initializing all elements of dp to 0
    
    for(int i =1 ; i < (n+1); i++ ){
        for(int j =1; j < t2; j++){
            if (a[i-1] <= j){                                   //      if a[i-1] is less than or equal to j then we have two options either to include a[i-1] or not
                int x = a[i-1]+dp[i-1][j-a[i-1]];
                int y = dp[i-1][j];
                dp[i][j] = x > y ? x : y;                       //      we choose the maximum of both options
                
            }
            else{
                dp[i][j]=dp[i-1][j];                            //      if a[i-1] is greater than j then we have only one option that is not to include a[i-1]
            }
        }
    }
    
    
    
    
    int x = (sum1-2*dp[n][sum1/2]);                             //      x stores the difference between sum of subset of a[0...n-1] and 
                                                                //      sum of subset of a[0...n-1] such that sum of subset is less than or equal to sum1/2
    
    x = x>0 ? x: -x;                                            //      if x is negative then we take -x as minimum difference is always positive
    
    return x;
    
    

}

int main()
{
    int n;
    printf("enter number of elements:");
    scanf("%d",&n);
    int a[n];

    printf("Enter %d elements: ", n);

    for(int i=0; i<n; i++){
        scanf("%d",&a[i]);
    }

    int b = minDiff(a,n);

    printf("minimum difference between partion is %d",b);
    return 0;
}