C Programming - Partition Problem - Dynamic Programming Partition problem is to determine whether a given set can be partitioned into two subsets

Partition problem is to determine whether a given set can be partitioned into two subsets such that the sum of elements in both subsets is same.

Examples

arr[] = {1, 5, 11, 5}
Output: true
The array can be partitioned as {1, 5, 5} and {11}
arr[] = {1, 5, 3}
Output: false
The array cannot be partitioned into equal sum sets.

Following are the two main steps to solve this problem:

- Calculate sum of the array. If sum is odd, there can not be two subsets with equal sum, so return false.
- If sum of array elements is even, calculate sum/2 and find a subset of array with sum equal to sum/2.

The first step is simple. The second step is crucial, it can be solved either using recursion or Dynamic Programming.

**Recursive Solution**

Following is the recursive property of the second step mentioned above.

et isSubsetSum(arr, n, sum/2) be the function that returns true if
there is a subset of arr[0..n-1] with sum equal to sum/2
The isSubsetSum problem can be divided into two subproblems
a) isSubsetSum() without considering last element
(reducing n to n-1)
b) isSubsetSum considering the last element
(reducing sum/2 by arr[n-1] and n to n-1)
If any of the above the above subproblems return true, then return true.
isSubsetSum (arr, n, sum/2) = isSubsetSum (arr, n-1, sum/2) ||
isSubsetSum (arr, n-1, sum/2 - arr[n-1])

C

```
// A recursive C program for partition problem
#include <stdio.h>
// A utility function that returns true if there is
// a subset of arr[] with sun equal to given sum
bool isSubsetSum (int arr[], int n, int sum)
{
// Base Cases
if (sum == 0)
return true;
if (n == 0 && sum != 0)
return false;
// If last element is greater than sum, then
// ignore it
if (arr[n-1] > sum)
return isSubsetSum (arr, n-1, sum);
/* else, check if sum can be obtained by any of
the following
(a) including the last element
(b) excluding the last element
*/
return isSubsetSum (arr, n-1, sum) ||
isSubsetSum (arr, n-1, sum-arr[n-1]);
}
// Returns true if arr[] can be partitioned in two
// subsets of equal sum, otherwise false
bool findPartiion (int arr[], int n)
{
// Calculate sum of the elements in array
int sum = 0;
for (int i = 0; i < n; i++)
sum += arr[i];
// If sum is odd, there cannot be two subsets
// with equal sum
if (sum%2 != 0)
return false;
// Find if there is subset with sum equal to
// half of total sum
return isSubsetSum (arr, n, sum/2);
}
// Driver program to test above function
int main()
{
int arr[] = {3, 1, 5, 9, 12};
int n = sizeof(arr)/sizeof(arr[0]);
if (findPartiion(arr, n) == true)
printf("Can be divided into two subsets "
"of equal sum");
else
printf("Can not be divided into two subsets"
" of equal sum");
return 0;
}
```

**Output :**

Can be divided into two subsets of equal sum

Time Complexity: O(2^n) In worst case, this solution tries two possibilities (whether to include or exclude) for every element.

**Dynamic Programming Solution**

The problem can be solved using dynamic programming when the sum of the elements is not too big. We can create a 2D array part[][] of size (sum/2)*(n+1). And we can construct the solution in bottom up manner such that every filled entry has following property

part[i][j] = true if a subset of {arr[0], arr[1], ..arr[j-1]} has sum
equal to i, otherwise false

C

```
// A Dynamic Programming based C program to partition problem
#include <stdio.h>
// Returns true if arr[] can be partitioned in two subsets of
// equal sum, otherwise false
bool findPartiion (int arr[], int n)
{
int sum = 0;
int i, j;
// Caculcate sun of all elements
for (i = 0; i < n; i++)
sum += arr[i];
if (sum%2 != 0)
return false;
bool part[sum/2+1][n+1];
// initialize top row as true
for (i = 0; i <= n; i++)
part[0][i] = true;
// initialize leftmost column, except part[0][0], as 0
for (i = 1; i <= sum/2; i++)
part[i][0] = false;
// Fill the partition table in botton up manner
for (i = 1; i <= sum/2; i++)
{
for (j = 1; j <= n; j++)
{
part[i][j] = part[i][j-1];
if (i >= arr[j-1])
part[i][j] = part[i][j] || part[i - arr[j-1]][j-1];
}
}
/* // uncomment this part to print table
for (i = 0; i <= sum/2; i++)
{
for (j = 0; j <= n; j++)
printf ("%4d", part[i][j]);
printf("\n");
} */
return part[sum/2][n];
}
// Driver program to test above funtion
int main()
{
int arr[] = {3, 1, 1, 2, 2, 1};
int n = sizeof(arr)/sizeof(arr[0]);
if (findPartiion(arr, n) == true)
printf("Can be divided into two subsets of equal sum");
else
printf("Can not be divided into two subsets of equal sum");
getchar();
return 0;
}
```

**Output :**

Can be divided into two subsets of equal sum

Following diagram shows the values in partition table.