# Partition problem

For this problem, a given set can be partitioned in such a way, that sum of each subset is equal.

At first, we have to find the sum of the given set. If it is even, then there is a chance to divide it into two sets. Otherwise, it cannot be divided.

For even value of the sum, then we will create a table named partTable, now use the following condition to solve the problem.

partTable[i, j] is true, when subset of array[0] to array[j-1] has sum equal to i, otherwise it is false.

## Input and Output

Input:
A set of integers. {3, 1, 1, 2, 2, 1}
Output:
True if the set can be partitioned into two parts with equal sum.
Here the answer is true. One pair of the partitions are: {3, 1, 1}, {2, 2, 1}

## Algorithm

checkPartition(set, n)

Input − The given set, the number of elements in the set.

Output − True when partitioning is possible to make two subsets of the equal sum.

Begin
sum := sum of all elements in the set
if sum is odd, then
return

define partTable of order (sum/2 + 1 x n+1)
set all elements in the 0th row to true
set all elements in the 0th column to false

for i in range 1 to sum/2, do
for j in range 1 to n, do
partTab[i, j] := partTab[i, j-1]
if i >= set[j-1], then
partTab[i, j] := partTab[i, j] or with
partTab[i – set[j-1], j-1]
done
done

return partTab[sum/2, n]
End

## Example

#include <iostream>
using namespace std;

bool checkPartition (int set[], int n) {
int sum = 0;

for (int i = 0; i < n; i++)    //find the sum of all elements of set
sum += set[i];

if (sum%2 != 0)     //when sum is odd, it is not divisible into two set
return false;

bool partTab[sum/2+1][n+1];    //create partition table
for (int i = 0; i <= n; i++)
partTab[0][i] = true;    //for set of zero element, all values are true

for (int i = 1; i <= sum/2; i++)
partTab[i][0] = false;    //as first column holds empty set, it is false

// Fill the partition table in botton up manner
for (int i = 1; i <= sum/2; i++)  {
for (int j = 1; j <= n; j++)  {
partTab[i][j] = partTab[i][j-1];
if (i >= set[j-1])
partTab[i][j] = partTab[i][j] || partTab[i - set[j-1]][j-1];
}
}
return partTab[sum/2][n];
}

int main() {
int set[] = {3, 1, 1, 2, 2, 1};
int n = 6;

if (checkPartition(set, n))
cout << "Given Set can be divided into two subsets of equal sum.";
else
cout << "Given Set can not be divided into two subsets of equal sum.";
} 

## Output

Given Set can be divided into two subsets of equal sum.
karthikeya Boyini

I love programming (: That's all I know