Dynamic Programming | Set 18 (Partition problem) - GeeksforGeeks

Dynamic Programming | Set 18 (Partition problem) - GeeksforGeeks

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.

Following are the two main steps to solve this problem:
1) Calculate sum of the array. If sum is odd, there can not be two subsets with equal sum, so return false.
2) If sum of array elements is even, calculate sum/2 and find a subset of array with sum equal to sum/2.

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

Let 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])

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

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];

       }       

     }          

     return part[sum/2][n];

}   

https://shanzi.gitbooks.io/algorithm-notes/content/problem_solutions/partition_problem.html
This is a NP-Complete problem and similar to knapsack problem, it has a pseudo-polynomial DP solution for integers when the problem's scale is not too large.

Recursive Solution

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]);

}

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);

}

Also refer to Subset sum problem
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