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.

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

Pre-check, exit early before time consuming operation.

O(2^n) 

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.

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

Time Complexity: O(sum*n)
Auxiliary Space: O(sum*n)

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];

       }       

     }   

    /* // 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];

}

Recursive Version: O(2^n)

// 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) // this only works if all is non-negative.

     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);

}

Also check http://en.wikipedia.org/wiki/Partition_problem

The greedy algorithm[edit]

One approach to the problem, imitating the way children choose teams for a game, is the greedy algorithm, which iterates through the numbers in descending order, assigning each of them to whichever subset has the smaller sum. This works well when the numbers in the set are of about the same size as its cardinality or less. This approach has a running time of . An example of a set upon which this heuristic "breaks" is:

S = {5, 5, 4, 3, 3}

For the above input, the greedy approach would build sets S₁ = {5, 4} and S₂ = {5, 3, 3} which are not a solution to the partition problem. The solution is S₁ = {5, 5} and S₂ = {4, 3, 3}.

This greedy approach is known to give a 4/3-approximation to the optimal solution of the optimization version

Also refer to Subset sum problem
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