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.
We strongly recommend that you click here and practice it, before moving on to the solution.
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.
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.
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])
C/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; }
Java
// A recursive Java solution for partition problem import java.io.*; class Partition { // A utility function that returns true if there is a // subset of arr[] with sun equal to given sum static boolean 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 static boolean findPartition (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 function to check for above function*/ public static void main (String[] args) { int arr[] = {3, 1, 5, 9, 12}; int n = arr.length; if (findPartition(arr, n) == true) System.out.println("Can be divided into two "+ "subsets of equal sum"); else System.out.println("Can not be divided into " + "two subsets of equal sum"); } } /* This code is contributed by Devesh Agrawal */
Python3
# A recursive Python3 program for # partition problem # A utility function that returns # true if there is a subset of # arr[] with sun equal to given sum def isSubsetSum (arr, n, sum): # Base Cases if sum == 0: return True if n == 0 and 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) or isSubsetSum (arr, n-1, sum-arr[n-1]) # Returns true if arr[] can be partitioned in two # subsets of equal sum, otherwise false def findPartion (arr, n): # Calculate sum of the elements in array sum = 0 for i in range(0, n): 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 arr = [3, 1, 5, 9, 12] n = len(arr) if findPartion(arr, n) == True: print ("Can be divided into two subsets of equal sum") else: print ("Can not be divided into two subsets of equal sum") # This code is contributed by shreyanshi_arun.
C#
// A recursive C# solution for partition problem using System; class GFG { // A utility function that returns true if there is a // subset of arr[] with sun equal to given sum static 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 static bool findPartition (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 function public static void Main () { int []arr = {3, 1, 5, 9, 12}; int n = arr.Length; if (findPartition(arr, n) == true) Console.Write("Can be divided into two "+ "subsets of equal sum"); else Console.Write("Can not be divided into " + "two subsets of equal sum"); } } // This code is contributed by Sam007
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/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; }
Java
// A dynamic programming based Java program for partition problem import java.io.*; class Partition { // Returns true if arr[] can be partitioned in two subsets of // equal sum, otherwise false static boolean findPartition (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; boolean part[][]=new boolean[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 function to check for above function*/ public static void main (String[] args) { int arr[] = {3, 1, 1, 2, 2,1}; int n = arr.length; if (findPartition(arr, n) == true) System.out.println("Can be divided into two " "subsets of equal sum"); else System.out.println("Can not be divided into" " two subsets of equal sum"); } } /* This code is contributed by Devesh Agrawal */
C#
// A dynamic programming based C# program // for partition problem using System; class GFG { // Returns true if arr[] can be partitioned // in two subsets of equal sum, otherwise // false static bool findPartition (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=new bool[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 public static void Main () { int []arr = {3, 1, 1, 2, 2,1}; int n = arr.Length; if (findPartition(arr, n) == true) Console.Write("Can be divided" + " into two subsets of" + " equal sum"); else Console.Write("Can not be " + "divided into two subsets" + " of equal sum"); } } // This code is contributed by Sam007.
Output:
Can be divided into two subsets of equal sum
Following diagram shows the values in partition table.
Time Complexity: O(sum*n)
Auxiliary Space: O(sum*n)
Please note that this solution will not be feasible for arrays with big sum.
References:
http://en.wikipedia.org/wiki/Partition_problem
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Recommended Posts:
- Print equal sum sets of array (Partition problem)
- Dynamic Programming | Set 19 (Word Wrap Problem)
- Dynamic Programming | Set 25 (Subset Sum Problem)
- Partition a set into two subsets such that the difference of subset sums is minimum
- Dynamic Programming | Set 21 (Variations of LIS)
- Dynamic Programming | Set 17 (Palindrome Partitioning)
- Dynamic Programming | Wildcard Pattern Matching | Linear Time and Constant Space
- Maximize array elements upto given number
- Multistage Graph (Shortest Path)
- Moser-de Bruijn Sequence
Writing code in comment? Please use ide.geeksforgeeks.org, generate link and share the link here.