Partition problem | DP-18

• Difficulty Level : Medium
• Last Updated : 07 Jul, 2021

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 the 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 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])

Below is the implementation of the above code:

C++

 // A recursive C++ program for partition problem#include using namespace std; // A utility function that returns true if there is// a subset of arr[] with sun equal to given sumbool 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 falsebool 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 codeint main(){    int arr[] = { 3, 1, 5, 9, 12 };    int n = sizeof(arr) / sizeof(arr);       // Function call    if (findPartiion(arr, n) == true)        cout << "Can be divided into two subsets "                "of equal sum";    else        cout << "Can not be divided into two subsets"                " of equal sum";    return 0;} // This code is contributed by rathbhupendra

C

 // A recursive C program for partition problem#include #include  // A utility function that returns true if there is// a subset of arr[] with sun equal to given sumbool 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 falsebool 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 codeint main(){    int arr[] = { 3, 1, 5, 9, 12 };    int n = sizeof(arr) / sizeof(arr);       // Function call    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 problemimport 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 code    public static void main(String[] args)    {         int arr[] = { 3, 1, 5, 9, 12 };        int n = arr.length;               // Function call        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 codearr = [3, 1, 5, 9, 12]n = len(arr) # Function callif 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 problemusing 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 code    public static void Main()    {         int[] arr = { 3, 1, 5, 9, 12 };        int n = arr.Length;               // Function call        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

PHP

 \$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 falsefunction findPartiion (\$arr, \$n){    // Calculate sum of the elements    // in array    \$sum = 0;    for (\$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 Code\$arr = array(3, 1, 5, 9, 12);\$n = count(\$arr); // Function callif (findPartiion(\$arr, \$n) == true)    echo "Can be divided into two subsets of equal sum";else    echo "Can not be divided into two subsets of equal sum"; // This code is contributed by rathbhupendra?>

Javascript


Output
Can be divided into two subsets of equal sum

Time Complexity: O(2^n) In the 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 + 1)*(n+1). And we can construct the solution in a bottom-up manner such that every filled entry has the following property

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

C++

 // A Dynamic Programming based// C++ program to partition problem#include using namespace std; // Returns true if arr[] can be partitioned// in two subsets of equal sum, otherwise falsebool findPartiion(int arr[], int n){    int sum = 0;    int i, j;     // Calculate sum 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[i] = true;     // initialize leftmost column,    // except part, as 0    for (i = 1; i <= sum / 2; i++)        part[i] = false;     // Fill the partition table in bottom 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++)        cout<

C

 // A Dynamic Programming based C program to partition// problem#include  // Returns true if arr[] can be partitioned in two subsets// of equal sum, otherwise falsebool findPartiion(int arr[], int n){    int sum = 0;    int i, j;     // Calculate sum 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[i] = true;     // initialize leftmost column, except part, as 0    for (i = 1; i <= sum / 2; i++)        part[i] = false;     // Fill the partition table in bottom 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 codeint main(){    int arr[] = { 3, 1, 1, 2, 2, 1 };    int n = sizeof(arr) / sizeof(arr);       // Function call    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// problemimport 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;         // Calculate sum 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[i] = true;         // initialize leftmost column, except part, as        // 0        for (i = 1; i <= sum / 2; i++)            part[i] = false;         // Fill the partition table in bottom 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 code    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 */

Python3

 # Dynamic Programming based python# program to partition problem # Returns true if arr[] can be# partitioned in two subsets of# equal sum, otherwise false  def findPartition(arr, n):    sum = 0    i, j = 0, 0     # calculate sum of all elements    for i in range(n):        sum += arr[i]     if sum % 2 != 0:        return false     part = [[True for i in range(n + 1)]            for j in range(sum // 2 + 1)]     # initialize top row as true    for i in range(0, n + 1):        part[i] = True     # initialize leftmost column,    # except part, as 0    for i in range(1, sum // 2 + 1):        part[i] = False     # fill the partition table in    # bottom up manner    for i in range(1, sum // 2 + 1):         for j in range(1, n + 1):            part[i][j] = part[i][j - 1]             if i >= arr[j - 1]:                part[i][j] = (part[i][j] or                              part[i - arr[j - 1]][j - 1])     return part[sum // 2][n]  # Driver Codearr = [3, 1, 1, 2, 2, 1]n = len(arr) # Function callif findPartition(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 mohit kumar 29

C#

 // A dynamic programming based C# program// for partition problemusing 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;         // Calculate sum 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, as 0        for (i = 1; i <= sum / 2; i++)            part[i, 0] = false;         // Fill the partition table in bottom        // 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 - 1]                          || 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 code    public static void Main()    {        int[] arr = { 3, 1, 1, 2, 2, 1 };        int n = arr.Length;         // Function call        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.

Javascript


Output
Can be divided into two subsets of equal sum

Following diagram shows the values in the 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.

Dynamic Programming Solution (Space Complexity Optimized)

Instead of creating a 2-D array of size (sum/2 + 1)*(n  + 1), we can solve this problem using an array of size (sum/2 + 1 ) only.

part[j] = true if there is a subset with sum equal to j, otherwise false.

Below is the implementation of the above approach:

C++

 // A Dynamic Programming based// C++ program to partition problem#include using namespace std; // Returns true if arr[] can be partitioned// in two subsets of equal sum, otherwise falsebool findPartiion(int arr[], int n){    int sum = 0;    int i, j;     // Calculate sum of all elements    for (i = 0; i < n; i++)        sum += arr[i];     if (sum % 2 != 0)        return false;     bool part[sum / 2 + 1];     // Initialize the part array    // as 0    for (i = 0; i <= sum / 2; i++) {        part[i] = 0;    }     // Fill the partition table in bottom up manner     for (i = 0; i < n; i++) {        // the element to be included        // in the sum cannot be        // greater than the sum        for (j = sum / 2; j >= arr[i];             j--) { // check if sum - arr[i]            // could be formed            // from a subset            // using elements            // before index i            if (part[j - arr[i]] == 1 || j == arr[i])                part[j] = 1;        }    }     return part[sum / 2];} // Driver Codeint main(){    int arr[] = { 1, 3, 3, 2, 3, 2 };    int n = sizeof(arr) / sizeof(arr);       // Function call    if (findPartiion(arr, n) == true)        cout << "Can be divided into two subsets of equal "                "sum";    else        cout << "Can not be divided into"             << " two subsets of equal sum";    return 0;}

Java

 // A Dynamic Programming based// Java program to partition problemimport java.io.*; class GFG{     // Returns true if arr[] can be partitioned// in two subsets of equal sum, otherwise falsepublic static boolean findPartiion(int arr[], int n){    int sum = 0;    int i, j;     // Calculate sum 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];     // Initialize the part array    // as 0    for(i = 0; i <= sum / 2; i++)    {        part[i] = false;    }     // Fill the partition table in    // bottom up manner    for(i = 0; i < n; i++)    {                 // The element to be included        // in the sum cannot be        // greater than the sum        for(j = sum / 2; j >= arr[i]; j--)        {                         // Check if sum - arr[i] could be            // formed from a subset using elements            // before index i            if (part[j - arr[i]] == true || j == arr[i])                part[j] = true;        }    }    return part[sum / 2];} // Driver codepublic static void main(String[] args){    int arr[] = { 1, 3, 3, 2, 3, 2 };    int n = 6;     // Function call    if (findPartiion(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 RohitOberoi

Python3

 # A Dynamic Programming based# Python3 program to partition problem # Returns true if arr[] can be partitioned# in two subsets of equal sum, otherwise falsedef findPartiion(arr, n) :    Sum = 0     # Calculate sum of all elements    for i in range(n) :        Sum += arr[i]    if (Sum % 2 != 0) :        return 0    part =  * ((Sum // 2) + 1)     # Initialize the part array as 0    for i in range((Sum // 2) + 1) :        part[i] = 0     # Fill the partition table in bottom up manner    for i in range(n) :               # the element to be included        # in the sum cannot be        # greater than the sum        for j in range(Sum // 2, arr[i] - 1, -1) :                       # check if sum - arr[i]            # could be formed            # from a subset            # using elements            # before index i            if (part[j - arr[i]] == 1 or j == arr[i]) :                part[j] = 1     return part[Sum // 2] # Drive code arr = [ 1, 3, 3, 2, 3, 2 ]n = len(arr) # Function callif (findPartiion(arr, n) == 1) :    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 divyeshrabadiya07

C#

 // A Dynamic Programming based// C# program to partition problemusing System;class GFG{         // Returns true if arr[] can be partitioned    // in two subsets of equal sum, otherwise false    static bool findPartiion(int[] arr, int n)    {        int sum = 0;        int i, j;              // Calculate sum 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];              // Initialize the part array        // as 0        for(i = 0; i <= sum / 2; i++)        {            part[i] = false;        }              // Fill the partition table in        // bottom up manner        for(i = 0; i < n; i++)        {                          // The element to be included            // in the sum cannot be            // greater than the sum            for(j = sum / 2; j >= arr[i]; j--)            {                                  // Check if sum - arr[i] could be                // formed from a subset using elements                // before index i                if (part[j - arr[i]] == true || j == arr[i])                    part[j] = true;            }        }        return part[sum / 2];    }     // Driver code  static void Main()  {    int[] arr = { 1, 3, 3, 2, 3, 2 };    int n = 6;      // Function call    if (findPartiion(arr, n) == true)        Console.WriteLine("Can be divided into two " +                           "subsets of equal sum");    else        Console.WriteLine("Can not be divided into " +                           "two subsets of equal sum");  }} // This code is contributed by divyesh072019

Javascript


Output
Can be divided into two subsets of equal sum

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

Please note that this solution will not be feasible for arrays with big sum.
References:
http://en.wikipedia.org/wiki/Partition_problem