Subset Sum Problem in O(sum) space - GeeksforGeeks

14 Mar, 2022
Given an array of non-negative integers and a value sum, determine if there is a subset of the given set with sum equal to given sum.
Examples: 
Input : arr[] = {4, 1, 10, 12, 5, 2}, 
          sum = 9
Output : TRUE
{4, 5} is a subset with sum 9.

Input : arr[] = {1, 8, 2, 5}, 
          sum = 4
Output : FALSE 
There exists no subset with sum 4.
We have discussed a Dynamic Programming based solution in below post. Dynamic Programming | Set 25 (Subset Sum Problem)The solution discussed above requires O(n * sum) space and O(n * sum) time. We can optimize space. We create a boolean 2D array subset[2][sum+1]. Using bottom up manner we can fill up this table. The idea behind using 2 in “subset[2][sum+1]” is that for filling a row only the values from previous row is required. So alternate rows are used either making the first one as current and second as previous or the first as previous and second as current. 
C++
C
Java
Python
C#
PHP
Javascript
// Returns true if there exists a subset// with given sum in arr[]#include <iostream>using namespace std; bool isSubsetSum(int arr[], int n, int sum){       // The value of subset[i%2][j] will be true    // if there exists a subset of sum j in    // arr[0, 1, ...., i-1]    bool subset[2][sum + 1];     for (int i = 0; i <= n; i++) {        for (int j = 0; j <= sum; j++) {             // A subset with sum 0 is always possible            if (j == 0)                subset[i % 2][j] = true;             // If there exists no element no sum            // is possible            else if (i == 0)                subset[i % 2][j] = false;            else if (arr[i - 1] <= j)                subset[i % 2][j] = subset[(i + 1) % 2]             [j - arr[i - 1]] || subset[(i + 1) % 2][j];            else                subset[i % 2][j] = subset[(i + 1) % 2][j];        }    }     return subset[n % 2][sum];} // Driver codeint main(){    int arr[] = { 6, 2, 5 };    int sum = 7;    int n = sizeof(arr) / sizeof(arr[0]);    if (isSubsetSum(arr, n, sum) == true)        cout <<"There exists a subset with given sum";    else        cout <<"No subset exists with given sum";    return 0;} // This code is contributed by shivanisinghss2110
// Returns true if there exists a subset// with given sum in arr[]#include <stdio.h>#include <stdbool.h> bool isSubsetSum(int arr[], int n, int sum){    // The value of subset[i%2][j] will be true    // if there exists a subset of sum j in    // arr[0, 1, ...., i-1]    bool subset[2][sum + 1];     for (int i = 0; i <= n; i++) {        for (int j = 0; j <= sum; j++) {             // A subset with sum 0 is always possible            if (j == 0)                subset[i % 2][j] = true;             // If there exists no element no sum            // is possible            else if (i == 0)                subset[i % 2][j] = false;            else if (arr[i - 1] <= j)                subset[i % 2][j] = subset[(i + 1) % 2]             [j - arr[i - 1]] || subset[(i + 1) % 2][j];            else                subset[i % 2][j] = subset[(i + 1) % 2][j];        }    }     return subset[n % 2][sum];} // Driver codeint main(){    int arr[] = { 6, 2, 5 };    int sum = 7;    int n = sizeof(arr) / sizeof(arr[0]);    if (isSubsetSum(arr, n, sum) == true)        printf("There exists a subset with given sum");    else        printf("No subset exists with given sum");    return 0;}
// Java Program to get a subset with a// with a sum provided by the userpublic class Subset_sum {         // Returns true if there exists a subset    // with given sum in arr[]    static boolean isSubsetSum(int arr[], int n, int sum)    {        // The value of subset[i%2][j] will be true        // if there exists a subset of sum j in        // arr[0, 1, ...., i-1]        boolean subset[][] = new boolean[2][sum + 1];              for (int i = 0; i <= n; i++) {            for (int j = 0; j <= sum; j++) {                      // A subset with sum 0 is always possible                if (j == 0)                    subset[i % 2][j] = true;                      // If there exists no element no sum                // is possible                else if (i == 0)                    subset[i % 2][j] = false;                else if (arr[i - 1] <= j)                    subset[i % 2][j] = subset[(i + 1) % 2]                 [j - arr[i - 1]] || subset[(i + 1) % 2][j];                else                    subset[i % 2][j] = subset[(i + 1) % 2][j];            }        }              return subset[n % 2][sum];    }          // Driver code    public static void main(String args[])    {        int arr[] = { 1, 2, 5 };        int sum = 7;        int n = arr.length;        if (isSubsetSum(arr, n, sum) == true)            System.out.println("There exists a subset with" +                                              "given sum");        else            System.out.println("No subset exists with" +                                           "given sum");    }}// This code is contributed by Sumit Ghosh
# Returns true if there exists a subset# with given sum in arr[] def isSubsetSum(arr, n, sum):        # The value of subset[i%2][j] will be true    # if there exists a subset of sum j in    # arr[0, 1, ...., i-1]    subset = [ [False for j in range(sum + 1)] for i in range(3) ]      for i in range(n + 1):        for j in range(sum + 1):            # A subset with sum 0 is always possible            if (j == 0):                subset[i % 2][j] = True              # If there exists no element no sum            # is possible            else if (i == 0):                subset[i % 2][j] = False            else if (arr[i - 1] <= j):                subset[i % 2][j] = subset[(i + 1) % 2][j - arr[i - 1]] or subset[(i + 1)                                                                               % 2][j]            else:                subset[i % 2][j] = subset[(i + 1) % 2][j]                     return subset[n % 2][sum]  # Driver codearr = [ 6, 2, 5 ]sum = 7n = len(arr)if (isSubsetSum(arr, n, sum) == True):    print ("There exists a subset with given sum")else:    print ("No subset exists with given sum")     # This code is contributed by Sachin Bisht
// C# Program to get a subset with a// with a sum provided by the user using System; public class Subset_sum {         // Returns true if there exists a subset    // with given sum in arr[]    static bool isSubsetSum(int []arr, int n, int sum)    {        // The value of subset[i%2][j] will be true        // if there exists a subset of sum j in        // arr[0, 1, ...., i-1]        bool [,]subset = new bool[2,sum + 1];             for (int i = 0; i <= n; i++) {            for (int j = 0; j <= sum; j++) {                     // A subset with sum 0 is always possible                if (j == 0)                    subset[i % 2,j] = true;                     // If there exists no element no sum                // is possible                else if (i == 0)                    subset[i % 2,j] = false;                else if (arr[i - 1] <= j)                    subset[i % 2,j] = subset[(i + 1) % 2,j - arr[i - 1]] || subset[(i + 1) % 2,j];                else                    subset[i % 2,j] = subset[(i + 1) % 2,j];            }        }             return subset[n % 2,sum];    }         // Driver code    public static void Main()    {        int []arr = { 1, 2, 5 };        int sum = 7;        int n = arr.Length;        if (isSubsetSum(arr, n, sum) == true)            Console.WriteLine("There exists a subset with" +                                         "given sum");        else            Console.WriteLine("No subset exists with" +                                        "given sum");    }}// This code is contributed by Ryuga
<?php// Returns true if there exists a subset// with given sum in arr[] function isSubsetSum($arr, $n, $sum){    // The value of subset[i%2][j] will be    // true if there exists a subset of    // sum j in arr[0, 1, ...., i-1]    $subset[2][$sum + 1] = array();     for ($i = 0; $i <= $n; $i++)    {        for ($j = 0; $j <= $sum; $j++)        {             // A subset with sum 0 is            // always possible            if ($j == 0)                $subset[$i % 2][$j] = true;             // If there exists no element no            // sum is possible            else if ($i == 0)                $subset[$i % 2][$j] = false;            else if ($arr[$i - 1] <= $j)                $subset[$i % 2][$j] = $subset[($i + 1) % 2]                                             [$j - $arr[$i - 1]] ||                                      $subset[($i + 1) % 2][$j];            else                $subset[$i % 2][$j] = $subset[($i + 1) % 2][$j];        }    }     return $subset[$n % 2][$sum];} // Driver code$arr = array( 6, 2, 5 );$sum = 7;$n = sizeof($arr);if (isSubsetSum($arr, $n, $sum) == true)    echo ("There exists a subset with given sum");else    echo ("No subset exists with given sum"); // This code is contributed by Sach_Code?>
<script> // Javascript Program to get a subset with a// with a sum provided by the user     // Returns true if there exists a subset    // with given sum in arr[]    function isSubsetSum(arr, n, sum)    {        // The value of subset[i%2][j] will be true        // if there exists a subset of sum j in        // arr[0, 1, ...., i-1]        let subset = new Array(2);                 // Loop to create 2D array using 1D array        for (var i = 0; i < subset.length; i++) {            subset[i] = new Array(2);        }                for (let i = 0; i <= n; i++) {            for (let j = 0; j <= sum; j++) {                        // A subset with sum 0 is always possible                if (j == 0)                    subset[i % 2][j] = true;                        // If there exists no element no sum                // is possible                else if (i == 0)                    subset[i % 2][j] = false;                else if (arr[i - 1] <= j)                    subset[i % 2][j] = subset[(i + 1) % 2]                 [j - arr[i - 1]] || subset[(i + 1) % 2][j];                else                    subset[i % 2][j] = subset[(i + 1) % 2][j];            }        }                return subset[n % 2][sum];    } // driver program        let arr = [ 1, 2, 5 ];        let sum = 7;        let n = arr.length;        if (isSubsetSum(arr, n, sum) == true)            document.write("There exists a subset with" +                                              "given sum");        else            document.write("No subset exists with" +                                           "given sum"); // This code is contributed by code_hunt.</script>
There exists a subset with given sum
Another Approach: To further reduce space complexity, we create a boolean 1D array subset[sum+1]. Using bottom up manner we can fill up this table. The idea is that we can check if the sum till position “i” is possible then if the current element in the array at position j is x, then sum i+x is also possible. We traverse the sum array from back to front so that we don’t count any element twice. 
Here’s the code for the given approach:
C++
Java
Python3
C#
Javascript
#include <iostream>using namespace std; bool isPossible(int elements[], int sum, int n){    int dp[sum + 1];         // Initializing with 1 as sum 0 is    // always possible    dp[0] = 1;         // Loop to go through every element of    // the elements array    for(int i = 0; i < n; i++)    {                 // To change the values of all possible sum        // values to 1        for(int j = sum; j >= elements[i]; j--)        {            if (dp[j - elements[i]] == 1)                dp[j] = 1;        }    }         // If sum is possible then return 1    if (dp[sum] == 1)        return true;             return false;} // Driver codeint main(){    int elements[] = { 6, 2, 5 };    int n = sizeof(elements) / sizeof(elements[0]);    int sum = 7;         if (isPossible(elements, sum, n))        cout << ("YES");    else        cout << ("NO");     return 0;} // This code is contributed by Potta Lokesh
import java.io.*;import java.util.*;class GFG {    static boolean isPossible(int elements[], int sum)    {        int dp[] = new int[sum + 1];        // initializing with 1 as sum 0 is always possible        dp[0] = 1;        // loop to go through every element of the elements        // array        for (int i = 0; i < elements.length; i++) {            // to change the values of all possible sum            // values to 1            for (int j = sum; j >= elements[i]; j--) {                if (dp[j - elements[i]] == 1)                    dp[j] = 1;            }        }        // if sum is possible then return 1        if (dp[sum] == 1)            return true;        return false;    }    public static void main(String[] args) throws Exception    {        int elements[] = { 6, 2, 5 };        int sum = 7;        if (isPossible(elements, sum))            System.out.println("YES");        else            System.out.println("NO");    }}
def isPossible(elements, target):     dp = [False]*(target+1)     # initializing with 1 as sum 0 is always possible    dp[0] = True     # loop to go through every element of the elements array    for ele in elements:               # to change the value o all possible sum values to True        for j in range(target, ele - 1, -1):            if dp[j - ele]:                dp[j] = True     # If target is possible return True else False    return dp[target] # Driver codearr = [6, 2, 5]target = 7 if isPossible(arr, target):    print("YES")else:    print("NO") # The code is contributed by Arpan.
using System; class GFG {    static Boolean isPossible(int []elements, int sum)    {        int []dp = new int[sum + 1];               // initializing with 1 as sum 0 is always possible        dp[0] = 1;               // loop to go through every element of the elements        // array        for (int i = 0; i < elements.Length; i++)        {                       // to change the values of all possible sum            // values to 1            for (int j = sum; j >= elements[i]; j--) {                if (dp[j - elements[i]] == 1)                    dp[j] = 1;            }        }               // if sum is possible then return 1        if (dp[sum] == 1)            return true;        return false;    }     // Driver code    public static void Main(String[] args)    {        int []elements = { 6, 2, 5 };        int sum = 7;        if (isPossible(elements, sum))            Console.Write("YES");        else            Console.Write("NO");    }} // This code is contributed by shivanisinghss2110
<script>function isPossible(elements, sum)    {        var dp =  [sum + 1];                 // initializing with 1 as sum 0 is always possible        dp[0] = 1;                 // loop to go through every element of the elements        // array        for (var i = 0; i < elements.length; i++)        {                     // to change the values of all possible sum            // values to 1            for (var j = sum; j >= elements[i]; j--) {                if (dp[j - elements[i]] == 1)                    dp[j] = 1;            }        }                 // if sum is possible then return 1        if (dp[sum] == 1)            return true;        return false;    }        var elements = [ 6, 2, 5 ];        var sum = 7;        if (isPossible(elements, sum))            document.write("YES");        else            document.write("NO");             // This code is contributed by shivanisinghss2110</script>
YES
Time Complexity: O(N*K) where N is the number of elements in the array and K is total sum.Space Complexity: O(K)
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