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Subset Sum Problem in O(sum) space

  • Difficulty Level : Medium
  • Last Updated : 13 Jul, 2021

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++




// 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 code
int 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




// Java Program to get a subset with a
// with a sum provided by the user
public 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

Python




# 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
            elif (i == 0):
                subset[i % 2][j] = False
            elif (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 code
arr = [ 6, 2, 5 ]
sum = 7
n = 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#




// 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




<?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
?>

Javascript




<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>
Output
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++




#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 code
int 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

Java




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");
    }
}
Output
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)

This article is contributed by Neelesh (Neelesh_Sinha). If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 

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