Check if is possible to get given sum from a given set of elements

Given array of numbers and a integer x. Find whether it is possible or not to get x by adding elements of given array, we may pick a single element multiple times. For a given array, there can be many sum queries.

Examples:

Input : arr[] = { 2, 3}
         q[]  = {8, 7}
Output : Yes Yes
Explanation : 
2 + 2 + 2 + 2 = 8
2 + 2 + 3 = 7

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The idea is to first sort the given array and then use the concept similar to Sieve of Eratosthenes. First take a large sized array ( which is maximum size of x). Initially keep zero in all it’s indexes. Make 1 at zero index ( we can get zero whatever the array is) . Now, traverse through the whole array and make all possible values as 1.

C++

// CPP program to find if we can get given 
// sum using elements of given array. 
#include <bits/stdc++.h> 
using namespace std; 
  
// maximum size of x 
const int MAX = 1000; 
  
// to check whether x is possible or not 
int ispossible[MAX]; 
  
void check(int arr[], int N) 
{ 
    ispossible[0] = 1; 
    sort(arr, arr + N); 
  
    for (int i = 0; i < N; ++i) { 
        int val = arr[i]; 
  
        // if it is already possible 
        if (ispossible[val]) 
            continue; 
  
        // make 1 to all it's next elements 
        for (int j = 0; j + val < MAX; ++j) 
            if (ispossible[j]) 
                ispossible[j + val] = 1; 
    } 
} 
  
// Driver code 
int main() 
{ 
    int arr[] = { 2, 3 }; 
    int N = sizeof(arr) / sizeof(arr[0]); 
    check(arr, N); 
    int x = 7; 
    if (ispossible[x]) 
        cout << x << " is possible."; 
    else
        cout << x << " is not possible."; 
    return 0; 
} 

Java

// Java program to find if we can get given 
// sum using elements of given array. 
import java.util.*; 
  
class solution 
{ 
  
// maximum size of x 
int MAX = 1000; 
  
// to check whether x is possible or not 
static int []ispossible = new int[1000]; 
static void check(int[] arr, int N) 
{ 
      
    ispossible[0] = 1; 
    Arrays.sort(arr); 
  
    for (int i = 0; i < N; ++i) { 
        int val = arr[i]; 
  
        // if it is already possible 
        if (ispossible[val] == 1) 
            continue; 
  
        // make 1 to all it's next elements 
        for (int j = 0; j + val < 1000; ++j) 
            if (ispossible[j]== 1) 
                ispossible[j + val] = 1; 
    } 
} 
  
// Driver code 
public static void main(String args[]) 
{ 
    int[] arr = { 2, 3 }; 
    int N = arr.length; 
    check(arr, N); 
    int x = 7; 
    if (ispossible[x]== 1) 
        System.out.println(x+" is possible."); 
    else
        System.out.println(x+" is not possible."); 
  
} 
} 
// This code is contributed by 
// Shashank_Sharma 

Python3

# Python3 program to find if we can get given  
# sum using elements of the given array.  
def check(arr, N):  
  
    ispossible[0] = 1
    arr.sort()  
  
    for i in range(0, N):  
        val = arr[i]  
  
        # if it is already possible  
        if ispossible[val]:  
            continue
  
        # make 1 to all it's next elements  
        for j in range(0, MAX - val):  
            if ispossible[j]:  
                ispossible[j + val] = 1
  
# Driver code  
if __name__ == "__main__":  
  
    arr = [2, 3]  
    N = len(arr)  
      
    # maximum size of x  
    MAX = 1000
      
    # to check whether x is possible or not  
    ispossible = [0] * MAX
  
    check(arr, N)  
    x = 7
      
    if ispossible[x]:  
        print(x, "is possible.")  
    else: 
        print(x, "is not possible.")  
      
# This code is contributed by  
# Rituraj Jain 

C#

// C# program to find if we can get given 
// sum using elements of given array. 
using System; 
  
class GFG 
{ 
      
// to check whether x is possible or not 
static int []ispossible = new int[1000]; 
static void check(int[] arr, int N) 
{ 
      
    ispossible[0] = 1; 
    Array.Sort(arr); 
  
    for (int i = 0; i < N; ++i)  
    { 
        int val = arr[i]; 
  
        // if it is already possible 
        if (ispossible[val] == 1) 
            continue; 
  
        // make 1 to all it's next elements 
        for (int j = 0; j + val < 1000; ++j) 
            if (ispossible[j] == 1) 
                ispossible[j + val] = 1; 
    } 
} 
  
// Driver code 
public static void Main() 
{ 
    int[] arr = { 2, 3 }; 
    int N = arr.Length; 
    check(arr, N); 
    int x = 7; 
    if (ispossible[x]== 1) 
        Console.WriteLine(x + " is possible."); 
    else
        Console.WriteLine(x + " is not possible."); 
} 
} 
  
// This code is contributed by 
// Akanksha Rai 

PHP

<?php 
// PHP program to find if we can get given 
// sum using elements of given array. 
  
// maximum size of x 
$MAX = 1000; 
  
// to check whether x is possible or not 
$ispossible[$MAX] = array(); 
  
function check($arr, $N) 
{ 
    global $MAX; 
    $ispossible[0] = 1; 
    sort($arr, 0); 
  
    for ($i = 0; $i < $N; ++$i) 
    { 
        $val = $arr[$i]; 
  
        // if it is already possible 
        if ($ispossible[$val]) 
            continue; 
  
        // make 1 to all it's next elements 
        for ($j = 0; $j + $val < $MAX; ++$j) 
            if ($ispossible[$j]) 
                $ispossible[$j + $val] = 1; 
    } 
} 
  
// Driver code 
$arr = array( 2, 3 ); 
$N = sizeof($arr); 
check($arr, $N); 
$x = 7; 
if (ispossible[$x]) 
    echo $x . " is possible."; 
else
    echo $x . " is not possible."; 
  
// This code is contributed  
// by Akanksha Rai 
?> 

Output:

7 is possible.

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Improved By : Shashank_Sharma, rituraj_jain, Akanksha_Rai

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