Check if is possible to get given sum from a given set of elements Last Updated : 11 Jul, 2025 Comments Improve Suggest changes Like Article Like Report 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 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. Implementation: 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; } C // C program to find if we can get given // sum using elements of given array. #include <stdio.h> // maximum size of x #define MAX 1000 // to check whether x is possible or not int ispossible[MAX]; void check(int arr[], int N) { ispossible[0] = 1; int temp; for (int i = 0; i < N; i++) { for (int j = i+1; j < N; j++) { if(arr[i] > arr[j]) { temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; } } } 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]) printf("%d is possible.",x); else printf("%d is not possible.",x); return 0; } // This code is contributed by kothavvsaakash. 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 ?> JavaScript <script> // Javascript program to find if we can get given // sum using elements of given array. // maximum size of x let MAX = 1000; // to check whether x is possible or not let ispossible = new Array(MAX); function check(arr, N) { ispossible[0] = 1; arr.sort(); for (let 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 (let j = 0; j + val < MAX; ++j) if (ispossible[j]) ispossible[j + val] = 1; } } // Driver code let arr = new Array( 2, 3 ); let N = arr.length; check(arr, N); let x = 7; if (ispossible[x]) document.write(x + " is possible."); else document.write(x + " is not possible."); // This code is contributed // by _sauraahb_jaiswal </script> Output7 is possible. Time Complexity: O(nlogn) + O(n*x) - For sorting and loop. Space Complexity: O(x) - Value of x. 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