Check if sum Y can be obtained from the Array by the given operations

Given an array of integers arr[] and two integers X and Y, the task is to check if it is possible to obtain a sequence having sum X such that the sum of each element of the subsequence multiplied by an array element is equal to Y. 
Note: Here X is always less than Y.

Examples:

Input: arr[] = {1, 2, 7, 9, 10}, X = 11, Y = 13 
Output: Yes 
Explanation:
The given value of X( = 11) can be split into a sequence {9, 2} such that 9 * 1(= arr[0] + 2 * 2(= arr[1]) = 13( = Y)
Input: arr[] ={1, 3, 5, 7}, X = 27, Y = 34 
Output: No

Approach: Follow the steps below in order to solve the problem: 

• Calculate the difference between Y and X.
• For every array element arr[i], which is > 1, update (Y – X) % (arr[i] – 1).
• If the difference is reduced 0, print “Yes“. Otherwise, print “No”.

Below is the implementation of the above approach:

C++

// C++ Program to implement
// the above approach
#include <iostream>
using namespace std;

// Function to check if it is possible
// to obtain sum Y from a sequence of
// sum X from the array arr[]
void solve(int arr[], int n, int X, int Y)
{

    // Store the difference
    int diff = Y - X;

    // Iterate over the array
    for (int i = 0; i < n; i++) {

        if (arr[i] != 1) {
            diff = diff % (arr[i] - 1);
        }
    }

    // If diff reduced to 0
    if (diff == 0)
        cout << "Yes";
    else
        cout << "No";
}

// Driver Code
int main()
{
    int arr[] = { 1, 2, 7, 9, 10 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int X = 11, Y = 13;
    solve(arr, n, X, Y);

    return 0;
}

Java

// Java program to implement
// the above approach
class GFG{

// Function to check if it is possible
// to obtain sum Y from a sequence of
// sum X from the array arr[]
static void solve(int arr[], int n,
                  int X, int Y)
{

    // Store the difference
    int diff = Y - X;

    // Iterate over the array
    for(int i = 0; i < n; i++)
    {
        if (arr[i] != 1)
        {
            diff = diff % (arr[i] - 1);
        }
    }

    // If diff reduced to 0
    if (diff == 0)
        System.out.print( "Yes");
    else
        System.out.print("No");
}

// Driver Code
public static void main (String []args)
{
    int arr[] = { 1, 2, 7, 9, 10 };
    int n = arr.length;
    int X = 11, Y = 13;

    solve(arr, n, X, Y);
}
}

// This code is contributed by chitranayal

Python3

# Python3 program to implement
# the above approach

# Function to check if it is possible
# to obtain sum Y from a sequence of
# sum X from the array arr[]
def solve(arr, n, X, Y):

    # Store the difference
    diff = Y - X

    # Iterate over the array
    for i in range(n):
        if(arr[i] != 1):
            diff = diff % (arr[i] - 1)

    # If diff reduced to 0
    if(diff == 0):
        print("Yes")
    else:
        print("No")

# Driver Code
arr = [ 1, 2, 7, 9, 10 ]
n = len(arr)
X, Y = 11, 13

# Function call
solve(arr, n, X, Y)

# This code is contributed by Shivam Singh

C#

// C# program to implement
// the above approach
using System;

class GFG{

// Function to check if it is possible
// to obtain sum Y from a sequence of
// sum X from the array []arr
static void solve(int []arr, int n,
                  int X, int Y)
{

    // Store the difference
    int diff = Y - X;

    // Iterate over the array
    for(int i = 0; i < n; i++)
    {
        if (arr[i] != 1)
        {
            diff = diff % (arr[i] - 1);
        }
    }

    // If diff reduced to 0
    if (diff == 0)
        Console.Write("Yes");
    else
        Console.Write("No");
}

// Driver Code
public static void Main(String []args)
{
    int []arr = { 1, 2, 7, 9, 10 };
    int n = arr.Length;
    int X = 11, Y = 13;

    solve(arr, n, X, Y);
}
}

// This code is contributed by Amit Katiyar

Javascript

<script>

// Javascript program to implement
// the above approach

// Function to check if it is possible
// to obtain sum Y from a sequence of
// sum X from the array arr[]
function solve(arr, n, X, Y)
{

    // Store the difference
    var diff = Y - X;

    // Iterate over the array
    for(var i = 0; i < n; i++)
    {

        if (arr[i] != 1)
        {
            diff = diff % (arr[i] - 1);
        }
    }

    // If diff reduced to 0
    if (diff == 0)
        document.write("Yes");
    else
        document.write("No");
}

// Driver Code
var arr = [ 1, 2, 7, 9, 10 ];
var n = arr.length;
var X = 11, Y = 13;

solve(arr, n, X, Y);

// This code is contributed by rutvik_56

</script>

Output:

Yes

Time Complexity: O(N) 
Auxiliary Space: O(1)