Related Articles

Related Articles

Check if sum Y can be obtained from the Array by the given operations
  • Last Updated : 24 Aug, 2020

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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


Output: 

Yes

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up
Recommended Articles
Page :