Skip to content
Related Articles

Related Articles

Improve Article

Check if each element of an Array is the Sum of any two elements of another Array

  • Difficulty Level : Expert
  • Last Updated : 06 Jun, 2021

Given two arrays A[] and B[] consisting of N integers, the task is to check if each element of array B[] can be formed by adding any two elements of array A[]. If it is possible, then print “Yes”. Otherwise, print “No”.

Examples:

Input: A[] = {3, 5, 1, 4, 2}, B[] = {3, 4, 5, 6, 7} 
Output: Yes 
Explanation: 
B[0] = 3 = (1 + 2) = A[2] + A[4], 
B[1] = 4 = (1 + 3) = A[2] + A[0], 
B[2] = 5 = (3 + 2) = A[0] + A[4], 
B[3] = 6 = (2 + 4) = A[4] + A[3], 
B[4] = 7 = (3 + 4) = A[0] + A[3]

Input: A[] = {1, 2, 3, 4, 5}, B[] = {1, 2, 3, 4, 5} 
Output: No 
 

Approach: 
Follow the steps below to solve the problem: 



  • Store each element of B[] in a Set.
  • For each pair of indices (i, j) of the array A[], check if A[i] + A[j] is present in the set. If found to be true, remove A[i] + A[j] from the set.
  • If the set becomes empty, then print “Yes”. Otherwise, print “No”.

Below is the implementation of the above approach: 

C++




// C++ program to implement
// the above approach
#include 
using namespace std;
 
// Function to check if each element
// of B[] can be formed by adding two
// elements of array A[]
string checkPossible(int A[], int B[], int n)
{
    // Store each element of B[]
    unordered_set values;
 
    for (int i = 0; i < n; i++) {
        values.insert(B[i]);
    }
 
    // Traverse all possible pairs of array
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
 
            // If A[i] + A[j] is present in
            // the set
            if (values.find(A[i] + A[j])
                != values.end()) {
 
                // Remove A[i] + A[j] from the set
                values.erase(A[i] + A[j]);
 
                if (values.empty())
                    break;
            }
        }
    }
 
    // If set is empty
    if (values.size() == 0)
        return "Yes";
 
    // Otherwise
    else
        return "No";
}
 
// Driver Code
int main()
{
    int N = 5;
 
    int A[] = { 3, 5, 1, 4, 2 };
    int B[] = { 3, 4, 5, 6, 7 };
 
    cout << checkPossible(A, B, N);
}

Java




// Java program to implement
// the above approach
import java.io.*;
import java.util.*;
 
class GFG{
     
// Function to check if each element
// of B[] can be formed by adding two
// elements of array A[]
static String checkPossible(int A[], int B[],
                            int n)
{
     
    // Store each element of B[]
    Set values = new HashSet();
 
    for(int i = 0; i < n; i++)
    {
        values.add(B[i]);
    }
 
    // Traverse all possible pairs of array
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < n; j++)
        {
 
            // If A[i] + A[j] is present in
            // the set
            if (values.contains(A[i] + A[j]))
            {
                 
                // Remove A[i] + A[j] from the set
                values.remove(A[i] + A[j]);
 
                if (values.size() == 0)
                    break;
            }
        }
    }
 
    // If set is empty
    if (values.size() == 0)
        return "Yes";
 
    // Otherwise
    else
        return "No";
}
 
// Driver Code
public static void main(String args[])
{
    int N = 5;
    int A[] = { 3, 5, 1, 4, 2 };
    int B[] = { 3, 4, 5, 6, 7 };
     
    System.out.print(checkPossible(A, B, N));
}
}
 
// This code is contributed by offbeat

Python3




# Python3 program to implement
# the above approach
 
# Function to check if each element
# of B[] can be formed by adding two
# elements of array A[]
def checkPossible(A, B, n):
 
    # Store each element of B[]
    values = set([])
 
    for i in range (n):
        values.add(B[i])
     
    # Traverse all possible
    # pairs of array
    for i in range (n):
        for j in range (n):
 
            # If A[i] + A[j] is present in
            # the set
            if ((A[i] + A[j]) in values):
 
                # Remove A[i] + A[j] from the set
                values.remove(A[i] + A[j])
 
                if (len(values) == 0):
                    break
 
    # If set is empty
    if (len(values) == 0):
        return "Yes"
 
    # Otherwise
    else:
        return "No"
 
# Driver Code
if __name__ == "__main__":
   
  N = 5
 
  A = [3, 5, 1, 4, 2]
  B = [3, 4, 5, 6, 7]
 
  print (checkPossible(A, B, N))
 
# This code is contributed by Chitranayal

C#




// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG{
     
// Function to check if each element
// of []B can be formed by adding two
// elements of array []A
static String checkPossible(int []A, int []B,
                            int n)
{
 
  // Store each element of []B
  HashSet values = new HashSet();
 
  for(int i = 0; i < n; i++)
  {
    values.Add(B[i]);
  }
 
  // Traverse all possible pairs of array
  for(int i = 0; i < n; i++)
  {
    for(int j = 0; j < n; j++)
    {
      // If A[i] + A[j] is present in
      // the set
      if (values.Contains(A[i] + A[j]))
      {                
        // Remove A[i] + A[j] from the set
        values.Remove(A[i] + A[j]);
 
        if (values.Count == 0)
          break;
      }
    }
  }
 
  // If set is empty
  if (values.Count == 0)
    return "Yes";
 
  // Otherwise
  else
    return "No";
}
 
// Driver Code
public static void Main(String []args)
{
  int N = 5;
  int []A = {3, 5, 1, 4, 2};
  int []B = {3, 4, 5, 6, 7};
 
  Console.Write(checkPossible(A, B, N));
}
}
 
// This code is contributed by Amit Katiyar

Javascript




<script>
 
// Javascript program to implement
// the above approach
 
// Function to check if each element
// of B[] can be formed by adding two
// elements of array A[]
function checkPossible(A, B, n)
{
     
    // Store each element of B[]
    var values = new Set();
 
    for(var i = 0; i < n; i++)
    {
        values.add(B[i]);
    }
 
    // Traverse all possible pairs of array
    for(var i = 0; i < n; i++)
    {
        for(var j = 0; j < n; j++)
        {
             
            // If A[i] + A[j] is present in
            // the set
            if (values.has(A[i] + A[j]))
            {
                 
                // Remove A[i] + A[j] from the set
                values.delete(A[i] + A[j]);
 
                if (values.size == 0)
                    break;
            }
        }
    }
 
    // If set is empty
    if (values.size == 0)
        return "Yes";
 
    // Otherwise
    else
        return "No";
}
 
// Driver Code
var N = 5;
var A = [ 3, 5, 1, 4, 2 ];
var B = [ 3, 4, 5, 6, 7 ];
 
document.write(checkPossible(A, B, N));
 
// This code is contributed by itsok
 
</script>
Output: 
Yes

 

Time Complexity: O(N2
Auxiliary Space: O(N) 

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.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :