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

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# 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)

Form an array of distinct elements with each element as sum of an element from each array

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

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