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# 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

 ``
Output:
`Yes`

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

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