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