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Check if an array element is concatenation of two elements from another array
  • Last Updated : 04 Mar, 2021

Given two arrays arr[] and brr[] consisting of N and M positive integers respectively, the task is to find all the elements from the array brr[] which are equal to the concatenation of any two elements from the array arr[]. If no such element exists, then print “-1”.

Examples:

Input: arr[] = {2, 34, 4, 5}, brr[] = {26, 24, 345, 4, 22}
Output: 24 345 22
Explanation:
The elements from the array brr[] which are concatenation of any two elements from the array arr[] are: 

  1. 24 is concatenation of 2 and 4.
  2. 345 is concatenation of 34 and 5.
  3. 22 is concatenation of 2 and 2.

Input: arr[] = {1, 2, 3}, brr[] = {1, 23}
Output: 23

Naive Approach: The simplest approach to solve the problem is to generate all possible pairs from the given array and check if the concatenation of pairs of elements from the array arr[] is present in the array brr[] or not. If found to be true, then print the concatenated number formed. 



Time Complexity: O(M * N2)
Auxiliary Space: O(N2)

Efficient Approach: The above approach can be optimized by checking for each element in the array brr[], whether brr[i] can be divided into 2 parts left and right such that both the parts exists in the array arr[].

Consider a number, b[i] = 2365
All possible combinations of left and right are:
Left              Right
2                  365
23                65
236              5

Follow the steps below to solve the problem:

  • Initialize a HashMap M and store all elements present in the array arr[].
  • Traverse the array brr[] and perform the following steps:
    • Generate all possible combinations of left and right parts, such that their concatenation results to brr[i].
    • If both the left and the right parts are present in Map M in one of the above combinations, then print the value of brr[i]. Otherwise, continue to the next iteration.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find elements present in
// the array b[] which are concatenation
// of any pair of elements in the array a[]
void findConcatenatedNumbers(vector<int> a,
                             vector<int> b)
{
    // Stores if there doesn't any such
    // element in the array brr[]
    bool ans = true;
 
    // Stored the size of both the arrays
    int n1 = a.size();
    int n2 = b.size();
 
    // Store the presence of an element
    // of array a[]
    unordered_map<int, int> cnt;
 
    // Traverse the array a[]
    for (int i = 0; i < n1; i++) {
        cnt[a[i]] = 1;
    }
 
    // Traverse the array b[]
    for (int i = 0; i < n2; i++) {
 
        int left = b[i];
        int right = 0;
        int mul = 1;
 
        // Traverse over all possible
        // concatenations of b[i]
        while (left > 9) {
 
            // Update right and left parts
            right += (left % 10) * mul;
            left /= 10;
            mul *= 10;
 
            // Check if both left and right
            // parts are present in a[]
            if (cnt[left] == 1
                && cnt[right] == 1) {
                ans = false;
                cout << b[i] << " ";
            }
        }
    }
 
    if (ans)
        cout << "-1";
}
 
// Driver Code
int main()
{
    vector<int> a = { 2, 34, 4, 5 };
    vector<int> b = { 26, 24, 345, 4, 22 };
    findConcatenatedNumbers(a, b);
 
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
import java.util.*;
class GFG
{
 
  // Function to find elements present in
  // the array b[] which are concatenation
  // of any pair of elements in the array a[]
  static void findConcatenatedNumbers(int[] a,
                                      int[] b)
  {
    // Stores if there doesn't any such
    // element in the array brr[]
    boolean ans = true;
 
    // Stored the size of both the arrays
    int n1 = a.length;
    int n2 = b.length;
 
    // Store the presence of an element
    // of array a[]
    int cnt[] = new int[100000];
 
    // Traverse the array
    for (int i = 0; i < n1; i++)
    {
      cnt[a[i]] = 1;
    }
 
    // Traverse the array b[]
    for (int i = 0; i < n2; i++) {
 
      int left = b[i];
      int right = 0;
      int mul = 1;
 
      // Traverse over all possible
      // concatenations of b[i]
      while (left > 9) {
 
        // Update right and left parts
        right += (left % 10) * mul;
        left /= 10;
        mul *= 10;
 
        // Check if both left and right
        // parts are present in a[]
        if (cnt[left] == 1
            && cnt[right] == 1) {
          ans = false;
          System.out.print(b[i] + " ");
        }
      }
    }
 
    if (ans)
      System.out.print("-1");
  }
 
 
  // Driver code
  public static void main(String[] args)
  {
    int[] a = { 2, 34, 4, 5 };
    int[] b = { 26, 24, 345, 4, 22 };
    findConcatenatedNumbers(a, b);
  }
}
 
// This code is contributed by sanjoy_62.

Python3




# Python3 program for the above approach
from collections import defaultdict
 
# Function to find elements present in
# the array b[] which are concatenation
# of any pair of elements in the array a[]
def findConcatenatedNumbers(a, b):
     
    # Stores if there doesn't any such
    # element in the array brr[]
    ans = True
 
    # Stored the size of both the arrays
    n1 = len(a)
    n2 = len(b)
 
    # Store the presence of an element
    # of array a[]
    cnt = defaultdict(int)
 
    # Traverse the array a[]
    for i in range(n1):
        cnt[a[i]] = 1
 
    # Traverse the array b[]
    for i in range(n2):
        left = b[i]
        right = 0
        mul = 1
 
        # Traverse over all possible
        # concatenations of b[i]
        while (left > 9):
 
            # Update right and left parts
            right += (left % 10) * mul
            left //= 10
            mul *= 10
 
            # Check if both left and right
            # parts are present in a[]
            if (cnt[left] == 1 and cnt[right] == 1):
                ans = False
                print(b[i], end = " ")
 
    if (ans):
        print("-1")
 
# Driver Code
if __name__ == "__main__":
 
    a = [ 2, 34, 4, 5 ]
    b = [ 26, 24, 345, 4, 22 ]
     
    findConcatenatedNumbers(a, b)
 
# This code is contributed by chitranayal

C#




// C# program for the above approach
using System;
class GFG
{
 
  // Function to find elements present in
  // the array b[] which are concatenation
  // of any pair of elements in the array a[]
  static void findConcatenatedNumbers(int[] a,
                                      int[] b)
  {
     
    // Stores if there doesn't any such
    // element in the array brr[]
    bool ans = true;
 
    // Stored the size of both the arrays
    int n1 = a.Length;
    int n2 = b.Length;
 
    // Store the presence of an element
    // of array a[]
    int []cnt = new int[100000];
 
    // Traverse the array
    for (int i = 0; i < n1; i++)
    {
      cnt[a[i]] = 1;
    }
 
    // Traverse the array b[]
    for (int i = 0; i < n2; i++) {
 
      int left = b[i];
      int right = 0;
      int mul = 1;
 
      // Traverse over all possible
      // concatenations of b[i]
      while (left > 9) {
 
        // Update right and left parts
        right += (left % 10) * mul;
        left /= 10;
        mul *= 10;
 
        // Check if both left and right
        // parts are present in a[]
        if (cnt[left] == 1
            && cnt[right] == 1) {
          ans = false;
          Console.Write(b[i] + " ");
        }
      }
    }
 
    if (ans)
      Console.Write("-1");
  }
 
  // Driver code
  public static void Main(String[] args)
  {
    int[] a = { 2, 34, 4, 5 };
    int[] b = { 26, 24, 345, 4, 22 };
    findConcatenatedNumbers(a, b);
  }
}
 
// This code is contributed by shivani
Output: 
24 345 22

 

Time Complexity: O(M*log(X)), where X is the largest element in the array brr[].
Auxiliary Space: O(N)

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