Check whether a binary string can be formed by concatenating given N numbers sequentially

Given a sequence of ‘n’ numbers (without leading zeros), the task is to find whether it is possible to create a binary string by concatenating these numbers sequentially.
If possible, then print the binary string formed, otherwise print “-1”.

Examples :

Input: arr[] = {10, 11, 1, 0, 10}
Output: 10111010
All the numbers contain the digits ‘1’ and ‘0’ only. So it is possible to form a binary string by concatenating
these numbers sequentially which is 10111010.

Input: arr[] = {1, 2, 11, 10}
Output: -1
One of the numbers contains the digit ‘2’ which cannot be a part of any binary string.
So, the output is -1.

Approach: The main observation is that we can only concatenate those numbers which contain the digits ‘1’ and ‘0’ only. Otherwise, it is impossible to form a binary string.



Below is the implementation of the above approach :

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function that returns false if
// the number passed as argument contains
// digit(s) other than '0' or '1'
bool isBinary(int n)
{
    while (n != 0) {
        int temp = n % 10;
        if (temp != 0 && temp != 1) {
            return false;
        }
        n = n / 10;
    }
    return true;
}
  
//Function that checks whether the 
//binary string can be formed or not
void formBinaryStr(int n, int a[])
{
    bool flag = true;
  
    // Empty string for storing
    // the binary number
    string s = "";
  
    for (int i = 0; i < n; i++) {
  
        // check if a[i] can be a
        // part of the binary string
        if (isBinary(a[i]))
  
            // Conversion of int into string
            s += to_string(a[i]);
        else {
  
            // if a[i] can't be a part
            // then break the loop
            flag = false;
            break;
        }
    }
  
    // possible to create binary string
    if (flag)
        cout << s << "\n";
  
    // impossible to create binary string
    else
        cout << "-1\n";
}
  
// Driver code
int main()
{
  
    int a[] = { 10, 1, 0, 11, 10 };
    int N = sizeof(a) / sizeof(a[0]);
  
    formBinaryStr(N, a);
  
    return 0;
}
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// Java  implementation of the approach
import java.util.*;
class Solution
{
// Function that returns false if
// the number passed as argument contains
// digit(s) other than '0' or '1'
static boolean isBinary(int n)
{
    while (n != 0) {
        int temp = n % 10;
        if (temp != 0 && temp != 1) {
            return false;
        }
        n = n / 10;
    }
    return true;
}
  
//Function that checks whether the 
//binary String can be formed or not
static void formBinaryStr(int n, int a[])
{
    boolean flag = true;
  
    // Empty String for storing
    // the binary number
    String s = "";
  
    for (int i = 0; i < n; i++) {
  
        // check if a[i] can be a
        // part of the binary String
        if (isBinary(a[i]))
  
            // Conversion of int into String
            s += ""+a[i];
        else {
  
            // if a[i] can't be a part
            // then break the loop
            flag = false;
            break;
        }
    }
  
    // possible to create binary String
    if (flag)
        System.out.print( s + "\n");
  
    // impossible to create binary String
    else
        System.out.print( "-1\n");
}
  
// Driver code
public static void main(String args[])
{
  
    int a[] = { 10, 1, 0, 11, 10 };
    int N = a.length;
  
    formBinaryStr(N, a);
}
}
//contributed by Arnab Kundu
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# Python3 implementation of the approach 
  
# Function that returns false if the 
# number passed as argument contains 
# digit(s) other than '0' or '1' 
def isBinary(n): 
  
    while n != 0
        temp = n % 10
        if temp != 0 and temp != 1
            return False
          
        n = n // 10
      
    return True
  
# Function that checks whether the 
# binary string can be formed or not 
def formBinaryStr(n, a):
  
    flag = True
  
    # Empty string for storing 
    # the binary number 
    s = "" 
    for i in range(0, n): 
  
        # check if a[i] can be a 
        # part of the binary string 
        if isBinary(a[i]) == True
              
            # Conversion of int into string 
            s += str(a[i]) 
          
        else
            # if a[i] can't be a part 
            # then break the loop 
            flag = False
            break
  
    # possible to create binary string 
    if flag == True:
        print(s) 
  
    # impossible to create binary string 
    else:
        cout << "-1\n"
  
# Driver code 
if __name__ == "__main__"
  
    a = [10, 1, 0, 11, 10
    N = len(a) 
  
    formBinaryStr(N, a) 
  
# This code is contributed by Rituraj Jain 
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// C#  implementation of the approach
using System;
  
public class Solution
{
// Function that returns false if 
// the number passed as argument contains 
// digit(s) other than '0' or '1' 
public static bool isBinary(int n)
{
    while (n != 0)
    {
        int temp = n % 10;
        if (temp != 0 && temp != 1)
        {
            return false;
        }
        n = n / 10;
    }
    return true;
}
  
//Function that checks whether the  
//binary String can be formed or not 
public static void formBinaryStr(int n, int[] a)
{
    bool flag = true;
  
    // Empty String for storing 
    // the binary number 
    string s = "";
  
    for (int i = 0; i < n; i++)
    {
  
        // check if a[i] can be a 
        // part of the binary String 
        if (isBinary(a[i]))
        {
  
            // Conversion of int into String 
            s += "" + a[i];
        }
        else
        {
  
            // if a[i] can't be a part 
            // then break the loop 
            flag = false;
            break;
        }
    }
  
    // possible to create binary String 
    if (flag)
    {
        Console.Write(s + "\n");
    }
  
    // impossible to create binary String 
    else
    {
        Console.Write("-1\n");
    }
}
  
// Driver code 
public static void Main(string[] args)
{
  
    int[] a = new int[] {10, 1, 0, 11, 10};
    int N = a.Length;
  
    formBinaryStr(N, a);
}
}
  
// This code is contributed by Shrikant13
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<?php
// PHP implementation of the approach
  
// Function that returns false if the
// number passed as argument contains
// digit(s) other than '0' or '1'
function isBinary($n)
{
    while ($n != 0) 
    {
        $temp = $n % 10;
        if ($temp != 0 && $temp != 1) 
        {
            return false;
        }
        $n = intval($n / 10);
    }
    return true;
}
  
// Function that checks whether the 
// binary string can be formed or not
function formBinaryStr($n, &$a)
{
    $flag = true;
  
    // Empty string for storing
    // the binary number
    $s = "";
  
    for ($i = 0; $i < $n; $i++)
    {
  
        // check if a[i] can be a
        // part of the binary string
        if (isBinary($a[$i]))
  
            // Conversion of int into string
            $s = $s.strval($a[$i]);
        else 
        {
  
            // if a[i] can't be a part
            // then break the loop
            $flag = false;
            break;
        }
    }
  
    // possible to create binary string
    if ($flag)
        echo $s . "\n";
  
    // impossible to create binary string
    else
        echo "-1\n";
}
  
// Driver code
$a = array( 10, 1, 0, 11, 10 );
$N = sizeof($a) / sizeof($a[0]);
  
formBinaryStr($N, $a);
  
// This code is contributed by ita_c
?>
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Output:
10101110

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