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 :

C++

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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














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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














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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#














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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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Output:


10101110






          
          
          
          

            

    

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