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.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

// 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; 
} 

Java

// 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 

Python3

# 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  

C#

// 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 

PHP

<?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 
?> 

Output:

10101110

My Personal Notes

Suggest a Topic

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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