Skip to content
Related Articles

Related Articles

Table of Contents

Improve Article
Save Article
Like Article

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

  • Last Updated : 13 May, 2021

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.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

 

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
?>
Javascript



<script>
 
// Javascript  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) {
            var temp = n % 10;
            if (temp != 0 && temp != 1) {
                return false;
            }
            n = parseInt(n / 10);
        }
        return true;
    }
 
    // Function that checks whether the
    // binary String can be formed or not
    function formBinaryStr(n , a) {
        var flag = true;
 
        // Empty String for storing
        // the binary number
        var 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 var 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)
            document.write(s + "\n");
 
        // impossible to create binary String
        else
            document.write("-1\n");
    }
 
    // Driver code
     
 
        var a = [ 10, 1, 0, 11, 10 ];
        var N = a.length;
 
        formBinaryStr(N, a);
 
// This code contributed by Rajput-Ji
 
</script>
Output: 
10101110
 



My Personal Notes
arrow_drop_up
Previous
is_standard_layout template in C++
Next

How to run Java RMI Application
Recommended Articles
Page :
String formed with middle character of every right substring followed by left sequentially
30, Jun 20
Check whether given string can be generated after concatenating given strings
09, May 19
Check if a Palindromic String can be formed by concatenating Substrings of two given Strings
23, Jul 20
Check whether second string can be formed from characters of first string
13, Dec 17
Check whether second string can be formed from characters of first string used any number of times
22, Jul 21
Check if a palindromic string can be obtained by concatenating substrings split from same indices of two given strings
16, Oct 20
Check if the number formed by concatenating all array elements is a Harshad number or not
23, Feb 21
Longest Palindrome in a String formed by concatenating its prefix and suffix
10, Apr 20
Longest palindrome formed by concatenating and reordering strings of equal length
10, Apr 20
Check if a string can be formed from another string using given constraints
21, Jun 18
N-th character in the string made by concatenating natural numbers
03, Nov 18
Check if a string can be formed from another string by at most X circular clockwise shifts
10, May 20
Check if string S1 can be formed using repeated insertions of another string S2
20, Aug 21
Count of palindromes that can be obtained by concatenating equal length prefix and substrings
15, Jan 21
Longest palindromic string possible by concatenating strings from a given array
14, Oct 20
Check if the given string of words can be formed from words present in the dictionary
04, Jul 18
Check if a given string can be formed using characters of adjacent cells of a Matrix
29, May 20
Check if K palindromic strings can be formed from a given string
04, Jun 20
Check if given string can be formed by two other strings or their permutations
26, Jun 19
Check whether the string S1 can be made equal to S2 with the given operation
13, Nov 19
Minimum count of Inversion Pairs possible by concatenating N binary strings in any order
25, Nov 21
Check whether the binary equivalent of a number ends with given string or not
30, Jun 20
Find the starting indices of the substrings in string (S) which is made by concatenating all words from a list(L)
31, Dec 17
Lexicographically smallest string obtained after concatenating array
28, Apr 18
Article Contributed By :
https://media.geeksforgeeks.org/auth/avatar.png
souradeep
@souradeep
Vote for difficulty





Improved By :
Article Tags :
Practice Tags :
Report Issue

We use cookies to ensure you have the best browsing experience on our website. By using our site, you
acknowledge that you have read and understood our
Cookie Policy &
        Privacy Policy

Lightbox