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++
// 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
// 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
# 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#
// 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
Output:
10101110
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