Skip to content
Related Articles

Related Articles

Save Article
Improve Article
Save Article
Like Article

Find any permutation of Binary String of given size not present in Array

  • Last Updated : 25 Oct, 2021

Given an array arr[] of N distinct binary strings each having N characters, the task is to find any binary string having N characters such that it doesn’t occur in the given array arr[].

Example:

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.

Input: arr[] = {“10”, “01”}
Output: 00
Explanation: String “00” does not appear in array arr[]. Another valid string can be “11” which does not occur in the array arr[] as well.



Input: arr[] {“111”, “011”, “001”}
Output: 101
Explanation: String “101” does not appear in array arr[]. Another valid strings are “000”, “010”, “100”, and “110”.

Naive Approach: The given problem can be solved by generating all the binary strings having N bits and returning the 1st string such that it does not occur in the given array arr[].

Time Complexity: O(2N * N2)
Auxiliary Space: O(N)

Efficient Approach: The given problem can be optimized by using a method similar to Cantor’s Diagonal Argument. It can be observed that the resulting string must have at least one bit that is different from all the strings in arr[]. Therefore, the resulting binary string can be constructed by taking the complement of the 1st element of the 1st string as the 1st element, the complement of the 2nd element of the 2nd string as the 2nd element, and so on. Below are the steps to follow:

  • Create a string ans, which stores the resulting string. Initially, it is empty.
  • Iterate through the given strings in a diagonal order, i.e, 1st element of 1st string, 2nd element of 2nd string, and so on, and append the compliment of the value at the current index into the string ans.
  • The string stored in ans after iterating through the complete array arr[] is one of the valid strings required.

Below is the implementation of the above approach:

C++




// C++ Program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find a binary string of
// N bits that does not occur in the
// given array arr[]
string findString(vector<string>& arr, int N)
{
    // Stores the resultant string
    string ans = "";
 
    // Loop to iterate over all the given
    // strings in a diagonal order
    for (int i = 0; i < N; i++) {
 
        // Append the complement of element
        // at current index into ans
        ans += arr[i][i] == '0' ? '1' : '0';
    }
 
    // Return Answer
    return ans;
}
 
// Driver code
int main()
{
    vector<string> arr{ "111", "011", "001" };
    int N = arr.size();
 
    cout << findString(arr, N);
 
    return 0;
}

Java




// Java Program for the above approach
 
import java.io.*;
 
class GFG {
   
      // Function to find a binary string of
    // N bits that does not occur in the
    // givrn array arr[]
    static String findString(String arr[], int N)
    {
        // Stores the resultant string
        String ans = "";
 
        // Loop to iterate over all the given
        // strings in a diagonal order
        for (int i = 0; i < N; i++) {
 
            // Append the complement of element
            // at current index into ans
            ans += arr[i].charAt(i) == '0' ? '1' : '0';
        }
 
        // Return Answer
        return ans;
    }
 
    // Driver code
    public static void main (String[] args) {
           String arr[] = { "111", "011", "001" };
        int N = arr.length;
 
        System.out.println(findString(arr, N));
    }
}
 
// This code is contributed by Dharanendra L V.

Python3




# Python 3 Program for the above approach
 
# Function to find a binary string of
# N bits that does not occur in the
# givrn array arr[]
def findString(arr, N):
   
    # Stores the resultant string
    ans = ""
 
    # Loop to iterate over all the given
    # strings in a diagonal order
    for i in range(N):
       
        # Append the complement of element
        # at current index into ans
        ans += '1' if arr[i][i] == '0' else '0'
 
    # Return Answer
    return ans
 
# Driver code
if __name__ == '__main__':
    arr = ["111", "011", "001"]
    N = len(arr)
 
    print(findString(arr, N))
 
    # This code is contributed by bgangwar59.

C#




// C# Program for the above approach
using System;
 
class GFG {
 
    // Function to find a binary string of
    // N bits that does not occur in the
    // givrn array arr[]
    static string findString(string[] arr, int N)
    {
       
        // Stores the resultant string
        string ans = "";
 
        // Loop to iterate over all the given
        // strings in a diagonal order
        for (int i = 0; i < N; i++) {
 
            // Append the complement of element
            // at current index into ans
            ans += arr[i][i] == '0' ? '1' : '0';
        }
 
        // Return Answer
        return ans;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        string[] arr = { "111", "011", "001" };
        int N = arr.Length;
 
        Console.WriteLine(findString(arr, N));
    }
}
 
// This code is contributed by ukasp.

Javascript




<script>
        // JavaScript Program to implement
        // the above approach
 
        // Function to find a binary string of
        // N bits that does not occur in the
        // givrn array arr[]
        function findString(arr, N)
        {
         
            // Stores the resultant string
            let ans = "";
 
            // Loop to iterate over all the given
            // strings in a diagonal order
            for (let i = 0; i < N; i++) {
 
                // Append the complement of element
                // at current index into ans
                ans += arr[i][i] == '0' ? '1' : '0';
            }
 
            // Return Answer
            return ans;
        }
 
        // Driver code
        let arr = ["111", "011", "001"];
        let N = arr.length;
 
        document.write(findString(arr, N));
 
     // This code is contributed by Potta Lokesh
 
    </script>
Output: 
000

 

Time Complexity: O(N)
Auxiliary Space: O(1)




My Personal Notes arrow_drop_up
Recommended Articles
Page :