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Count of binary strings of given length consisting of at least one 1

  • Difficulty Level : Medium
  • Last Updated : 03 Aug, 2021

Given an integer N, the task is to print the number of binary strings of length N which at least one ‘1’.

Examples:  

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Input:
Output:
Explanation: 
“01”, “10” and “11” are the possible strings



Input:
Output:
Explanation: 
“001”, “011”, “010”, “100”, “101”, “110” and “111” are the possible strings 
 

Approach: 
We can observe that: 

Only one string of length N does not contain any 1, the one filled with only 0’s. 
Since 2N strings are possible of length N, the required answer is 2N – 1
 

Follow the steps below to solve the problem: 

  • Initialize X = 1.
  • Compute upto 2N by performing bitwise left shift on X, N-1 times.
  • Finally, print X – 1 as the required answer.

Below is the implementation of our approach: 

C++




// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return
// the count of strings
long count_strings(long n)
{
    int x = 1;
 
    // Calculate pow(2, n)
    for (int i = 1; i < n; i++) {
        x = (1 << x);
    }
 
    // Return pow(2, n) - 1
    return x - 1;
}
 
// Driver Code
int main()
{
    long n = 3;
 
    cout << count_strings(n);
 
    return 0;
}

Java




// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
 
// Function to return
// the count of Strings
static long count_Strings(long n)
{
    int x = 1;
 
    // Calculate Math.pow(2, n)
    for(int i = 1; i < n; i++)
    {
       x = (1 << x);
    }
 
    // Return Math.pow(2, n) - 1
    return x - 1;
}
 
// Driver Code
public static void main(String[] args)
{
    long n = 3;
 
    System.out.print(count_Strings(n));
}
}
 
// This code is contributed by Amit Katiyar

Python3




# Python3 program to implement
# the above approach
 
# Function to return
# the count of Strings
def count_Strings(n):
     
    x = 1;
 
    # Calculate pow(2, n)
    for i in range(1, n):
        x = (1 << x);
 
    # Return pow(2, n) - 1
    return x - 1;
 
# Driver Code
if __name__ == '__main__':
     
    n = 3;
 
    print(count_Strings(n));
 
# This code is contributed by Princi Singh

C#




// C# program to implement
// the above approach
using System;
 
class GFG{
 
// Function to return
// the count of Strings
static long count_Strings(long n)
{
    int x = 1;
 
    // Calculate Math.Pow(2, n)
    for(int i = 1; i < n; i++)
    {
       x = (1 << x);
    }
     
    // Return Math.Pow(2, n) - 1
    return x - 1;
}
 
// Driver Code
public static void Main(String[] args)
{
    long n = 3;
 
    Console.Write(count_Strings(n));
}
}
 
// This code is contributed by Amit Katiyar

Javascript




<script>
 
// Javascript program to implement
// the above approach
 
    // Function to return
    // the count of Strings
    function count_Strings(n)
    {
        var x = 1;
 
        // Calculate Math.pow(2, n)
        for (i = 1; i < n; i++) {
            x = (1 << x);
        }
 
        // Return Math.pow(2, n) - 1
        return x - 1;
    }
 
    // Driver Code
     
        var n = 3;
 
        document.write(count_Strings(n));
 
// This code is contributed by todaysgaurav
 
</script>
Output: 
3

 

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




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