Given an integer N, the task is to print the number of binary strings of length N which at least one ‘1’.
Examples:
Input: 2
Output: 3
Explanation:
“01”, “10” and “11” are the possible stringsInput: 3
Output: 7
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 stringslong 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 Codeint main(){ long n = 3; cout << count_strings(n); return 0;} |
Java
// Java program to implement// the above approachimport java.util.*;class GFG{// Function to return// the count of Stringsstatic 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 Codepublic 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 Stringsdef 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 Codeif __name__ == '__main__': n = 3; print(count_Strings(n));# This code is contributed by Princi Singh |
C#
// C# program to implement// the above approachusing System;class GFG{// Function to return// the count of Stringsstatic 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 Codepublic 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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