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 strings

Input:3

Output:7

Expalnation:

“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.

Since2strings are possible of length N, the required answer is^{N}2.^{N}– 1

Follow the steps below to solve the problem:

- Initialize X = 1.
- Compute upto
**2**by performing bitwise left shift on X, N-1 times.^{N} - 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; ` `} ` |

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## 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 ` |

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## 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 ` |

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## 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 ` |

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**Output:**

3

**Time Complexity:** O(N)

**Auxiliary Space:** O(1)

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