Smallest odd number with N digits

Given a number N. The task is to find the smallest N digit ODD number.
Examples:

Input: N = 1
Output: 1

Input: N = 3
Output: 101 

Approach: There can be two cases depending on the value of N. 
Case 1 : If N = 1 then answer will be 1. 
Case 2 : If N != 1 then answer will be (10^(n-1)) + 1 because the series of the smallest odd numbers will go on like: 1, 11, 101, 1001, 10001, 100001, ….
Below is the implementation of the above approach:

C++

// C++ implementation of the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to return smallest odd
// with n digits
int smallestOdd(int n)
{
    if (n == 1)
        return 1;

    return pow(10, n - 1) + 1;
}

// Driver Code
int main()
{
    int n = 4;
    cout << smallestOdd(n);

    return 0;
}

Java

// Java implementation of the approach
class Solution {

    // Function to return smallest odd with n digits
    static int smallestOdd(int n)
    {
        if (n == 1)
            return 0;
        return Math.pow(10, n - 1) + 1;
    }

    // Driver code
    public static void main(String args[])
    {
        int n = 4;

        System.out.println(smallestOdd(n));
    }
}

Python3

# Python3 implementation of the approach

# Function to return smallest even
# number with n digits
def smallestOdd(n) :

    if (n == 1):
        return 1
    return pow(10, n - 1) + 1

# Driver Code
n = 4
print(smallestOdd(n))

# This code is contributed by ihritik.

C#

// C# implementation of the approach
using System;
class Solution {

    // Function to return smallest odd with n digits
    static int smallestOdd(int n)
    {
        if (n == 1)
            return 0;

        return Math.pow(10, n - 1) + 1;
    }

    // Driver code
    public static void Main()
    {
        int n = 4;

        Console.Write(smallestOdd(n));
    }
}

PHP

<?php
// PHP implementation of the approach

// Function to return smallest even
// number with n digits
function smallestOdd($n)
{
    if ($n == 1)
        return 1;
    return pow(10, $n - 1) + 1;
}

// Driver Code
$n = 4;
echo smallestOdd($n);

// This code is contributed by ihritik
?>

Javascript

<script>

  // Javascript implementation of the above approach

  // Function to return smallest odd
  // with n digits
  function smallestOdd(n) {
    if (n == 1)
      return 1;

    return Math.pow(10, n - 1) + 1;
  }

  // Driver Code
  var n = 4;
  document.write(smallestOdd(n));

  // This code is contributed by rrrtnx.
</script>

Output:

1001

Time Complexity: O(log n).
Auxiliary Space: O(1)