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++

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// 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;
}

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Java

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// 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));
    }
}

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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#

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// 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));
    }
}

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PHP

Output:

1001

Time Complexity: O(1).



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Improved By : ihritik