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 digitsint smallestOdd(int n){ if (n == 1) return 1; return pow(10, n - 1) + 1;}// Driver Codeint main(){ int n = 4; cout << smallestOdd(n); return 0;} |
Java
// Java implementation of the approachclass 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 digitsdef smallestOdd(n) : if (n == 1): return 1 return pow(10, n - 1) + 1# Driver Coden = 4print(smallestOdd(n))# This code is contributed by ihritik. |
C#
// C# implementation of the approachusing 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 digitsfunction 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> |
1001
Time Complexity: O(log n).
Auxiliary Space: O(1)
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