Smallest odd digits number not less than N

Given a number N, the task is to find the smallest number not less than N, which has all digits odd. 

Examples:

Input: N = 1345  
Output: 1351
1351 is the smallest number not less than N, whose all digits are odd. 

Input: N = 2397 
Output: 3111 
3111 is the smallest number not less than N, whose all digits are odd.

Naive approach: A naive approach is to keep iterating from N until we find a number with all digits odd. 

Below is the implementation of the above approach:  

C++

// CPP program to print the smallest
// integer not less than N with all odd digits
#include <bits/stdc++.h>
using namespace std;

// function to check if all digits
// are odd of a given number
int check_digits(int n)
{
    // iterate for all digits
    while (n) {
        if ((n % 10) % 2 == 0) // if digit is even
            return 0;

        n /= 10;
    }

    // all digits are odd
    return 1;
}

// function to return the smallest number
// with all digits odd
int smallest_number(int n)
{
    // iterate till we find a
    // number with all digits odd
    for (int i = n;; i++)
        if (check_digits(i))
            return i;
}

// Driver Code
int main()
{
    int N = 2397;
    cout << smallest_number(N);

    return 0;
}

Java

// Java program to print the smallest
// integer not less than N with all odd digits
import java.io.*;
class Geeks {

// function to check if all digits
// are odd of a given number
static int check_digits(int n)
{
    // iterate for all digits
    while (n > 0) {
        if ((n % 10) % 2 == 0) // if digit is even
            return 0;

        n /= 10;
    }

    // all digits are odd
    return 1;
}

// function to return the smallest number
// with all digits odd
static int smallest_number(int n)
{
    // iterate till we find a
    // number with all digits odd
    for (int i = n;; i++)
        if (check_digits(i) > 0)
            return i;
}

// Driver Code
public static void main(String args[])
{
    int N = 2397;
    System.out.println(smallest_number(N));
}
}

// This code is contributed by Kirti_Mangal

Python3

# Python 3 program to print the smallest
# integer not less than N with all odd digits

# function to check if all digits
# are odd of a given number
def check_digits(n):

    # iterate for all digits
    while (n):

        # if digit is even
        if ((n % 10) % 2 == 0):
            return 0

        n = int(n / 10)

    # all digits are odd
    return 1

# function to return the smallest
# number with all digits odd
def smallest_number(n):

    # iterate till we find a
    # number with all digits odd
    i = n
    while(1):
        if (check_digits(i)):
            return i

        i += 1

# Driver Code
if __name__ == '__main__':
    N = 2397
    print(smallest_number(N))

# This code is contributed by
# Sanjit_Prasad

C#

// C# program to print the smallest
// integer not less than N with all
// odd digits
using System;

class GFG
{
// function to check if all digits
// are odd of a given number
static int check_digits(int n)
{
    // iterate for all digits
    while (n != 0)
    {
        if ((n % 10) % 2 == 0) // if digit is even
            return 0;

        n /= 10;
    }

    // all digits are odd
    return 1;
}

// function to return the smallest
// number with all digits odd
static int smallest_number(int n)
{
    // iterate till we find a
    // number with all digits odd
    for (int i = n;; i++)
        if (check_digits(i) == 1)
            return i;
}

// Driver Code
static void Main()
{
    int N = 2397;
    Console.WriteLine(smallest_number(N));
}
}

// This code is contributed by ANKITRAI1

PHP

<?php
// PHP program to print the smallest
// integer not less than N with all
// odd digits

// function to check if all digits
// are odd of a given number
function check_digits($n)
{
    // iterate for all digits
    while ($n > 1)
    {
        // if digit is even
        if (($n % 10) % 2 == 0)
            return 0;

        $n = (int)$n / 10;
    }

    // all digits are odd
    return 1;
}

// function to return the smallest
// number with all digits odd
function smallest_number( $n)
{
    // iterate till we find a
    // number with all digits odd
    for ($i = $n;; $i++)
        if (check_digits($i))
            return $i;
}

// Driver Code
$N = 2397;
echo smallest_number($N);

// This code is contributed by ajit
?>

Javascript

<script>
// java script program to print the smallest
// integer not less than N with all
// odd digits

// function to check if all digits
// are odd of a given number
function check_digits(n)
{

    // iterate for all digits
    while (n > 1)
    {

        // if digit is even
        if ((n % 10) % 2 == 0)
            return 0;

         n = parseInt(n / 10);
    }

    // all digits are odd
    return 1;
}

// function to return the smallest
// number with all digits odd
function smallest_number( n)
{

    // iterate till we find a
    // number with all digits odd
    for (i = n;; i++)
        if (check_digits(i))
            return i;
}

// Driver Code
let N = 2397;
document.write( smallest_number(N));

// This code is contributed by sravan kumar (vignan)
</script>

Output

3111

Complexity Analysis:

• Time Complexity: O(N)
• Auxiliary Space: O(1)

Efficient Approach: 

We can find the number by increasing the first even digit in N by one and replacing all digits to the right of that odd digit with the smallest odd digit (i.e. 1). If there are no even digits in N, then N is the smallest number itself. For example, consider N = 213. Increment first even digit in N i.e., 2 to 3 and replace all digits right to it by 1. So, our required number will be 311.

Below is the implementation of the efficient approach:  

C++

// CPP program to print the smallest
// integer not less than N with all odd digits
#include <bits/stdc++.h>
using namespace std;

// function to return the smallest number
// with all digits odd
int smallestNumber(int n)
{
    int num = 0;
    string s = "";

    int duplicate = n;
    // convert the number to string to
    // perform operations
    while (n) {
        s.push_back(((n % 10) + '0'));
        n /= 10;
    }
    reverse(s.begin(), s.end());

    int index = -1;

    // find out the first even number
    for (int i = 0; i < s.length(); i++) {
        int digit = s[i] - '0';
        if ((digit & 1) == 0) {
            index = i;
            break;
        }
    }

    // if no even numbers are there, than n is the answer
    if (index == -1)
        return duplicate;

    // add all digits till first even
    for (int i = 0; i < index; i++)
        num = num * 10 + (s[i] - '0');

    // increase the even digit by 1
    num = num * 10 + (s[index] - '0' + 1);

    // add 1 to the right of the even number
    for (int i = index + 1; i < s.length(); i++)
        num = num * 10 + 1;

    return num;
}

// Driver Code
int main()
{
    int N = 2397;
    cout << smallestNumber(N);

    return 0;
}

Java

//Java program to print the smallest
// integer not less than N with all odd digits
import java.io.*;
public class GFG {

// function to return the smallest number
// with all digits odd
    static int smallestNumber(int n) {
        int num = 0;
        String s = "";

        int duplicate = n;
        // convert the number to string to
        // perform operations
        while (n > 0) {
            s = (char) (n % 10 + 48) + s;
            n /= 10;
        }

        int index = -1;

        // find out the first even number
        for (int i = 0; i < s.length(); i++) {
            int digit = s.charAt(i) - '0';
            if ((digit & 1) == 0) {
                index = i;
                break;
            }
        }

        // if no even numbers are there, than n is the answer
        if (index == -1) {
            return duplicate;
        }

        // add all digits till first even
        for (int i = 0; i < index; i++) {
            num = num * 10 + (s.charAt(i) - '0');
        }

        // increase the even digit by 1
        num = num * 10 + (s.charAt(index) - '0' + 1);

        // add 1 to the right of the even number
        for (int i = index + 1; i < s.length(); i++) {
            num = num * 10 + 1;
        }

        return num;
    }

// Driver Code
    static public void main(String[] args) {
        int N = 2397;
        System.out.println(smallestNumber(N));
    }
}

/*This code is contributed by PrinciRaj1992*/

Python 3

# Python 3 program to print the smallest
# integer not less than N with all odd digits

# function to return the smallest
# number with all digits odd
def smallestNumber(n):

    num = 0
    s = ""

    duplicate = n

    # convert the number to string to
    # perform operations
    while (n):
        s = chr(n % 10 + 48) + s
        n //= 10

    index = -1

    # find out the first even number
    for i in range(len( s)):
        digit = ord(s[i]) - ord('0')
        if ((digit & 1) == 0) :
            index = i
            break

    # if no even numbers are there,
    # than n is the answer
    if (index == -1):
        return duplicate

    # add all digits till first even
    for i in range( index):
        num = num * 10 + (ord(s[i]) -
                          ord('0'))

    # increase the even digit by 1
    num = num * 10 + (ord(s[index]) -
                      ord('0') + 1)

    # add 1 to the right of the
    # even number
    for i in range(index + 1 , len(s)):
        num = num * 10 + 1

    return num

# Driver Code
if __name__ == "__main__":

    N = 2397
    print(smallestNumber(N))

# This code is contributed by ita_c

C#

// C# program to print the smallest
// integer not less than N with all odd digits

using System;
public class GFG{


// function to return the smallest number
// with all digits odd
    static int smallestNumber(int n) {
        int num = 0;
        String s = "";

        int duplicate = n;
        // convert the number to string to
        // perform operations
        while (n > 0) {
            s = (char) (n % 10 + 48) + s;
            n /= 10;
        }

        int index = -1;

        // find out the first even number
        for (int i = 0; i < s.Length; i++) {
            int digit = s[i] - '0';
            if ((digit & 1) == 0) {
                index = i;
                break;
            }
        }

        // if no even numbers are there, than n is the answer
        if (index == -1) {
            return duplicate;
        }

        // add all digits till first even
        for (int i = 0; i < index; i++) {
            num = num * 10 + (s[i] - '0');
        }

        // increase the even digit by 1
        num = num * 10 + (s[index] - '0' + 1);

        // add 1 to the right of the even number
        for (int i = index + 1; i < s.Length; i++) {
            num = num * 10 + 1;
        }

        return num;
    }

// Driver Code
    static public void Main() {
        int N = 2397;
        Console.WriteLine(smallestNumber(N));
    }
}

// This code is contributed by PrinciRaj1992

Javascript

<script>

// Javascript program to print the smallest
// integer not less than N with all odd digits

// function to return the smallest number
// with all digits odd
function smallestNumber(n)
{
    var num = 0;
    var s = "";

    var duplicate = n;
    // convert the number to string to
    // perform operations
    while (n) {
        s = String.fromCharCode(n % 10 + 48) + s;
        n = parseInt(n/10);
    }

    var index = -1;

    // find out the first even number
    for (var i = 0; i < s.length; i++) {
        var digit = s[i].charCodeAt(0) - '0'.charCodeAt(0);
        if ((digit & 1) == 0) {
            index = i;
            break;
        }
    }

    // if no even numbers are there, than n is the answer
    if (index == -1)
        return duplicate;

    // add all digits till first even
    for (var i = 0; i < index; i++)
        num = num * 10 + (s[i].charCodeAt(0) - '0'.charCodeAt(0));

    // increase the even digit by 1
    num = num * 10 + (s[index].charCodeAt(0) - '0'.charCodeAt(0) + 1);

    // add 1 to the right of the even number
    for (var i = index + 1; i < s.length; i++)
        num = num * 10 + 1;

    return num;
}

// Driver Code
var N = 2397;
document.write( smallestNumber(N));

</script>

Output

3111

Time Complexity: O(M), where M is the number of digits in N.
Auxiliary Space: O(M)

Another Approach:

Initialize a variable num to n.

If num is even, increment it by 1.

While num contains even digits:

a. Convert num to a string and check each digit.

b. If the current digit is even, add a power of 10 to num such that the resulting number has the same number of digits as num.

Return num.

C++

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

int main()
{
    int n = 123456;

    int num = n;

    // If num is even, increment it by 1
    if (num % 2 == 0) {
        num++;
    }

    // While num contains even digits,
    // add a power of 10 to num such that the resulting
    // number
    // has the same number of digits as num
    while (1) {
        int contains_even = 0;
        string str = to_string(num);
        int len = str.length();
        for (int i = 0; i < len; i++) {
            if ((str[i] - '0') % 2 == 0) {
                contains_even = 1;
                int pow10 = 1;
                for (int j = 0; j < len - i - 1; j++) {
                    pow10 *= 10;
                }
                num += pow10;
                break;
            }
        }
        if (!contains_even) {
            break;
        }
    }

    cout << "Smallest odd digits number not less than " << n
         << " is: " << num << endl;

    return 0;
}
// this code is contributed by thebeginner01(Akash Bankar)

C

#include <stdio.h>
#include <string.h>

int main()
{
    int n = 123456;

    int num = n;

    // If num is even, increment it by 1
    if (num % 2 == 0) {
        num++;
    }

    // While num contains even digits, add a power of 10 to
    // num such that the resulting number has the same
    // number of digits as num
    while (1) {
        int contains_even = 0;
        char str[20];
        sprintf(str, "%d", num);
        int len = strlen(str);
        for (int i = 0; i < len; i++) {
            if ((str[i] - '0') % 2 == 0) {
                contains_even = 1;
                int pow10 = 1;
                for (int j = 0; j < len - i - 1; j++) {
                    pow10 *= 10;
                }
                num += pow10;
                break;
            }
        }
        if (!contains_even) {
            break;
        }
    }

    printf("Smallest odd digits number not less than %d "
           "is: %d\n",
           n, num);

    return 0;
}

Python3

# Function to get the smallest odd digits number not less than n
def GetSmallestOddDigitsNumber(n):
    num = n

    # If num is even, increment it by 1
    if num % 2 == 0:
        num += 1

    # While num contains even digits,
    # add a power of 10 to num such that the resulting number
    # has the same number of digits as num
    while True:
        contains_even = False
        str_num = str(num)
        len_num = len(str_num)
        for i in range(len_num):
            if int(str_num[i]) % 2 == 0:
                contains_even = True
                pow10 = 1
                for j in range(len_num - i - 1):
                    pow10 *= 10
                num += pow10
                break
        if not contains_even:
            break

    print("Smallest odd digits number not less than", n, "is:", num)


# Example usage
GetSmallestOddDigitsNumber(123456)

Javascript

let n = 123456;

let num = n;

// If num is even, increment it by 1
if (num % 2 == 0) {
    num++;
}

// While num contains even digits, add a power of 10 to num
// such that the resulting number has the same number of digits as num
while (1) {
    let contains_even = 0;
    let str = num.toString();
    let len = str.length;
    for (let i = 0; i < len; i++) {
        if ((str[i] - '0') % 2 == 0) {
            contains_even = 1;
            let pow10 = 1;
            for (let j = 0; j < len - i - 1; j++) {
                pow10 *= 10;
            }
            num += pow10;
            break;
        }
    }
    if (!contains_even) {
        break;
    }
}

console.log("Smallest odd digits number not less than " + n + " is: " + num);

Java

import java.io.*;
import java.lang.*;
import java.util.*;

class Main {
    public static void main(String[] args)
        throws java.lang.Exception
    {
        int n = 123456;
        int num = n;
        // If num is even, increment it by 1
        if (num % 2 == 0) {
            num++;
        }
        // While num contains even digits,
        // add a power of 10 to num such that the resulting
        // number
        // has the same number of digits as num
        while (true) {
            int contains_even = 0;
            String str = Integer.toString(num);
            int len = str.length();
            for (int i = 0; i < len; i++) {
                if ((str.charAt(i) - '0') % 2 == 0) {
                    contains_even = 1;
                    int pow10 = 1;
                    for (int j = 0; j < len - i - 1; j++) {
                        pow10 *= 10;
                    }
                    num += pow10;
                    break;
                }
            }
            if (contains_even == 0) {
                break;
            }
        }
        System.out.println(
            "Smallest odd digits number not less than " + n
            + " is: " + num);
    }
}

C#

using System;

public class Program {
    public static void Main()
    {
        int n = 123456;

        int num = n;

        // If num is even, increment it by 1
        if (num % 2 == 0) {
            num++;
        }

        // While num contains even digits,
        // add a power of 10 to num such that the resulting
        // number has the same number of digits as num
        while (true) {
            bool contains_even = false;
            string str = num.ToString();
            int len = str.Length;
            for (int i = 0; i < len; i++) {
                if ((str[i] - '0') % 2 == 0) {
                    contains_even = true;
                    int pow10 = 1;
                    for (int j = 0; j < len - i - 1; j++) {
                        pow10 *= 10;
                    }
                    num += pow10;
                    break;
                }
            }
            if (!contains_even) {
                break;
            }
        }

        Console.WriteLine(
            "Smallest odd digits number not less than " + n
            + " is: " + num);
    }
}

Output

Smallest odd digits number not less than 123456 is: 133557

time complexity of O(log N) , Where N is given in the problem statement

 space complexity of O(1)