Largest number not greater than N all the digits of which are odd

Given a number N, the task is to find the largest number not greater than N which has all of it’s digits as odd. 
Examples:

Input: N = 23 
Output: 19 
19 is the largest number less than 23 which has all odd digits. 
Input: N = 7236 
Output: 7199 

Naive Approach: Iterate from N to 0, and find the first number which has all of its digits as odd. This approach can still be optimized if the even numbers are skipped as every even number will have an even digit on it’s right.
Below is the implementation of the above approach: 

C++

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

// Function to check if all digits
// of a number are odd
bool allOddDigits(int n)
{
    // iterate for all digits
    while (n) {

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

    // all digits are odd
    return true;
}

// function to return the largest number
// with all digits odd
int largestNumber(int n)
{
    if (n % 2 == 0)
        n--;

    // iterate till we find a
    // number with all digits odd
    for (int i = n;; i -= 2)
        if (allOddDigits(i))
            return i;
}

// Driver Code
int main()
{
    int N = 23;
    cout << largestNumber(N);
    return 0;
}

Java

// Java program to print the largest integer
// not greater than N with all odd digits

public class GFG{

        // Function to check if all digits
        // of a number are odd
        static boolean allOddDigits(int n)
        {
            // iterate for all digits
            while (n != 0) {

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

            // all digits are odd
            return true;
        }

        // function to return the largest number
        // with all digits odd
        static int largestNumber(int n)
        {
            if (n % 2 == 0)
                n--;

            // iterate till we find a
            // number with all digits odd
            for (int i = n;; i -= 2)
                if (allOddDigits(i))
                    return i;
        }

     public static void main(String []args){

        int N = 23;
        System.out.println(largestNumber(N));

    }
    // This code is contributed by ANKITRAI1

}

Python3

# Python 3 program to print the largest
# integer not greater than N with all
# odd digits

# Function to check if all digits
# of a number are odd
def allOddDigits(n):

    # iterate for all digits
    while (n):

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

    # all digits are odd
    return True

# function to return the largest
# number with all digits odd
def largestNumber(n):
    if (n % 2 == 0):
        n -= 1

    # iterate till we find a
    # number with all digits odd
    i = n
    while(1):
        if (allOddDigits(i)):
            return i
        i -= 2

# Driver Code
if __name__ =='__main__':
    N = 23
    print(largestNumber(N))

# This code is contributed by
# Shashank_Sharma

C#

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

class GFG
{

// Function to check if all
// digits of a number are odd
static bool allOddDigits(int n)
{
    // iterate for all digits
    while (n != 0)
    {

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

    // all digits are odd
    return true;
}

// function to return the largest
// number with all digits odd
static int largestNumber(int n)
{
    if (n % 2 == 0)
        n--;

    // iterate till we find a
    // number with all digits odd
    for (int i = n;; i -= 2)
        if (allOddDigits(i))
            return i;
}

// Driver Code
public static void Main()
{
    int N = 23;

    Console.WriteLine(largestNumber(N));
}
}

// This code is contributed by anuj_67

PHP

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

// Function to check if all digits
// of a number are odd
function allOddDigits($n)
{
    // iterate for all digits
    while ($n > 1)
    {

        // if digit is even
        if (($n % 10) % 2 == 0)
            return false;
        $n = (int)$n / 10;
    }

    // all digits are odd
    return true;
}

// function to return the largest
// number with all digits odd
function largestNumber($n)
{
    if ($n % 2 == 0)
        $n--;

    // iterate till we find a
    // number with all digits odd
    for ($i = $n;; $i= ($i - 2))
        if (allOddDigits($i))
            return $i;
}

// Driver Code
$N = 23;
echo largestNumber($N);

// This code is contributed by ajit
?>

Javascript

<script>

// javascript program to print the largest integer
// not greater than N with all odd digits

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

    // iterate for all digits
    while (n != 0)
    {

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

    // all digits are odd
    return true;
}

// function to return the largest number
// with all digits odd
function largestNumber(n)
{
    if (parseInt(n % 2) == 0)
        n--;

    // iterate till we find a
    // number with all digits odd

    for (i = n; i > 0; i -= 2)
        if (allOddDigits(i))
            return i;
}

// Driver code
var N = 23;
document.write(largestNumber(N));

// This code is contributed by shikhasingrajput
</script>

Output:

19

Time Complexity: O(NlogN) 

Auxiliary Space: O(1)
Efficient Approach: We can obtain the required number by decreasing the first even digit in N by one and then replacing all the other digits with the largest odd digit i.e. 9. For example, if N = 24578 then X = 19999. If there are no even digits in N, then N is the number itself. 
Below is the implementation of the above approach: 

C++

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

// function to return the largest number
// with all digits odd
int largestNumber(int n)
{
    string s = "";
    int duplicate = n;

    // convert the number to a string for
    // easy operations
    while (n) {
        s = char(n % 10 + 48) + s;
        n /= 10;
    }

    int index = -1;

    // find first even digit
    for (int i = 0; i < s.length(); i++) {
        if (((s[i] - '0') % 2 & 1) == 0) {
            index = i;
            break;
        }
    }

    // if no even digit, then N is the answer
    if (index == -1)
        return duplicate;

    int num = 0;

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

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

    // add 9 in the rest of the digits
    for (int i = index + 1; i < s.length(); i++)
        num = num * 10 + 9;

    return num;
}

// Driver Code
int main()
{
    int N = 24578;

    cout << largestNumber(N);

    return 0;
}

Java

// Java program to print the largest integer
// not greater than N with all odd digits

class GFG
{

// function to return the largest number
// with all digits odd
static int largestNumber(int n)
{
    String s = "";
    int duplicate = n;

    // convert the number to a string for
    // easy operations
    while (n > 0)
    {
        s = (char)(n % 10 + 48) + s;
        n /= 10;
    }

    int index = -1;

    // find first even digit
    for (int i = 0; i < s.length(); i++)
    {
        if (((int)(s.charAt(i) - '0') % 2 & 1) == 0)
        {
            index = i;
            break;
        }
    }

    // if no even digit, then N is the answer
    if (index == -1)
        return duplicate;

    int num = 0;

    // till first even digit, add all odd numbers
    for (int i = 0; i < index; i++)
        num = num * 10 + (int)(s.charAt(i) - '0');

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

    // add 9 in the rest of the digits
    for (int i = index + 1; i < s.length(); i++)
        num = num * 10 + 9;

    return num;
}

// Driver Code
public static void main (String[] args)
{
    int N = 24578;

    System.out.println(largestNumber(N));
}
}

// This code is contributed by mits

Python3

# Python3 program to print the largest integer
# not greater than N with all odd digits

# function to return the largest number
# with all digits odd
def largestNumber(n):

    s = ""
    duplicate = n

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

    index = -1

    # find first even digit
    for i in range(len(s)):
        if (((ord(s[i]) -
              ord('0')) % 2 & 1) == 0):
            index = i
            break

    # if no even digit, then N is the answer
    if (index == -1):
        return duplicate

    num = 0

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

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

    # add 9 in the rest of the digits
    for i in range(index + 1, len(s)):
        num = num * 10 + 9

    return num

# Driver Code
N = 24578
print(largestNumber(N))

# This code is contributed by mohit kumar

C#

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

class GFG
{

// function to return the largest number
// with all digits odd
static int largestNumber(int n)
{
    string s = "";
    int duplicate = n;

    // convert the number to a string for
    // easy operations
    while (n > 0)
    {
        s = (char)(n % 10 + 48) + s;
        n /= 10;
    }

    int index = -1;

    // find first even digit
    for (int i = 0; i < s.Length; i++)
    {
        if (((int)(s[i] - '0') % 2 & 1) == 0)
        {
            index = i;
            break;
        }
    }

    // if no even digit, then N is the answer
    if (index == -1)
        return duplicate;

    int num = 0;

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

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

    // add 9 in the rest of the digits
    for (int i = index + 1; i < s.Length; i++)
        num = num * 10 + 9;

    return num;
}

// Driver Code
static void Main()
{
    int N = 24578;

    Console.WriteLine(largestNumber(N));
}
}

// This code is contributed by mits

Javascript

<script>

// javascript program to print the largest integer
// not greater than N with all odd digits
// function to return the largest number
// with all digits odd
function largestNumber(n)
{
    var s = "";
    var duplicate = n;

    // convert the number to a string for
    // easy operations
    while (n > 0)
    {
        s = String.fromCharCode(n % 10 + 48) + s;
        n = parseInt(n/10);
    }

    var index = -1;

    // find first even digit
    for (i = 0; i < s.length; i++)
    {
        if (((s.charAt(i).charCodeAt(0) -
        '0'.charCodeAt(0)) % 2 & 1) == 0)
        {
            index = i;
            break;
        }
    }

    // if no even digit, then N is the answer
    if (index == -1)
        return duplicate;

    var num = 0;

    // till first even digit, add all odd numbers
    for (i = 0; i < index; i++)
        num = num * 10 + (s.charAt(i).charCodeAt(0)
        - '0'.charCodeAt(0));

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

    // add 9 in the rest of the digits
    for (i = index + 1; i < s.length; i++)
        num = num * 10 + 9;

    return num;
}

// Driver Code
var N = 24578;

document.write(largestNumber(N));


// This code contributed by Princi Singh

</script>

Output:

19999

Time Complexity: O(M) where M is the number of digits

Auxiliary Space: O(1)