# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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

 ` 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 ` `?> `

Output:

```19
```

Time Complexity: O(N)

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:

 `// CPP program to print the largest integer ` `// not greater than N with all odd digits ` `#include ` `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; ` `} `

Output:

```19999
```

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

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