Given an integer **N**, the task is to find the number closest to** N** having only odd digits. If more than one such number exists, print the minimum.

**Examples:**

Input:N = 725Output:719Explanation:

Numbers 719 and 731 consists of odd digits only and are closest to 725. Since 719 is the minimum, it is the required answer.

Input :N = 111Output: 111

**Naive Approach**: The simplest approach to solve the problem is to iterate and find the closest numbers smaller and greater than N, having only odd numbers as its digits. Follow the steps below to solve the problem:

- If all the digits of
**N**are odd, then print**N**as the answer. - Find the closest smaller and greater numbers and return the one having the least difference with
**N**. If both are found to be equidistant from**N**, print the smaller number.

**Time Complexity**: O(10^{(len(N) – 1)})**Auxiliary Space:**O(1)

**Efficient Approach**: Follow the steps below to optimize the above approach:

- If all the digits of
**N**are odd, then return**N**. - Calculate the closest smaller number to
**N**efficiently by the following steps:- Find the position of first even digit in
**N**from left to right, i.e,**pos**. - If the first even digit is
**0**, then replace**0**with**9**and iterate over all preceding digits and for every digit, if found to be**1**, replace it with**9**. Otherwise, decrease the digit by 2 and return the number. - If no preceding digits are found to be exceeding
**1**, then replace the most significant digit with**0**. - If the first even digit is non-zero, then decrease that digit by 1. Replace all the succeeding digits with
**9**.

- Find the position of first even digit in
- Calculate the greater number closest to
**N**by the steps below:- Find the position of first even digit i.e
**pos**. - Increment the digits at
**pos**by 1. - Iterate over the succeeding digits and increment them by
**1.**

- Find the position of first even digit i.e
- Compare the absolute difference of the respective closest numbers obtained with
**N**and print the one having least difference. - If both the differences are found to be equal, print the smaller number.

Below is the implementation of the above approach:

## Python

`# Python program to implement ` `# the above approach ` ` ` `# Function to return the smaller ` `# number closest to N made up of ` `# only odd digits ` `def` `closest_smaller(N): ` ` ` ` ` `N ` `=` `str` `(N) ` ` ` ` ` `l ` `=` `list` `(` `map` `(` `int` `, N)) ` ` ` `length ` `=` `len` `(l) ` ` ` ` ` `# Stores the position of ` ` ` `# first even digit of N ` ` ` `pos ` `=` `-` `1` ` ` ` ` `# Iterate through each digit of N ` ` ` `for` `i ` `in` `range` `(length): ` ` ` ` ` `# Check for even digit. ` ` ` `if` `l[i] ` `%` `2` `=` `=` `0` `: ` ` ` `pos ` `=` `i ` ` ` `break` ` ` ` ` `# If the first even digit is 0 ` ` ` `if` `l[pos] ` `=` `=` `0` `: ` ` ` ` ` `# Replace 0 with 9 ` ` ` `l[pos] ` `=` `9` ` ` ` ` `# Iterate over preceding ` ` ` `# digits ` ` ` `for` `i ` `in` `range` `(pos ` `-` `1` `, ` `-` `1` `, ` `-` `1` `): ` ` ` ` ` `# If current digit is 1 ` ` ` `if` `l[i] ` `=` `=` `1` `: ` ` ` ` ` `# Check if it is the ` ` ` `# first digit or not ` ` ` `if` `i ` `=` `=` `0` `: ` ` ` ` ` `# Append leading 0's ` ` ` `l[i] ` `=` `0` ` ` `break` ` ` ` ` `# Otherwise, replace by 9 ` ` ` `l[i] ` `=` `9` ` ` ` ` `# Otherwise ` ` ` `else` `: ` ` ` ` ` `# Decrease its value by 2 ` ` ` `l[i] ` `-` `=` `2` ` ` `break` ` ` ` ` `# If the first even digit exceeds 0 ` ` ` `else` `: ` ` ` ` ` `# Reduce the digit by 1 ` ` ` `l[pos] ` `-` `=` `1` ` ` ` ` `# Replace all succeeding digits by 9 ` ` ` `for` `i ` `in` `range` `(pos ` `+` `1` `, length): ` ` ` `l[i] ` `=` `9` ` ` ` ` `# Remove leading 0s ` ` ` `if` `l[` `0` `] ` `=` `=` `0` `: ` ` ` `l.pop(` `0` `) ` ` ` ` ` ` ` `result ` `=` `''.join(` `map` `(` `str` `, l)) ` ` ` `return` `result ` ` ` `# Function to return the greater ` `# number closest to N made up of ` `# only odd digits ` `def` `closest_greater(N): ` ` ` ` ` `N ` `=` `str` `(N) ` ` ` ` ` `l ` `=` `list` `(` `map` `(` `int` `, N)) ` ` ` `length ` `=` `len` `(l) ` ` ` ` ` `# Stores the position of ` ` ` `# first even digit of N ` ` ` `pos ` `=` `-` `1` ` ` ` ` `# Iterate over each digit ` ` ` `# of N ` ` ` `for` `i ` `in` `range` `(length): ` ` ` ` ` `# If even digit is found ` ` ` `if` `l[i] ` `%` `2` `=` `=` `0` `: ` ` ` `pos ` `=` `i ` ` ` `break` ` ` ` ` `# Increase value of first ` ` ` `# even digit by 1 ` ` ` `l[pos] ` `+` `=` `1` ` ` ` ` `for` `i ` `in` `range` `(pos ` `+` `1` `, length): ` ` ` `l[i] ` `=` `1` ` ` ` ` `result ` `=` `''.join(` `map` `(` `str` `, l)) ` ` ` `return` `result ` ` ` `# Function to check if all ` `# digits of N are odd or not ` `def` `check_all_digits_odd(N): ` ` ` ` ` `N ` `=` `str` `(N) ` ` ` ` ` `l ` `=` `list` `(` `map` `(` `int` `, N)) ` ` ` `length ` `=` `len` `(l) ` ` ` ` ` `# Stores the position of ` ` ` `# first even digit of N ` ` ` `pos ` `=` `-` `1` ` ` ` ` `# Iterating over each digit ` ` ` `# of N ` ` ` `for` `i ` `in` `range` `(length): ` ` ` ` ` `# If even digit is found ` ` ` `if` `l[i] ` `%` `2` `=` `=` `0` `: ` ` ` `pos ` `=` `i ` ` ` `break` ` ` ` ` `# If no even digit is found ` ` ` `if` `pos ` `=` `=` `-` `1` `: ` ` ` `return` `True` ` ` `return` `False` ` ` `# Function to return the ` `# closest number to N ` `# having odd digits only ` `def` `closestNumber(N): ` ` ` ` ` `# If all digits of N are odd ` ` ` `if` `check_all_digits_odd(N): ` ` ` `print` `(N) ` ` ` `else` `: ` ` ` ` ` `# Find smaller number ` ` ` `# closest to N ` ` ` `l ` `=` `int` `(closest_smaller(N)) ` ` ` ` ` `# Find greater number ` ` ` `# closest to N ` ` ` `r ` `=` `int` `(closest_greater(N)) ` ` ` ` ` `# Print the number with least ` ` ` `# absolute difference ` ` ` `if` `abs` `(N ` `-` `l) <` `=` `abs` `(N ` `-` `r): ` ` ` `print` `(l) ` ` ` `else` `: ` ` ` `print` `(r) ` ` ` `# Driver Code. ` `if` `__name__ ` `=` `=` `'__main__'` `: ` ` ` ` ` `N ` `=` `110` ` ` `closestNumber(N) ` ` ` |

*chevron_right*

*filter_none*

**Output:**

111

**Time Complexity: **O(len(N))**Auxiliary Space:** O(len(N))

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