# Program to find nearest number of n where all digits are odd in python

Suppose we have a number n, we have to find the next closest value where all digits are odd. When there are two values tied for being closest to n, return the larger one.

So, if the input is like n = 243, then the output will be 199.

To solve this, we will follow these steps −

• first_even := -1
• s := n as string
• l := size of s
• for i in range 0 to l, do
• if s[i] is even, then
• first_even := i
• come out from the loop
• if first_even is same as -1, then
• return n
• big := 1 + numeric value of s[from index 0 to i]
• if s[i] is same as "0", then
• if s[i - 1] is same as "1", then
• small := numeric value of s[from index 0 to i] - 1
• otherwise,
• small := numeric value of s[from index 0 to i] - 11
• otherwise,
• small := numeric value of s[from index 0 to i] - 1
• for i in range i + 1 to l, do
• big := big concatenate "1"
• small := small concatenate "9"
• big := numeric value of big, small := numeric value of small
• d2 := big - n, d1 := n - small
• if d1 < d2, then
• return small
• otherwise when d1 >= d2, then
• return big

Let us see the following implementation to get better understanding −

## Example

Live Demo

class Solution:
def solve(self, n):
first_even = -1
s = str(n)
l = len(s)
for i in range(l):
if int(s[i]) % 2 == 0:
first_even = i
break
if first_even == -1:
return n
big = str(int(s[: i + 1]) + 1)
if s[i] == "0":
if s[i - 1] == "1":
small = str(int(s[: i + 1]) - 1)
else:
small = str(int(s[i : i + 1]) - 11)
else:
small = str(int(s[: i + 1]) - 1)

for i in range(i + 1, l):
big += "1"
small += "9"

big, small = int(big), int(small)
d2 = big - n
d1 = n - small
if d1 < d2:
return small
elif d1 >= d2:
return big

ob = Solution()
n = 243
print(ob.solve(n))

## Input

243

## Output

199