# 2843.Count Symmetric Integers

You are given two positive integers low and high.

An integer x consisting of ==2 * n== digits is symmetric if the sum of the first n digits of x is equal to the sum of the last n digits of x. Numbers with an odd number of digits are never symmetric.

Return the number of symmetric integers in the range ==[low, high]==.

* Example 1:

Input: low = 1, high = 100
Output: 9
Explanation: There are 9 symmetric integers between 1 and 100: 11, 22, 33, 44, 55, 66, 77, 88, and 99.

* Example 2:

Input: low = 1200, high = 1230
Output: 4
Explanation: There are 4 symmetric integers between 1200 and 1230: 1203, 1212, 1221, and 1230.

* Constraints:
==1 <= low <= high <= 10000==

## Answer

class Solution:
    def countSymmetricIntegers(self, low: int, high: int) -> int:
        def sum_digit(num):
                return num//10 + num%10
        ans = 0
        for i in range(low, high+1):
            if 9< i <100:
                if sum_digit(i//10) == sum_digit(i%10):
                    ans +=1
            if 999< i < 10000:
                if sum_digit(i//100) == sum_digit(i%100):
                    ans +=1
        return ans