Description
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 <= 104
Solution
Python3
class Solution:
def countSymmetricIntegers(self, low: int, high: int) -> int:
res = 0
def good(x):
A = []
while x > 0:
A.append(x % 10)
x //= 10
N = len(A)
return N % 2 == 0 and sum(A[:N//2]) == sum(A[N//2:])
for x in range(low, high + 1):
if good(x):
res += 1
return res