Count Symmetric Integers

Problem Source: Leetcode

Problem 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].

tests

def test(fn):
    low,high = 1,100
    expected =  9
    symmetric_count = fn(low,high) # 11, 22, 33, 44, 55, 66, 77, 88, and 99.
    assert symmetric_count == expected 

    low, high = 1200, 1230
    expected = 4
    symmetric_count = fn(low,high) # 1203, 1212, 1221, and 1230
    assert symmetric_count == expected 

    low, high = 0, 10000
    expected = 624
    symmetric_count = fn(low,high) 
    assert symmetric_count == expected 

Solution

• Time Complexity: O(n^2)
• Space Complexity: O(1)

def sum(arr):
    ttl = 0
    for a in arr:
        ttl += int(a)
    return ttl

def split_int(x):
    half = len(x)//2
    left, right = x[:half], x[-half:]
    return left, right

def isSymmetric(x, ):
    x = str(x)
    if len(x)%2 == 1: return False
    half = len(x)//2
    return sum(x[:half]) == sum(x[-half:])

def countSymmetricIntegers(low: int, high: int) -> int:
    symmetric_count = 0
    for num in range(low, high+1):
        symmetric_count += isSymmetric(num)
    return symmetric_count
    
test(countSymmetricIntegers)

import timeit
timeit.timeit(lambda: countSymmetricIntegers(0,10000),number=1000)