<p>You are given two positive integers <code>low</code> and <code>high</code>.</p>

<p>An integer <code>x</code> consisting of <code>2 * n</code> digits is <strong>symmetric</strong> if the sum of the first <code>n</code> digits of <code>x</code> is equal to the sum of the last <code>n</code> digits of <code>x</code>. Numbers with an odd number of digits are never symmetric.</p>

<p>Return <em>the <strong>number of symmetric</strong> integers in the range</em> <code>[low, high]</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

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

<p><strong class="example">Example 2:</strong></p>

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

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= low &lt;= high &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution:
    def countSymmetricIntegers(self, low: int, high: int) -> int:
        cnt = 0

        #Iterate from low to high and convert the number to a string for quick digit access. If it has an even # of digits, we must
        #check it. Run two loops that iterate through the two halves, incrementing sumLeft and sumRight. If they're the same value
        #it's symmetric. Checking 100 means we can skip 100-999
        for i in range(low, high + 1):
            strNum = str(i)
            n = len(strNum)
            
            if n % 2 == 0:
                sumLeft = 0
                sumRight = 0
                for j in range(n // 2):
                    sumLeft += (ord(strNum[j]) - ord("0"))
                
                for j in range(n // 2, n):
                    sumRight += (ord(strNum[j]) - ord("0"))
                
                if sumLeft == sumRight:
                    cnt += 1
            elif i == 100:
                i += 899

        return cnt
452