2843. Count Symmetric Integers

Problem

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 (Java)

class Solution {
    public int countSymmetricIntegers(int low, int high) {
        int ans = 0;
        for (int i = low; i <= high; ++i) {
            String s = Integer.toString(i);
            int t = 0, n = s.length();
            for (int j = 0; j < n / 2; ++j) {
                t += s.charAt(j) - s.charAt(n - j - 1);
            }
            if (n % 2 == 0 && t == 0)  ans++;
        }
        return ans;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).