2998. Count Symmetric Integers¶

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2998. Count Symmetric Integers

Easy

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¶

class Solution {
    public int countSymmetricIntegers(int low, int high) {
        int count = 0;
        for (int i = low; i <= high; i++) {
            if (check(i)) count++;
        }
        return count;
    }
    private boolean check(int n) {
        int count = 0;
        int temp = n;
        while (temp > 0) {
            count++;
            temp = temp / 10;
        }
        if (count % 2 == 1) return false;

        int sum = 0;
        temp = n;

        count = count / 2;
        while (count > 0) {
            sum += temp % 10;
            temp = temp / 10;
            count--;
        }
        while (temp > 0) {
            sum -= temp % 10;
            temp = temp / 10;
        }
        return sum == 0;
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here