README.md

2843. 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

Related Topics:
Math, Enumeration

Similar Questions:

• Palindrome Number (Easy)
• Sum of Digits in Base K (Easy)

Solution 1.

// OJ: https://leetcode.com/problems/count-symmetric-integers
// Author: github.com/lzl124631x
// Time: O((H - L) * lg(H))
// Space: O(lg(H))
class Solution {
    bool isSymmetric(int n) {
        auto s = to_string(n);
        if (s.size() % 2) return false;
        int N = s.size(), first = 0, second = 0;
        for (int i = 0; i < N; ++i) {
            if (i < N / 2) first += s[i] - '0';
            else second += s[i] - '0';
        }
        return first == second;
    }
public:
    int countSymmetricIntegers(int low, int high) {
        int ans = 0;
        for (int i = low ; i <= high; ++i) ans += isSymmetric(i);
        return ans;
    }
};