LeetCode-in-Java
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
Solution
public class Solution {
public int countSymmetricIntegers(int low, int high) {
int count = 0;
for (int i = low; i <= high; i++) {
if (isSymmetric(i)) {
count++;
}
}
return count;
}
private boolean isSymmetric(int num) {
String str = String.valueOf(num);
int n = str.length();
if (n % 2 != 0) {
return false;
}
int leftSum = 0;
int rightSum = 0;
for (int i = 0, j = n - 1; i < j; i++, j--) {
leftSum += str.charAt(i) - '0';
rightSum += str.charAt(j) - '0';
}
return leftSum == rightSum;
}
}