LeetCode in Kotlin
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
class Solution {
fun countSymmetricIntegers(low: Int, high: Int): Int {
var count = 0
for (i in low..high) {
if (isSymmetric(i)) {
count++
}
}
return count
}
private fun isSymmetric(num: Int): Boolean {
val str = num.toString()
val n = str.length
if (n % 2 != 0) {
return false
}
var leftSum = 0
var rightSum = 0
var i = 0
var j = n - 1
while (i < j) {
leftSum += str[i].code - '0'.code
rightSum += str[j].code - '0'.code
i++
j--
}
return leftSum == rightSum
}
}