LeetCode in Kotlin
3747. Count Distinct Integers After Removing Zeros
Medium
You are given a positive integer n.
For every integer x from 1 to n, we write down the integer obtained by removing all zeros from the decimal representation of x.
Return an integer denoting the number of distinct integers written down.
Example 1:
Input: n = 10
Output: 9
Explanation:
The integers we wrote down are 1, 2, 3, 4, 5, 6, 7, 8, 9, 1. There are 9 distinct integers (1, 2, 3, 4, 5, 6, 7, 8, 9).
Example 2:
Input: n = 3
Output: 3
Explanation:
The integers we wrote down are 1, 2, 3. There are 3 distinct integers (1, 2, 3).
Constraints:
1 <= n <= 1015
Solution
class Solution {
fun countDistinct(n: Long): Long {
val digits = n.toString()
val m = digits.length
val power9 = LongArray(m + 1)
power9[0] = 1
for (i in 1..m) {
power9[i] = power9[i - 1] * 9
}
var total: Long = 0
for (length in 1..<m) {
total += power9[length]
}
for (idx in 0..<m) {
val d = digits[idx].code - '0'.code
if (d == 0) {
return total
}
total += (d - 1) * power9[m - idx - 1]
}
return total + 1
}
}