LeetCode in Kotlin
1012. Numbers With Repeated Digits
Hard
Given an integer n, return the number of positive integers in the range [1, n] that have at least one repeated digit.
Example 1:
Input: n = 20
Output: 1
Explanation: The only positive number (<= 20) with at least 1 repeated digit is 11.
Example 2:
Input: n = 100
Output: 10
Explanation: The positive numbers (<= 100) with atleast 1 repeated digit are 11, 22, 33, 44, 55, 66, 77, 88, 99, and 100.
Example 3:
Input: n = 1000
Output: 262
Constraints:
1 <= n <= 109
Solution
import kotlin.math.pow
class Solution {
private var noRepeatCount = 0
fun numDupDigitsAtMostN(n: Int): Int {
val nStrLength = n.toString().length
val allNineLength: Int = if (n < 0 || nStrLength < 2) {
return 0
} else if (10.0.pow(nStrLength.toDouble()) - 1 == n.toDouble()) {
nStrLength
} else {
nStrLength - 1
}
for (numberOfDigits in 1..allNineLength) {
noRepeatCount += calcNumberOfNoRepeat(numberOfDigits)
}
if (10.0.pow(nStrLength.toDouble()) - 1 > n) {
val mutations = 10
val hs: HashSet<Int> = HashSet()
for (index1 in 0 until nStrLength) {
var noRepeatCountLocal = 0
hs.clear()
for (index2 in 0 until nStrLength) {
val index2Digit = (
n /
10.0.pow(n.toString().length - (index2 + 1.0)) %
10
).toInt()
if (index2 < index1) {
if (hs.contains(index2Digit)) {
noRepeatCountLocal = 0
break
} else {
hs.add(index2Digit)
}
} else if (index2 == index1) {
noRepeatCountLocal = if (index2 == 0) {
index2Digit - 1
} else {
var inIndex2Range = 0
for (j in hs) {
if (index2 < nStrLength - 1 && j <= index2Digit - 1 ||
index2 == nStrLength - 1 && j <= index2Digit
) {
inIndex2Range++
}
}
if (index2 == nStrLength - 1) {
index2Digit + 1 - inIndex2Range
} else {
index2Digit - inIndex2Range
}
}
} else {
noRepeatCountLocal *= mutations - index2
}
}
if (noRepeatCountLocal > 0) {
noRepeatCount += noRepeatCountLocal
}
}
}
return n - noRepeatCount
}
private fun calcNumberOfNoRepeat(numberOfDigits: Int): Int {
var repeatCount = 0
var mutations = 9
for (i in 0 until numberOfDigits) {
if (i == 0) {
repeatCount = mutations
} else {
repeatCount *= mutations--
}
}
return repeatCount
}
}