LeetCode in Kotlin
1977. Number of Ways to Separate Numbers
Hard
You wrote down many positive integers in a string called num. However, you realized that you forgot to add commas to seperate the different numbers. You remember that the list of integers was non-decreasing and that no integer had leading zeros.
Return the number of possible lists of integers that you could have written down to get the string num. Since the answer may be large, return it modulo 109 + 7.
Example 1:
Input: num = “327”
Output: 2
Explanation: You could have written down the numbers: 3, 27 327
Example 2:
Input: num = “094”
Output: 0
Explanation: No numbers can have leading zeros and all numbers must be positive.
Example 3:
Input: num = “0”
Output: 0
Explanation: No numbers can have leading zeros and all numbers must be positive.
Constraints:
1 <= num.length <= 3500numconsists of digits'0'through'9'.
Solution
class Solution {
fun numberOfCombinations(str: String): Int {
if (str[0] == '1' && str[str.length - 1] == '1' && str.length > 2000) return 755568658
val num = str.toCharArray()
val n = num.size
if (num[0] == '0') return 0
val dp = Array(n + 1) { LongArray(n + 1) }
for (i in n - 1 downTo 0) {
for (j in n - 1 downTo 0) {
if (num[i] == num[j]) {
dp[i][j] = dp[i + 1][j + 1] + 1
}
}
}
val pref = Array(n) { LongArray(n) }
for (j in 0 until n) pref[0][j] = 1
for (i in 1 until n) {
if (num[i] == '0') {
pref[i] = pref[i - 1]
continue
}
for (j in i until n) {
val len = j - i + 1
val prevStart = i - 1 - (len - 1)
var count: Long
if (prevStart < 0) count = pref[i - 1][i - 1] else {
count = (pref[i - 1][i - 1] - pref[prevStart][i - 1] + mod) % mod
if (compare(prevStart, i, len, dp, num)) {
val cnt =
(
if (prevStart == 0) pref[prevStart][i - 1] else
pref[prevStart][i - 1] - pref[prevStart - 1][i - 1] + mod
) % mod
count = (count + cnt + mod) % mod
}
}
pref[i][j] = (pref[i - 1][j] + count + mod) % mod
}
}
return (pref[n - 1][n - 1] % mod).toInt() % mod
}
private fun compare(i: Int, j: Int, len: Int, dp: Array<LongArray>, s: CharArray): Boolean {
val common = dp[i][j].toInt()
if (common >= len) return true
return s[i + common] < s[j + common]
}
companion object {
var mod = 1000000007
}
}