LeetCode in Kotlin
306. Additive Number
Medium
An additive number is a string whose digits can form an additive sequence.
A valid additive sequence should contain at least three numbers. Except for the first two numbers, each subsequent number in the sequence must be the sum of the preceding two.
Given a string containing only digits, return true if it is an additive number or false otherwise.
Note: Numbers in the additive sequence cannot have leading zeros, so sequence 1, 2, 03 or 1, 02, 3 is invalid.
Example 1:
Input: “112358”
Output: true
Explanation: The digits can form an additive sequence: 1, 1, 2, 3, 5, 8. 1 + 1 = 2, 1 + 2 = 3, 2 + 3 = 5, 3 + 5 = 8
Example 2:
Input: “199100199”
Output: true
Explanation: The additive sequence is: 1, 99, 100, 199. 1 + 99 = 100, 99 + 100 = 199
Constraints:
1 <= num.length <= 35numconsists only of digits.
Follow up: How would you handle overflow for very large input integers?
Solution
class Solution {
fun isAdditiveNumber(num: String): Boolean {
if (num.isEmpty() || num.length < 3) {
return false
}
fun isInvalid(s: String): Boolean {
return s[0] == '0' && s.length > 1
}
fun backtrack(first: Long, second: Long, startIndex: Int): Boolean {
val third = (first + second).toString()
if (num.substring(startIndex).startsWith(third)) {
if (third.length == num.length - startIndex) return true
return backtrack(second, third.toLong(), startIndex + third.length)
}
return false
}
for (i in 1 until num.length) {
val first = num.substring(0, i)
if (isInvalid(first)) break
for (j in i + 1 until num.length) {
val second = num.substring(i, j)
if (isInvalid(second)) break
if (backtrack(first.toLong(), second.toLong(), j)) return true
}
}
return false
}
}