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 <= 35
• num consists 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
    }
}

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