306. Additive Number

Problem

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 {
    public boolean isAdditiveNumber(String num) {
        int n = num.length();
        for (int i = 1; i <= n / 2; ++i) {
            for (int j = 1; Math.max(j, i) <= n - i - j; ++j) {
                if (isValid(i, j, num)) {
                    return true;
                }
            }
        }
        return false;
    }

    private boolean isValid(int i, int j, String num) {
        if (num.charAt(0) == '0' && i > 1) {
            return false;
        }
        if (num.charAt(i) == '0' && j > 1) {
            return false;
        }
        String sum;
        Long x1 = Long.parseLong(num.substring(0, i));
        Long x2 = Long.parseLong(num.substring(i, i + j));
        for (int start = i + j; start != num.length(); start += sum.length()) {
            x2 = x2 + x1;
            x1 = x2 - x1;
            sum = x2.toString();
            if (!num.startsWith(sum, start)) {
                return false;
            }
        }
        return true;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).