Additive Number LeetCode Solution

Problem – Additive Number LeetCode Solution

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?

Additive Number LeetCode Solution in Java

import java.math.BigInteger;

public class Solution {
    public boolean isAdditiveNumber(String num) {
        int n = num.length();
        for (int i = 1; i <= n / 2; ++i) {
            if (num.charAt(0) == '0' && i > 1) return false;
            BigInteger x1 = new BigInteger(num.substring(0, i));
            for (int j = 1; Math.max(j, i) <= n - i - j; ++j) {
                if (num.charAt(i) == '0' && j > 1) break;
                BigInteger x2 = new BigInteger(num.substring(i, i + j));
                if (isValid(x1, x2, j + i, num)) return true;
            }
        }
        return false;
    }
    private boolean isValid(BigInteger x1, BigInteger x2, int start, String num) {
        if (start == num.length()) return true;
        x2 = x2.add(x1);
        x1 = x2.subtract(x1);
        String sum = x2.toString();
        return num.startsWith(sum, start) && isValid(x1, x2, start + sum.length(), num);
    }
}

Additive Number LeetCode Solution in C++

class Solution {
public:
        bool isAdditiveNumber(string num) {
            for(int i=1; i<=num.size()/2; i++){
                for(int j=1; j<=(num.size()-i)/2; j++){
                    if(check(num.substr(0,i), num.substr(i,j), num.substr(i+j))) return true;
                }
            }
            return false;
        }
        bool check(string num1, string num2, string num){
            if(num1.size()>1 && num1[0]=='0' || num2.size()>1 && num2[0]=='0') return false;
            string sum=add(num1, num2);
            if(num==sum) return true;
            if(num.size()<=sum.size() || sum.compare(num.substr(0,sum.size()))!=0) return false;
            else return check(num2, sum, num.substr(sum.size()));
        } 
        string add(string n, string m){
            string res;
            int i=n.size()-1, j=m.size()-1, carry=0;
            while(i>=0 || j>=0){
                int sum=carry+(i>=0 ? (n[i--]-'0') : 0) + (j>=0?  (m[j--]-'0') : 0);
                res.push_back(sum%10+'0');
                carry=sum/10;
            }
            if(carry) res.push_back(carry+'0');
            reverse(res.begin(), res.end());
            return res;
        }
    };

Additive Number LeetCode Solution in Python

def isAdditiveNumber(self, num):
    n = len(num)
    for i, j in itertools.combinations(range(1, n), 2):
        a, b = num[:i], num[i:j]
        if b != str(int(b)):
            continue
        while j < n:
            c = str(int(a) + int(b))
            if not num.startswith(c, j):
                break
            j += len(c)
            a, b = b, c
        if j == n:
            return True
    return False

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