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## 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 {
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;
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:
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;
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 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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