Additive Number

Desicription

Additive number is a string whose digits can form 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 ‘0’-‘9’, write a function to determine if it’s an additive number.

Note: Numbers in the additive sequence cannot have leading zeros, so sequence 1, 2, 03 or 1, 02, 3 is invalid.

Example 1:

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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:

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Input: "199100199"
Output: true
Explanation: The additive sequence is: 1, 99, 100, 199.
1 + 99 = 100, 99 + 100 = 199

Follow up:

How would you handle overflow for very large input integers?

Solution

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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 != num.substr(0, sum.size())) {
return false;
}
return check(num2, sum, num.substr(sum.size()));
}

string add(string a, string b) {
int i = a.size() - 1;
int j = b.size() - 1;
int carry = 0;
string res;
while(i >= 0 || j >= 0) {
int sum = carry + (i >= 0 ? a[i--] - '0' : 0) + (j >= 0 ? b[j--] - '0' : 0);
res += sum % 10 + '0';
carry = sum / 10;
}
if(carry != 0) {
res += carry + '0';
}
reverse(res.begin(), res.end());
return res;
}
};