Leetcode Additive Number problem solution

class Solution:
    def isAdditiveNumber(self, num: str) -> bool:
        
        def ableToSeeTheLight(n0, n1, num_str):
            s = str(n0+n1)
            #print("n0", n0, "n1", n1, "s", s, "num", num_str)
            if not num_str and num.endswith(str(n1)):
                return True
            
            if num_str.startswith(s):
                return ableToSeeTheLight(n1, int(s), num_str[len(s):])
            else:
                return False

        for i in range(1, len(num)):
            if i > 1 and num[0] == '0':
                continue
            for j in range(i + 1, len(num)):
                if j - i > 1 and num[i] == '0':
                    continue
                
                n0 = int(num[0:i])
                n1 = int(num[i:j])
                if ableToSeeTheLight(n0, n1, num[j:]):
                    return True
        return False

class Solution {
    public boolean isAdditiveNumber(String num) {
        int n = num.length();
        if (n < 3) return false;
        for (int i = 1; i <= n / 2; i++) {
            if (i > 1 && num.charAt(0) == '0') break;
            for (int j = i + 1; j < n; j++) {
                if (j > i + 1 && num.charAt(i) == '0') break;
                long first = Long.parseLong(num.substring(0, i));
                long second = Long.parseLong(num.substring(i, j));
                int k = j;
                while (k < n) {
                    long target = first + second;
                    String s = String.valueOf(target);
                    if (k + s.length() <= n && num.substring(k, k + s.length()).equals(s)) {
                        k += s.length();
                        first = second;
                        second = target;
                    }
                    else break;
                }
                if (k == n) return true;
            }
        }
        return false;
    }
}

class Solution {
public:
    vector<long long> path;
    bool dfs(string& s, int start) {
        int sz = path.size();
        if (start == s.length() && path.size() >= 3) {
            return true;
        }
        long long n = 0, max_len;
        max_len = (s[start] == '0' ? 1 : 15); 
        for (int i = start; i < s.length() && max_len -- ; i++) {
            n = n * 10 + (s[i] - '0');
            if (sz >= 2 &&  path[sz - 1] + path[sz - 2] < n) return false;
            else if (sz <= 1|| path[sz - 1] + path[sz - 2] == n) {
                path.push_back(n);
                if (dfs(s, i + 1)) return true;
                path.pop_back();
            }
        }
        return false;
    }
    
    bool isAdditiveNumber(string& num) {
        return dfs(num, 0) ;
    }
};

bool check(char* num, char* num1, char* num2, int last_idx){
    int num1_len = strlen(num1), num2_len = strlen(num2);
    if(num1_len>1 && num1[0] == '0' || num2_len>1 && num2[0] == '0' || last_idx > strlen(num)-1)
        return false;
    long long n1 = strtol(num1,NULL,10); 
    long long n2 = strtol(num2,NULL,10);
    long long sum = n1+n2;
    int sum_len = 0;
    while(sum>0){
        sum_len+=1;
        sum/=10;
    }
    if(last_idx+sum_len>strlen(num)-1)
        return false;
    char* sum1 = (char*)calloc(sum_len+1,sizeof(char));
    memcpy(sum1,num+last_idx+1,sizeof(char)*sum_len);
    sum1[sum_len] = '';
    sum = strtol(sum1,NULL,10);
    if(sum!=(n1+n2))
        return false;
    if(last_idx+sum_len == strlen(num)-1)
        return true;
    if(sum_len == 0)// for "000" test case
        sum_len = 1;
    return check(num,num2,sum1,last_idx+sum_len);
}

bool isAdditiveNumber(char * num){
    int len = strlen(num);
    for(int i=1;i<=len/2;i++){
        char* num1 = (char*)calloc(i+1,sizeof(char));
        memcpy(num1,num,sizeof(char)*i);
        num1[i] = '';
        for(int j=1;j<=(len-i)/2;j++){
            char* num2 = (char*)calloc(j+1,sizeof(char));
            memcpy(num2,num+i,sizeof(char)*j);
            num2[j] = '';
            if(check(num,num1,num2,i+j-1))
                return true;
        }
    }
    return false;
}