C++ solution using find string | 92.48% Time | 98.77% Space | Brute Force
This is a brute force solution that just tries to find all of the strings by checking each string if it can be made, one letter at a time using the find functio
This is a brute force solution that just tries to find all of the strings by checking each string if it can be made, one letter at a time using the find function. Although the time complexity is O(mn), in practice it is not bad, as a lot of the times, the string in words will not have to go all the way through, and there are only ever 50 string in words at max. However, I am not exactly sure what this does so well to be completely honest. \n![image](https://assets.leetcode.com/users/images/581353d3-8db6-4214-b159-e73d04d4ab3f_1658280885.95077.png)\n```\nbool hasMatches(string &curr, string &s) {\n        int pos = 1;\n        int i = s.find(curr[0]);\n        if (i == -1)\n            return false;\n        while (pos < curr.length()) {\n            i = s.find(curr[pos], i + 1);\n            if (i == -1)\n                return false;\n            pos++;\n        }\n        return true;\n    }\n    \n    int numMatchingSubseq(string s, vector<string>& words) {\n        int count = 0;\n        for (string str : words)\n            if (hasMatches(str, s))\n                count++;\n        \n        return count;\n    }
['C']