Solution

Method 1 – Simple Substring Check

Intuition

For each pattern, check if it appears as a substring in the given word. This is a direct application of the substring operation and is efficient given the constraints.

Approach

Initialize a counter ans to 0.
For each string in patterns, check if it is a substring of word.
If yes, increment ans.
Return ans.

Code

[{"language":"C++","code":"class Solution {\npublic:\n    int numOfStrings(vector<string>& patterns, string word) {\n        int ans = 0;\n        for (const auto& p : patterns) {\n            if (word.find(p) != string::npos) ++ans;\n        }\n        return ans;\n    }\n};","lang":"cpp","html":"<pre class=\"shiki shiki-themes github-light github-dark\" style=\"background-color:#fff;--shiki-dark-bg:#24292e;color:#24292e;--shiki-dark:#e1e4e8\" tabindex=\"0\"><code><span class=\"line\"><span style=\"color:#D73A49;--shiki-dar

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