gpt-o1-mini
Intro\n\ngpt-o1-mini generator\n\n\n\n\n# Code\nc []\nint minValidStrings(char** words, int wordsSize, char* target) {\n    int lenTarget = strlen(target);\n   
# Intro\n\ngpt-o1-mini generator\n\n![\u622A\u5C4F2024-09-17 21.23.42.png](https://assets.leetcode.com/users/images/c24f2720-4709-45ae-82e7-3d144c5c1ef7_1726579457.184184.png)\n\n\n# Code\n```c []\nint minValidStrings(char** words, int wordsSize, char* target) {\n    int lenTarget = strlen(target);\n    // Using uint16_t for best to store prefix lengths efficiently\n    uint16_t best[lenTarget];\n    memset(best, 0, lenTarget * sizeof(uint16_t));\n    \n    // Find the length of the longest word to allocate kmp array on the stack once\n    int lenLongestWord = 0;\n    for (int i = 0; i < wordsSize; i++) {\n        int lenWord = strlen(words[i]);\n        if (lenWord > lenLongestWord) lenLongestWord = lenWord;\n    }\n    \n    // Allocate kmp array on the stack\n    uint16_t kmp[lenLongestWord + 1];\n    \n    for (int w = 0; w < wordsSize; w++) {\n        char* word = words[w];\n        int lenWord = strlen(word);\n        \n        // Build KMP failure function for the current word\n        kmp[0] = 0;\n        int j = 0;\n        for (int i = 1; i < lenWord; i++) {\n            while (j > 0 && word[i] != word[j]) {\n                j = kmp[j - 1];\n            }\n            if (word[i] == word[j]) {\n                j++;\n            }\n            kmp[i] = j;\n        }\n        \n        // Perform KMP search to find prefixes in the target\n        j = 0;\n        for (int i = 0; i < lenTarget;) {\n            if (word[j] == target[i]) {\n                j++;\n                i++;\n                if (j > best[i - 1]) {\n                    best[i - 1] = j;\n                }\n                if (j == lenWord) {\n                    j = kmp[j - 1];\n                }\n            } else {\n                if (j > 0) {\n                    j = kmp[j - 1];\n                } else {\n                    i++;\n                }\n            }\n        }\n    }\n    \n    // Greedy approach to determine the minimum number of valid strings\n    int steps = 0;\n    int pos = lenTarget;\n    while (pos > 0) {\n        if (best[pos - 1] == 0) {\n            return -1;\n        }\n        pos -= best[pos - 1];\n        steps++;\n    }\n    \n    return steps;\n}\n```
['C']