C# Solution for Count the Number of Consistent Strings Problem
Intuition\n Describe your first thoughts on how to solve this problem. \nThe problem asks us to check if each word in the words array contains only characters f
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem asks us to check if each word in the words array contains only characters from the allowed string. In this C# solution, we take advantage of bit manipulation to efficiently track which characters are allowed. Each lowercase letter can be represented by a bit in an integer (since there are 26 lowercase letters, a 32-bit integer is sufficient). This enables fast comparisons to check if a word is \u201Cconsistent\u201D (i.e., composed only of allowed characters).\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1.\tBitmask Representation:\n\t\u2022\tEach character from \u2018a\u2019 to \u2018z\u2019 can be mapped to a bit in a 32-bit integer.\n\t\u2022\tFor the allowed string, we create a bitmask where each bit represents the presence of a particular character. For example, if allowed = "abc", then the bitmask will have the first three bits set to 1, corresponding to \u2018a\u2019, \u2018b\u2019, and \u2018c\u2019.\n2.\tGenerate Bitmask for Each Word:\n\t\u2022\tFor each word in the words array, we generate its bitmask by setting bits for each character in the word.\n3.\tCheck Consistency:\n\t\u2022\tA word is consistent if all the characters in the word are also in the allowed string. In terms of bit manipulation, this means the word\u2019s bitmask should be a subset of the allowed bitmask. We check this using the condition (wordMask & allowedMask) == wordMask, where wordMask represents the bitmask of the word.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(n * m) where:\n\t\u2022\tn is the number of words in the words array.\n\t\u2022\tm is the average length of each word.\nExplanation:\n\t\u2022\tCreating the allowedMask takes O(k), where k is the length of the allowed string. Since k is at most 26 (the number of letters in the alphabet), this step is O(1).\n\t\u2022\tFor each word, we compute its bitmask by iterating through its characters, which takes O(m) time where m is the length of the word. We repeat this for all n words, resulting in O(n * m) total time.\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nO(1): The space complexity is constant because:\n\t\u2022\tWe only use a few integer variables (e.g., allowedMask, wordMask) to store the bitmasks, which do not scale with the input size.\n\t\u2022\tThere are no additional data structures (like arrays or sets) that grow with the input size.\n\n# Code\n```csharp []\npublic class Solution {\n    public int CountConsistentStrings(string allowed, string[] words) {\n        // Step 1: Generate bitmask for allowed characters\n        int allowedMask = 0;\n        foreach (char c in allowed) {\n            allowedMask |= (1 << (c - \'a\'));  // Set the bit for the character\n        }\n\n        int consistentCount = 0;\n\n        // Step 2: Iterate through each word\n        foreach (string word in words) {\n            int wordMask = 0;\n\n            // Step 3: Generate bitmask for the current word\n            foreach (char c in word) {\n                wordMask |= (1 << (c - \'a\'));  // Set the bit for the character\n            }\n\n            // Step 4: Check if the word is consistent\n            if ((wordMask & allowedMask) == wordMask) {\n                consistentCount++;  // The word is consistent\n            }\n        }\n\n        return consistentCount;\n    }\n}\n```
['C#']