01. Panagram Checking

The problem can be found at the following link: Question Link

My Approach

1. Vector Initialization: Create a vector f of size 26, initialized with zeros. This vector is used to keep track of the occurrence of each letter in the alphabet.

   std::vector<int> f(26, 0);

2. Iterate Through the String: Iterate through each character in the input string.

   for (auto i : s) {

3. Convert to Lowercase: Convert the current character to lowercase.

   char cur = std::tolower(i);

4. Check if Character is a Lowercase Letter: If the character is a lowercase letter, update the corresponding index in the vector f to 1.

   if (cur >= 'a' && cur <= 'z') {
       f[cur - 'a'] = 1;
   }

5. Check Pangram Condition: Check if all elements in the vector f are set to 1. This indicates that all lowercase letters are present in the input string.

   return std::accumulate(f.begin(), f.end(), 0) == 26;

Time and Auxiliary Space Complexity

• Time Complexity: The time complexity is O(n), where n is the length of the input string. This is because each character is processed once.

• Auxiliary Space Complexity: The auxiliary space complexity is O(1) since the size of the vector f is constant (26).

Code (C++)

class Solution {
public:
    // Function to check if a string is Pangram or not.
    bool checkPangram(std::string s) {
        std::vector<int> f(26, 0);

        for (auto i : s) {
            char cur = std::tolower(i);

            if (cur >= 'a' && cur <= 'z') {
                f[cur - 'a'] = 1;
            }
        }

        return std::accumulate(f.begin(), f.end(), 0) == 26;
    }
};

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