/* In this challenge, you will be given a list of letter heights in the alphabet and a string. Using the letter heights given, determine the area of the rectangle highlight in assuming all letters are 1 mm wide. For example, the highlighted word = torn. Assume the heights of the letters are t = 2, r = 1, 0 = 1 and n = 1. The tallest letter is 2 high and there are 4 letters. The hightlighted area will be 2 * 4 = 8 mm^2 so the answer is 8. Function Description Complete the designerPdfViewer function in the editor below. It should return an integer representing the size of the highlighted area. designerPdfViewer has the following parameter(s): h: an array of integers representing the heights of each letter word: a string */ /* The implemented algorithm takes O(n) where n is the size of the word. It uses a very basic hashing technique to determine the height of the letters in the provided word. This look up takes constant time. If the word is way too long, the algorithm could be improved by first scanning the alphabet vector trying to find the highest letter. Once done, scan the word just like here, but if it finds the highest letter, then it can break out of the loop. Asympotically speaking it would still be O(n) though. */ #include #include #include int hash_char(char c) { return ((int)(c)) - 97; } // Complete the designerPdfViewer function below. int designerPdfViewer(std::vector& h, const std::string& word) { int max_height = h[hash_char(word[0])]; for (int i = 1; i < word.size(); ++i) { int height = h[hash_char(word[i])]; if (height > max_height) { max_height = height; } } int highlighted_space = word.size() * max_height; return highlighted_space; } int main() { std::vector alphabet = {1 ,3, 1, 3, 1, 4, 1, 3, 2, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 7 }; std::string word = "zaba"; int size = designerPdfViewer(alphabet, word); std::cout << "Size of highlighted word is: " << size << std::endl; return 0; }

Highlighted Word