"""
When you select a contiguous block of text in a PDF viewer, the selection is highlighted with a blue rectangle. In this PDF viewer, each word is highlighted independently. For example:

PDF-highighting.png

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  wide.

For example, the highlighted . Assume the heights of the letters are  and . The tallest letter is  high and there are  letters. The hightlighted area will be  so the answer is .

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
Input Format

The first line contains  space-separated integers describing the respective heights of each consecutive lowercase English letter, ascii[a-z].
The second line contains a single word, consisting of lowercase English alphabetic letters.

Constraints

, where  is an English lowercase letter.
 contains no more than  letters.
Output Format

Print a single integer denoting the area in  of highlighted rectangle when the given word is selected. Do not print units of measure.

Sample Input 0

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 5
abc
Sample Output 0

9
"""

#!/bin/python3

import math
import os
import random
import re
import sys
from typing import List

# Complete the designerPdfViewer function below.
def designerPdfViewer(h: List[int], word: str) -> int:
    maxHeight = 0
    for char in word:
        height = h[ord(char) - 97]
        if height > maxHeight:
            maxHeight = height
    return maxHeight * len(word)




if __name__ == '__main__':
    fptr = open(os.environ['OUTPUT_PATH'], 'w')

    h = list(map(int, input().rstrip().split()))

    word = input()

    result = designerPdfViewer(h, word)

    fptr.write(str(result) + '\n')

    fptr.close()

Solve by exploiting ascii values
Solve by exploiting ascii values
