**Designer PDF Viewer**

### Problem Statement :

When a contiguous block of text is selected in a PDF viewer, the selection is highlighted with a blue rectangle. In this PDF viewer, each word is highlighted independently. For example: There is a list of 26 character heights aligned by index to their letters. For example, 'a' is at index 0 and 'z' is at index 05. There will also be a string. Using the letter heights given, determine the area of the rectangle highlight in mm^2 assuming all letters are 1mm wide. Example h = [1, 3, 1, 3, 1, 4, 1, 3, 2, 5, 5, 5, 5, 1, 1, 5, 5, 1, 5, 2, 5, 5, 5, 5, 5, 5] word = 'torn' The heights are t = 2, o = 1, r=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. designerPdfViewer has the following parameter(s): int h[26]: the heights of each letter string word: a string Returns int: the size of the highlighted area Input Format The first line contains 26 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 1 <= h[?] <= 7, where ? is an English lowercase letter. word contains no more than 10 letters.

### Solution :

` ````
Solution in C :
python 3 :
#!/bin/python3
import sys
h = [int(h_temp) for h_temp in input().strip().split(' ')]
word = input().strip()
max_height = 0
for i in range (len(word)):
num = ord(word[i])-97
if (h[num] >= max_height):
max_height = h[num]
area = len(word)*max_height
print(area)
Java :
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = 26;
int h[] = new int[n];
for(int h_i=0; h_i < n; h_i++){
h[h_i] = in.nextInt();
}
String word = in.next();
int maxHeight = 0;
for(char ch = 'a'; ch < 'z'; ch++) {
if(word.contains(ch + "") && h[ch - 'a'] > maxHeight)
maxHeight = h[ch - 'a'];
}
System.out.println(maxHeight * word.length());
}
}
C ++ :
#include <bits/stdc++.h>
using namespace std;
int a[42];
char s[1231212];
int main() {
for (int i = 0; i < 26; i++) {
scanf("%d", a + i);
}
scanf("%s", s);
int h = 0;
int w = 0;
for (int i = 0; s[i]; i++) {
w++;
h = max(h, a[s[i] - 'a']);
}
printf("%d\n", h * w);
return 0;
}
C :
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>
int main(){
int n=26;
int *h = malloc(sizeof(int) * n);
for(int h_i = 0; h_i < n; h_i++){
scanf("%d",&h[h_i]);
}
char* word = (char *)malloc(512000 * sizeof(char));
scanf("%s",word);
int l=strlen(word);
int i;
int max=0;
for(i=0;i<l;i++)
{
if(h[(int)word[i]-'a']>max)
max=h[(int)word[i]-'a'];
}
printf("%d",max*l);
return 0;
}
```

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