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.

 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


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.


1 <= h[?] <= 7, where ? is an English lowercase letter.
word contains no more than 10 letters.

Solution :


                            Solution in C :

python 3  :


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

Java  :

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(;
        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 =;
        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++) {
    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++){
    char* word = (char *)malloc(512000 * sizeof(char));
    int l=strlen(word);
    int i;
    int max=0;
    return 0;

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