Dada una array arr[] que contiene el costo de agregar cada alfabeto de (a – z) y una string str que puede o no ser un Panagram . La tarea es encontrar el costo total para hacer el Panagrama de cuerdas.
Ejemplos:
Entrada: arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22 , 23, 24, 25, 26},
str = “abcdefghijklmopqrstuvwz ”
Salida: 63
n, xey son los únicos caracteres que faltan para convertir la string en un pangrama.
Y sus costes son 14, 24 y 25 respectivamente.
Por lo tanto, costo total = 14 + 24 + 25 = 63Entrada: arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22 , 23, 24, 25, 26},
str = “thequickbrownfoxjumpsoverthelazydog”
Salida: 0
La string ya es un Pangram ya que contiene todos los caracteres de (a – z).
Enfoque: marque todos los alfabetos desde (a – z) que ocurrieron en str y luego encuentre los alfabetos que faltan en la string. Sume el costo de estos alfabetos que faltan para obtener el costo total requerido para hacer el pangrama de cuerdas.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to find the cost to make a string Panagram
#include <bits/stdc++.h>
using namespace std;
// Function to return the total cost required
// to make the string Pangram
int pangramCost(int arr[], string str)
{
int cost = 0;
bool occurred[26] = { false };
// Mark all the alphabets that occurred in the string
for (int i = 0; i < str.size(); i++)
occurred[str[i] - 'a'] = true;
// Calculate the total cost for the missing alphabets
for (int i = 0; i < 26; i++) {
if (!occurred[i])
cost += arr[i];
}
return cost;
}
// Driver Code
int main()
{
int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15,
16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26 };
string str = "abcdefghijklmopqrstuvwz";
cout << pangramCost(arr, str);
return 0;
}
Java
// Java program to find the cost to make a string Panagram
class GFG
{
// Function to return the total cost required
// to make the string Pangram
static int pangramCost(int arr[], String str)
{
int cost = 0;
boolean []occurred=new boolean[26];
for(int i=0;i<26;i++)
occurred[i]=false;
// Mark all the alphabets that occurred in the string
for (int i = 0; i < str.length(); i++)
occurred[str.charAt(i) - 'a'] = true;
// Calculate the total cost for the missing alphabets
for (int i = 0; i < 26; i++) {
if (occurred[i]==false)
cost += arr[i];
}
return cost;
}
// Driver Code
public static void main(String []args)
{
int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15,
16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26 };
String str = "abcdefghijklmopqrstuvwz";
System.out.println(pangramCost(arr, str));
}
}
// This code is contributed by ihritik
Python3
# Python3 program to find the cost
# to make a string Panagram
# Function to return the total cost
# required to make the string Pangram
def pangramCost(arr, string) :
cost = 0
occurred = [False] * 26
# Mark all the alphabets that
# occurred in the string
for i in range(len(string)) :
occurred[ord(string[i]) - ord('a')] = True
# Calculate the total cost for
# the missing alphabets
for i in range(26) :
if (not occurred[i]) :
cost += arr[i]
return cost
# Driver Code
if __name__ == "__main__" :
arr = [ 1, 2, 3, 4, 5, 6, 7, 8, 9,
10, 11, 12, 13, 14, 15, 16,
17, 18, 19, 20, 21, 22, 23,
24, 25, 26 ]
string = "abcdefghijklmopqrstuvwz"
print(pangramCost(arr, string))
# This code is contributed by Ryuga
C#
// C# program to find the cost to make a string Panagram
using System;
class GFG
{
// Function to return the total cost required
// to make the string Pangram
static int pangramCost(int []arr, string str)
{
int cost = 0;
bool []occurred=new bool[26];
for(int i=0;i<26;i++)
occurred[i]=false;
// Mark all the alphabets that occurred in the string
for (int i = 0; i < str.Length; i++)
occurred[str[i] - 'a'] = true;
// Calculate the total cost for the missing alphabets
for (int i = 0; i < 26; i++) {
if (occurred[i]==false)
cost += arr[i];
}
return cost;
}
// Driver Code
public static void Main(string []args)
{
int []arr = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15,
16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26 };
string str = "abcdefghijklmopqrstuvwz";
Console.WriteLine(pangramCost(arr, str));
}
}
// This code is contributed by ihritik
PHP
<?php
// PHP program to find the cost to make
// a string Panagram
// Function to return the total cost
// required to make the string Pangram
function pangramCost($arr, $str)
{
$cost = 0;
$occurred = array();
for($i = 0; $i < 26; $i++)
$occurred[$i] = false;
// Mark all the alphabets that occurred
// in the string
for($i = 0; $i < strlen($str); $i++)
{
$idx = ord($str[$i]) - 97;
$occurred[$idx] = true;
}
// Calculate the total cost for the
// missing alphabets
for($i = 0; $i < 26; $i++)
{
if ($occurred[$i] == false)
$cost += $arr[$i];
}
return $cost;
}
// Driver Code
$arr = array( 1, 2, 3, 4, 5, 6, 7, 8, 9, 10,
11, 12, 13, 14, 15, 16, 17, 18,
19, 20, 21, 22, 23, 24, 25, 26 );
$str = "abcdefghijklmopqrstuvwz";
echo pangramCost($arr, $str);
// This code is contributed by ihritik
?>
Javascript
<script>
// Javascript program to find the cost to
// make a string Panagram
// Function to return the total cost required
// to make the string Pangram
function pangramCost(arr, str)
{
var cost = 0;
var occurred = new Array(26);
for(let i = 0; i < 26; i++)
occurred[i] = false;
// Mark all the alphabets that occurred
// in the string
for(let i = 0; i < str.length; i++)
occurred[str[i].charCodeAt() -
'a'.charCodeAt()] = true;
// Calculate the total cost for
// the missing alphabets
for(let i = 0; i < 26; i++)
{
if (occurred[i] == false)
cost += arr[i];
}
return cost;
}
// Driver Code
var arr = [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10,
11, 12, 13, 14, 15, 16, 17, 18,
19, 20, 21, 22, 23, 24, 25, 26 ];
var str = "abcdefghijklmopqrstuvwz";
document.write(pangramCost(arr, str));
// This code is contributed by bunnyram19
</script>
63
Publicación traducida automáticamente
Artículo escrito por SURENDRA_GANGWAR y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA