Dada una string S que consta de N alfabetos en minúsculas, la tarea es modificar la string S reemplazando cada carácter con el alfabeto cuya distancia circular desde el carácter es igual a la frecuencia del carácter en S .
Ejemplos:
Entrada: S = “geeks”
Salida: hgglt
Explicación:
Las siguientes modificaciones se realizan en la string S:
- La frecuencia de ‘g’ en la string es 1. Por lo tanto, ‘g’ se reemplaza por ‘h’.
- La frecuencia de ‘e’ en la string es 2. Por lo tanto, ‘e’ se reemplaza por ‘g’.
- La frecuencia de ‘e’ en la string es 2. Por lo tanto, ‘e’ se reemplaza por ‘g’.
- La frecuencia de ‘k’ en la string es 1. Por lo tanto, ‘k’ se convierte en ‘k’ + 1 = ‘l’.
- La frecuencia de ‘s’ en la string es 1. Por lo tanto, ‘s’ se convierte en ‘s’ + 1 = ‘t’.
Por lo tanto, la string S modificada es «hgglt».
Entrada: S = “jazz”
Salida: “kbbb”
Enfoque: el problema dado se puede resolver manteniendo la array de frecuencias que almacena las ocurrencias de cada carácter en la string . Siga los pasos a continuación para resolver el problema:
- Inicialice una array , freq[26] inicialmente con todos los elementos como 0 para almacenar la frecuencia de cada carácter de la string .
- Recorre la string S dada e incrementa la frecuencia de cada carácter S[i] en 1 en la array freq[] .
- Recorra la string S usó la variable i y realizó los siguientes pasos:
- Almacene el valor que se agregará a S[i] en una variable, agregue como (freq[i] % 26) .
- Si, después de agregar el valor de add a S[i] , S[i] no excede el carácter z , entonces actualice S[i] a S[i] + add .
- De lo contrario, actualice el valor de add a (S[i] + add – z) y luego configure S[i] a (a + add – 1) .
- Después de completar los pasos anteriores, imprima la string modificada S .
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to modify string by replacing
// characters by the alphabet present at
// distance equal to frequency of the string
void addFrequencyToCharacter(string s)
{
// Stores frequency of characters
int frequency[26] = { 0 };
// Stores length of the string
int n = s.size();
// Traverse the given string S
for (int i = 0; i < n; i++) {
// Increment frequency of
// current character by 1
frequency[s[i] - 'a'] += 1;
}
// Traverse the string
for (int i = 0; i < n; i++) {
// Store the value to be added
// to the current character
int add = frequency[s[i] - 'a'] % 26;
// Check if after adding the
// frequency, the character is
// less than 'z' or not
if (int(s[i]) + add <= int('z'))
s[i] = char(int(s[i]) + add);
// Otherwise, update the value of
// add so that s[i] doesn't exceed 'z'
else {
add = (int(s[i]) + add) - (int('z'));
s[i] = char(int('a') + add - 1);
}
}
// Print the modified string
cout << s;
}
// Driver Code
int main()
{
string str = "geeks";
addFrequencyToCharacter(str);
return 0;
}
Java
// Java program for the above approach
class GFG{
// Function to modify string by replacing
// characters by the alphabet present at
// distance equal to frequency of the string
static void addFrequencyToCharacter(char[] s)
{
// Stores frequency of characters
int frequency[] = new int[26];
// Stores length of the string
int n = s.length;
// Traverse the given string S
for(int i = 0; i < n; i++)
{
// Increment frequency of
// current character by 1
frequency[s[i] - 'a'] += 1;
}
// Traverse the string
for(int i = 0; i < n; i++)
{
// Store the value to be added
// to the current character
int add = frequency[s[i] - 'a'] % 26;
// Check if after adding the
// frequency, the character is
// less than 'z' or not
if ((int)(s[i]) + add <= (int)('z'))
s[i] = (char)((int)(s[i]) + add);
// Otherwise, update the value of
// add so that s[i] doesn't exceed 'z'
else
{
add = ((int)(s[i]) + add) - ((int)('z'));
s[i] = (char)((int)('a') + add - 1);
}
}
// Print the modified string
System.out.println(s);
}
// Driver Code
public static void main(String[] args)
{
String str = "geeks";
addFrequencyToCharacter(str.toCharArray());
}
}
// This code is contributed by AnkThon
Python3
# Python3 program for the above approach
# Function to modify string by replacing
# characters by the alphabet present at
# distance equal to frequency of the string
def addFrequencyToCharacter(s):
# Stores frequency of characters
frequency = [0] * 26
# Stores length of the string
n = len(s)
# Traverse the given string S
for i in range(n):
# Increment frequency of
# current character by 1
frequency[ord(s[i]) - ord('a')] += 1
# Traverse the string
for i in range(n):
# Store the value to be added
# to the current character
add = frequency[ord(s[i]) - ord('a')] % 26
# Check if after adding the
# frequency, the character is
# less than 'z' or not
if (ord(s[i]) + add <= ord('z')):
s[i] = chr(ord(s[i]) + add)
# Otherwise, update the value of
# add so that s[i] doesn't exceed 'z'
else:
add = (ord(s[i]) + add) - (ord('z'))
s[i] = chr(ord('a') + add - 1)
# Print the modified string
print("".join(s))
# Driver Code
if __name__ == '__main__':
str = "geeks"
addFrequencyToCharacter([i for i in str])
# This code is contributed by mohit kumar 29
C#
// C# program for the above approach
using System;
class GFG{
// Function to modify string by replacing
// characters by the alphabet present at
// distance equal to frequency of the string
static void addFrequencyToCharacter(char[] s)
{
// Stores frequency of characters
int[] frequency = new int[26];
// Stores length of the string
int n = s.Length;
// Traverse the given string S
for(int i = 0; i < n; i++)
{
// Increment frequency of
// current character by 1
frequency[s[i] - 'a'] += 1;
}
// Traverse the string
for(int i = 0; i < n; i++)
{
// Store the value to be added
// to the current character
int add = frequency[s[i] - 'a'] % 26;
// Check if after adding the
// frequency, the character is
// less than 'z' or not
if ((int)(s[i]) + add <= (int)('z'))
s[i] = (char)((int)(s[i]) + add);
// Otherwise, update the value of
// add so that s[i] doesn't exceed 'z'
else
{
add = ((int)(s[i]) + add) - ((int)('z'));
s[i] = (char)((int)('a') + add - 1);
}
}
// Print the modified string
Console.WriteLine(s);
}
// Driver Code
public static void Main(string[] args)
{
string str = "geeks";
addFrequencyToCharacter(str.ToCharArray());
}
}
// This code is contributed by ukasp
Javascript
<script>
// JavaScript program for the above approach
// Function to modify string by replacing
// characters by the alphabet present at
// distance equal to frequency of the string
function addFrequencyToCharacter(s) {
// Stores frequency of characters
var frequency = new Array(26).fill(0);
// Stores length of the string
var n = s.length;
// Traverse the given string S
for (var i = 0; i < n; i++) {
// Increment frequency of
// current character by 1
frequency[s[i].charCodeAt(0) - "a".charCodeAt(0)] += 1;
}
// Traverse the string
for (var i = 0; i < n; i++) {
// Store the value to be added
// to the current character
var add = frequency[s[i].charCodeAt(0) - "a".charCodeAt(0)] % 26;
// Check if after adding the
// frequency, the character is
// less than 'z' or not
if (s[i].charCodeAt(0) + add <= "z".charCodeAt(0))
s[i] = String.fromCharCode(s[i].charCodeAt(0) + add);
// Otherwise, update the value of
// add so that s[i] doesn't exceed 'z'
else {
add = s[i].charCodeAt(0) + add - "z".charCodeAt(0);
s[i] = String.fromCharCode("a".charCodeAt(0) + add - 1);
}
}
// Print the modified string
document.write(s.join("") + "<br>");
}
// Driver Code
var str = "geeks";
addFrequencyToCharacter(str.split(""));
</script>
Producción:
hgglt
Complejidad temporal: O(N)
Espacio auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por coder_nero y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA