Dada una string ‘S’ que contiene vocales y consonantes de alfabetos ingleses en minúsculas. La tarea es encontrar el número de formas en que los caracteres de la palabra se pueden organizar de manera que las vocales ocupen solo las posiciones impares.
Ejemplos:
Entrada: geeks
Salida: 36
Entrada: publicar
Salida: 1440
Acercarse:
Primero encuentre el número total. de lugares impares y lugares pares en la palabra dada.
Número total de lugares pares = piso (longitud de palabra/2)
Número total de lugares impares = longitud de palabra – total de lugares pares
Consideremos la string «contribuir», entonces hay 10 letras en la palabra dada y hay 5 lugares impares, 5 pares lugares, 4 vocales y 6 consonantes.
Marquemos estas posiciones como abajo:
(1) (2) (3) (4) (5) (6) (7) (8) (9) (10)
Ahora, 4 vocales se pueden colocar en cualquiera de los cinco lugares, marcados 1, 3, 5, 7, 9.
El número de formas de ordenar las vocales = 5_P_4 = 5! = 120
Además, las 6 consonantes se pueden organizar en las 6 posiciones restantes.
Número de formas de estos arreglos = 6_P_6 = 6! = 720.
Número total de vías = (120 x 720) = 86400
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to find the number of ways // in which the characters of the word // can be arranged such that the vowels // occupy only the odd positions #include <bits/stdc++.h> using namespace std; // Function to return the // factorial of a number int fact(int n) { int f = 1; for (int i = 2; i <= n; i++) { f = f * i; } return f; } // calculating nPr int npr(int n, int r) { return fact(n) / fact(n - r); } // Function to find the number of ways // in which the characters of the word // can be arranged such that the vowels // occupy only the odd positions int countPermutations(string str) { // Get total even positions int even = floor(str.length() / 2); // Get total odd positions int odd = str.length() - even; int ways = 0; // Store frequency of each character of // the string int freq[26] = { 0 }; for (int i = 0; i < str.length(); i++) { ++freq[str[i] - 'a']; } // Count total number of vowels int nvowels = freq[0] + freq[4] + freq[8] + freq[14] + freq[20]; // Count total number of consonants int nconsonants = str.length() - nvowels; // Calculate the total number of ways ways = npr(odd, nvowels) * npr(nconsonants, nconsonants); return ways; } // Driver code int main() { string str = "geeks"; cout << countPermutations(str); return 0; }
Java
// Java program to find the number of ways // in which the characters of the word // can be arranged such that the vowels // occupy only the odd positions class GFG{ // Function to return the // factorial of a number static int fact(int n) { int f = 1; for (int i = 2; i <= n; i++) { f = f * i; } return f; } // calculating nPr static int npr(int n, int r) { return fact(n) / fact(n - r); } // Function to find the number of ways // in which the characters of the word // can be arranged such that the vowels // occupy only the odd positions static int countPermutations(String str) { // Get total even positions int even = (int)Math.floor((double)(str.length() / 2)); // Get total odd positions int odd = str.length() - even; int ways = 0; // Store frequency of each character of // the string int[] freq=new int[26]; for (int i = 0; i < str.length(); i++) { freq[(int)(str.charAt(i)-'a')]++; } // Count total number of vowels int nvowels= freq[0] + freq[4]+ freq[8] + freq[14]+ freq[20]; // Count total number of consonants int nconsonants= str.length() - nvowels; // Calculate the total number of ways ways = npr(odd, nvowels) * npr(nconsonants, nconsonants); return ways; } // Driver code public static void main(String[] args) { String str = "geeks"; System.out.println(countPermutations(str)); } } // This code is contributed by mits
Python3
# Python3 program to find the number # of ways in which the characters # of the word can be arranged such # that the vowels occupy only the # odd positions import math # Function to return the factorial # of a number def fact(n): f = 1; for i in range(2, n + 1): f = f * i; return f; # calculating nPr def npr(n, r): return fact(n) / fact(n - r); # Function to find the number of # ways in which the characters of # the word can be arranged such # that the vowels occupy only the # odd positions def countPermutations(str): # Get total even positions even = math.floor(len(str) / 2); # Get total odd positions odd = len(str) - even; ways = 0; # Store frequency of each # character of the string freq = [0] * 26; for i in range(len(str)): freq[ord(str[i]) - ord('a')] += 1; # Count total number of vowels nvowels = (freq[0] + freq[4] + freq[8] + freq[14] + freq[20]); # Count total number of consonants nconsonants = len(str) - nvowels; # Calculate the total number of ways ways = (npr(odd, nvowels) * npr(nconsonants, nconsonants)); return int(ways); # Driver code str = "geeks"; print(countPermutations(str)); # This code is contributed by mits
C#
// C# program to find the number of ways // in which the characters of the word // can be arranged such that the vowels // occupy only the odd positions using System; class GFG{ // Function to return the // factorial of a number static int fact(int n) { int f = 1; for (int i = 2; i <= n; i++) { f = f * i; } return f; } // calculating nPr static int npr(int n, int r) { return fact(n) / fact(n - r); } // Function to find the number of ways // in which the characters of the word // can be arranged such that the vowels // occupy only the odd positions static int countPermutations(String str) { // Get total even positions int even = (int)Math.Floor((double)(str.Length / 2)); // Get total odd positions int odd = str.Length - even; int ways = 0; // Store frequency of each character of // the string int[] freq=new int[26]; for (int i = 0; i < str.Length; i++) { freq[(int)(str[i]-'a')]++; } // Count total number of vowels int nvowels= freq[0] + freq[4]+ freq[8] + freq[14]+ freq[20]; // Count total number of consonants int nconsonants= str.Length - nvowels; // Calculate the total number of ways ways = npr(odd, nvowels) * npr(nconsonants, nconsonants); return ways; } // Driver code static void Main() { String str = "geeks"; Console.WriteLine(countPermutations(str)); } } // This code is contributed by mits
PHP
<?php // PHP program to find the number // of ways in which the characters // of the word can be arranged such // that the vowels occupy only the // odd positions // Function to return the // factorial of a number function fact($n) { $f = 1; for ($i = 2; $i <= $n; $i++) { $f = $f * $i; } return $f; } // calculating nPr function npr($n, $r) { return fact($n) / fact($n - $r); } // Function to find the number // of $ways in which the characters // of the word can be arranged such // that the vowels occupy only the // odd positions function countPermutations($str) { // Get total even positions $even = floor(strlen($str)/ 2); // Get total odd positions $odd = strlen($str) - $even; $ways = 0; // Store $frequency of each // character of the string $freq = array_fill(0, 26, 0); for ($i = 0; $i < strlen($str); $i++) { ++$freq[ord($str[$i]) - ord('a')]; } // Count total number of vowels $nvowels= $freq[0] + $freq[4] + $freq[8] + $freq[14] + $freq[20]; // Count total number of consonants $nconsonants= strlen($str) - $nvowels; // Calculate the total number of ways $ways = npr($odd, $nvowels) * npr($nconsonants, $nconsonants); return $ways; } // Driver code $str = "geeks"; echo countPermutations($str); // This code is contributed by mits ?>
Javascript
<script> // Javascript program to find the number // of ways in which the characters // of the word can be arranged such // that the vowels occupy only the // odd positions // Function to return the // factorial of a number function fact(n) { let f = 1; for (let i = 2; i <= n; i++) { f = f * i; } return f; } // calculating nPr function npr(n, r) { return fact(n) / fact(n - r); } // Function to find the number // of ways in which the characters // of the word can be arranged such // that the vowels occupy only the // odd positions function countPermutations(str) { // Get total even positions let even = Math.floor(str.length/ 2); // Get total odd positions let odd = str.length - even; let ways = 0; // Store frequency of each // character of the string let freq = new Array(26).fill(0); for (let i = 0; i < str.length; i++) { ++freq[str.charCodeAt(i) - 'a'.charCodeAt(0)]; } // Count total number of vowels let nvowels= freq[0] + freq[4] + freq[8] + freq[14] + freq[20]; // Count total number of consonants let nconsonants= str.length - nvowels; // Calculate the total number of ways ways = npr(odd, nvowels) * npr(nconsonants, nconsonants); return ways; } // Driver code let str = "geeks"; document.write(countPermutations(str)); // This code is contributed by gfgking </script>
36
Complejidad de tiempo: O(n) donde n es la longitud de la string
Espacio Auxiliar: O(26)
Publicación traducida automáticamente
Artículo escrito por SURENDRA_GANGWAR y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA