Given a string ‘S’ containing vowels and consonants of lowercase English alphabets. The task is 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.
Examples:
Input: neveropen
Output: 36Input: publish
Output: 1440
Approach:
- First, find the total no. of odd places and even places in the given word.
- Total number of even places = floor(word length/2)
- Total number of odd places = word length – total even places
- Let’s consider the string “contribute” then there are 10 letters in the given word and there are 5 odd places, 5 even places, 4 vowels and 6 consonants.
- Let us mark these positions as under:
- (1) (2) (3) (4) (5) (6) (7) (8) (9) (10)
- Now, 4 vowels can be placed at any of the five places, marked 1, 3, 5, 7, 9.
- The number of ways of arranging the vowels = 5_P_4 = 5! = 120
- Also, the 6 consonants can be arranged at the remaining 6 positions.
- Number of ways of these arrangements = 6_P_6 = 6! = 720.
- Total number of ways = (120 x 720) = 86400
Below is the implementation of the above approach:
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 numberint fact(int n){ int f = 1; for (int i = 2; i <= n; i++) { f = f * i; } return f;}// calculating nPrint 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 positionsint 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 codeint main(){ string str = "neveropen"; 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 positionsclass GFG{// Function to return the// factorial of a numberstatic int fact(int n){ int f = 1; for (int i = 2; i <= n; i++) { f = f * i; } return f;}// calculating nPrstatic 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 positionsstatic 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 codepublic static void main(String[] args){ String str = "neveropen"; 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 positionsimport math# Function to return the factorial# of a numberdef fact(n): f = 1; for i in range(2, n + 1): f = f * i; return f;# calculating nPrdef 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 positionsdef 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 codestr = "neveropen";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 positionsusing System;class GFG{// Function to return the// factorial of a numberstatic int fact(int n){ int f = 1; for (int i = 2; i <= n; i++) { f = f * i; } return f;}// calculating nPrstatic 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 positionsstatic 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 codestatic void Main(){ String str = "neveropen"; 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 numberfunction fact($n){ $f = 1; for ($i = 2; $i <= $n; $i++) { $f = $f * $i; } return $f;}// calculating nPrfunction npr($n, $r){ return fact($n) / fact($n - $r);}// Function to find the number // of $ways in which the characters // of the word can be arranged such // that the vowels occupy only the // odd positionsfunction countPermutations($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 = "neveropen";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 numberfunction fact(n){ let f = 1; for (let i = 2; i <= n; i++) { f = f * i; } return f;}// calculating nPrfunction 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 positionsfunction 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 codelet str = "neveropen";document.write(countPermutations(str)); // This code is contributed by gfgking</script> |
Output
36
Complexity Analysis:
- Time Complexity: O(n) where n is the length of the string
- Auxiliary Space: O(26)
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