Given a string str consisting of lowercase English alphabets, the task is to find the count of all possible string of maximum length that can be formed using the characters of str such that no two characters in the generated string are same.
Examples:
Input: str = “aba”
Output: 2
“ab” and “ba” are the only valid strings.
Input: str = “neveropen”
Output: 5040
Approach: First, count the number of distinct characters in the string say cnt as no two characters can be same in the resultant string. Now, the total number of strings that can be formed with cnt number of characters is cnt! as every character of str has to be present in the generated string in order to maximise the length and no character should appear more than once.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the factorial of nint fact(int n){ int fact = 1; for (int i = 1; i <= n; i++) fact *= i; return fact;}// Function to return the count of all// possible strings that can be formed// with the characters of the given string// without repeating charactersint countStrings(string str, int n){ // To store the distinct characters // of the string str set<char> distinct_char; for (int i = 0; i < n; i++) { distinct_char.insert(str[i]); } return fact(distinct_char.size());}// Driver codeint main(){ string str = "neveropen"; int n = str.length(); cout << countStrings(str, n); return 0;} |
Java
// Java implementation of the approach import java.util.*;class GFG {// Function to return the factorial of nstatic int fact(int n){ int fact = 1; for (int i = 1; i <= n; i++) fact *= i; return fact;}// Function to return the count of all// possible strings that can be formed// with the characters of the given string// without repeating charactersstatic int countStrings(String str, int n){ // To store the distinct characters // of the string str Set<Character> distinct_char = new HashSet<>(); for (int i = 0; i < n; i++) { distinct_char.add(str.charAt(i)); } return fact(distinct_char.size());}// Driver codepublic static void main(String[] args){ String str = "neveropen"; int n = str.length(); System.out.println(countStrings(str, n));}}// This code is contributed by PrinciRaj1992 |
Python3
# Python3 implementation of the approach # Function to return the factorial of n def fact(n) : fact = 1; for i in range(1, n + 1) : fact *= i; return fact; # Function to return the count of all # possible strings that can be formed # with the characters of the given string # without repeating characters def countStrings(string, n) : # To store the distinct characters # of the string str distinct_char = set(); for i in range(n) : distinct_char.add(string[i]); return fact(len(distinct_char)); # Driver code if __name__ == "__main__" : string = "neveropen"; n = len(string); print(countStrings(string, n)); # This code is contributed by AnkitRai01 |
C#
// C# implementation of the approachusing System;using System.Collections.Generic;class GFG {// Function to return the factorial of nstatic int fact(int n){ int fact = 1; for (int i = 1; i <= n; i++) fact *= i; return fact;}// Function to return the count of all// possible strings that can be formed// with the characters of the given string// without repeating charactersstatic int countStrings(String str, int n){ // To store the distinct characters // of the string str HashSet<char> distinct_char = new HashSet<char>(); for (int i = 0; i < n; i++) { distinct_char.Add(str[i]); } return fact(distinct_char.Count);}// Driver codepublic static void Main(String[] args){ String str = "neveropen"; int n = str.Length; Console.WriteLine(countStrings(str, n));}}// This code is contributed by 29AjayKumar |
Javascript
<script> // Javascript implementation of the approach // Function to return the factorial of n function fact(n) { let fact = 1; for (let i = 1; i <= n; i++) fact *= i; return fact; } // Function to return the count of all // possible strings that can be formed // with the characters of the given string // without repeating characters function countStrings(str, n) { // To store the distinct characters // of the string str let distinct_char = new Set(); for (let i = 0; i < n; i++) { distinct_char.add(str[i]); } return fact(distinct_char.size); } let str = "neveropen"; let n = str.length; document.write(countStrings(str, n)); </script> |
Output:
5040
Time Complexity: O(N^2), because it contains a loop with N iterations
Auxiliary Space: O(M), where M is the number of distinct characters in the input string.
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