You are given a string, find its rank among all its permutations sorted lexicographically.
Examples:
Input : str[] = "acb"
Output : Rank = 2
Input : str[] = "string"
Output : Rank = 598
Input : str[] = "cba"
Output : Rank = 6
We have already discussed solutions to find Lexicographic rank of string In this post, we use the STL function “next_permutation ()” to generate all possible permutations of the given string and, as it gives us permutations in lexicographic order, we will put an iterator to find the rank of each string. While iterating when Our permuted string becomes identical to the original input string, we break from the loop and the iterator value for the last iteration is our required result.
Implementation:
// C++ program to print rank of // string using next_permute() #include <bits/stdc++.h> using namespace std;
// Function to print rank of string // using next_permute() int findRank(string str)
{ // store original string
string orgStr = str;
// Sort the string in lexicographically
// ascending order
sort(str.begin(), str.end());
// Keep iterating until
// we reach equality condition
long int i = 1;
do {
// check for nth iteration
if (str == orgStr)
break ;
i++;
} while (next_permutation(str.begin(), str.end()));
// return iterator value
return i;
} // Driver code int main()
{ string str = "cba" ;
cout << findRank(str);
return 0;
} |
# Python program to print rank of # string using itertools.permutations import itertools
# Function to print rank of string # using itertools.permutations def findRank( str ):
# Store original string and
# Sort the string in lexicographically
# ascending order
orgStr = ''.join( sorted ( str ))
# Create an dictionary to handle
# repeating characters in Python
# while using itertools.permutations
d = []
# Keep iterating until we reach equality condition
i = 1
for permutation in itertools.permutations(orgStr):
if permutation not in d:
d.append(permutation)
# check for nth iteration
if ''.join(permutation) = = str :
break
i + = 1
# Return iterator value
return i
# Driver code if __name__ = = '__main__' :
str = 'bca'
print (findRank( str ))
# This code is contributed by Utkarsh. |
using System;
using System.Collections.Generic;
using System.Linq;
class Program
{ // Function to print rank of string
// using itertools.permutations
static int FindRank( string str)
{
// Store original string and
// Sort the string in lexicographically
// ascending order
string orgStr = new string (str.OrderBy(c => c).ToArray());
// Create a set to handle
// repeating characters in C#
// while using itertools.permutations
HashSet< string > set = new HashSet< string >();
// Keep iterating until we reach equality condition
int i = 1;
foreach ( var permutation in Permutations(orgStr))
{
if (! set .Contains(permutation))
{
set .Add(permutation);
// check for nth iteration
if (permutation == str)
{
break ;
}
i++;
}
}
// Return iterator value
return i;
}
// Function to get all permutations of a string
static IEnumerable< string > Permutations( string str)
{
if (str.Length == 1)
{
yield return str;
}
else
{
foreach ( var ch in str)
{
foreach ( var permutation in Permutations(str.Remove(str.IndexOf(ch), 1)))
{
yield return ch + permutation;
}
}
}
}
// Driver code
static void Main( string [] args)
{
string str = "bca" ;
Console.WriteLine(FindRank(str));
}
} // This code is contributed by shiv1o43g |
function findRank(str) {
const MOD = 1000003;
const n = str.length;
let rank = 1;
let factorial = 1;
for (let i = 0; i < n; i++) {
let smaller = 0;
for (let j = i + 1; j < n; j++) {
if (str[j] < str[i]) {
smaller++;
}
}
rank = (rank + (smaller * factorial)) % MOD;
factorial = (factorial * (n - i)) % MOD;
}
return rank;
} // Driver code const str = "cba" ;
console.log(findRank(str)); |
6
Time Complexity: O(n log n)
Auxiliary Space: O(1)
This article is contributed by Aarti_Rathi and Shivam Pradhan (anuj_charm).