Lexicographic rank of a string with duplicate characters
Last Updated :
29 Mar, 2024
Given a string s that may have duplicate characters. Find out the lexicographic rank of s. s may consist of lower as well as upper case letters. We consider the lexicographic order of characters as their order of ASCII value. Hence the lexicographical order of characters will be 'A', 'B', 'C', ..., 'Y', 'Z', 'a', 'b', 'c', ..., 'y', 'z'.
Examples:
Input : "abab"
Output : 2
Explanation: The lexicographical order is: "aabb", "abab", "abba", "baab", "baba", "bbaa". Hence the rank of "abab" is 2.
Input: "settLe"
Output : 107
Prerequisite: Lexicographic rank of a string
Method:
The method here is a little different from the without repetition version. Here we have to take care of the duplicate characters also. Let's look at the string "settLe". It has repetition(2 'e' and 2 't') as well as upper case letter('L'). Total 6 characters and the total number of permutations are 6!/(2!*2!).
Now there are 3 characters(2 'e' and 1 'L') on the right side of 's' which come before 's' lexicographically. If there were no repetition then there would be 3*5! smaller strings which have the first character less than 's'. But starting from position 0, till the end there are 2 'e' and 2 't'(i.e. repetitions). Hence, the number of possible smaller permutations with the first letter smaller than 's' are (3*5!)/(2!*2!).
Similarly, if we fix 's' and look at the letters from index 1 to end then there is 1 character('L') lexicographically less than 'e'. And starting from position 1 there are 2 repeated characters(2 'e' and 2 't'). Hence, the number of possible smaller permutations with first letter 's' and second letter smaller than 'e' are (1*4!)/(2!*2!).
Similarly, we can form the following table:
WorkFlow:
1. Initialize t_count(total count) variable
to 1(as rank starts from 1).
2. Run a loop for every character of the string, string[i]:
(i) using a loop count less_than(number of smaller
characters on the right side of string[i]).
(ii) take one array d_count of size 52 and using a
loop count the frequency of characters starting
from string[i].
(iii) compute the product, d_fac(the product of
factorials of each element of d_count).
(iv) compute (less_than*fac(n-i-1))/(d_fac).
Add it to t_count.
3. return t_count
C++
// C++ program to find out lexicographic
// rank of a string which may have duplicate
// characters and upper case letters.
#include <iostream>
#include <vector>
using namespace std;
// Function to calculate factorial of a number.
long long fac(long long n)
{
if (n == 0 or n == 1)
return 1;
return n * fac(n - 1);
}
// Function to calculate rank of the string.
int lexRank(string s)
{
long long n = s.size();
// Initialize total count to 1.
long long t_count = 1;
// loop to calculate number of smaller strings.
for (int i = 0; i < n; i++)
{
// Count smaller characters than s[i].
int less_than = 0;
for (int j = i + 1; j < n; j++)
{
if (int(s[i]) > int(s[j]))
{
less_than += 1;
}
}
// Count frequency of duplicate characters.
vector<int> d_count(52, 0);
for (int j = i; j < n; j++)
{
// Check whether the character is upper
// or lower case and then increase the
// specific element of the array.
if ((int(s[j]) >= 'A') && int(s[j]) <= 'Z')
d_count[int(s[j]) - 'A'] += 1;
else
d_count[int(s[j]) - 'a' + 26] += 1;
}
// Compute the product of the factorials
// of frequency of characters.
long long d_fac = 1;
for (int ele : d_count)
d_fac *= fac(ele);
// add the number of smaller string
// possible from index i to total count.
t_count += (fac(n - i - 1) * less_than) / d_fac;
}
return (int)t_count;
}
// Driver Code
int main()
{
// Test case 1
string s1 = "abab";
cout << "Rank of " << s1 << " is: " << lexRank(s1)
<< endl;
// Test case 2
string s2 = "settLe";
cout << "Rank of " << s2 << " is: " << lexRank(s2)
<< endl;
return 0;
}
// Java program to find out lexicographic
// rank of a String which may have duplicate
// characters and upper case letters.
class GFG {
// Function to calculate
// factorial of a number.
static long fac(long n)
{
if (n == 0 || n == 1)
return 1;
return n * fac(n - 1);
}
// Function to calculate
// rank of the String.
static int lexRank(String s)
{
long n = s.length();
// Initialize total count to 1.
long t_count = 1;
// loop to calculate
// number of smaller Strings.
for (int i = 0; i < n; i++)
{
// Count smaller
// characters than s[i].
long less_than = 0;
for (int j = i + 1; j < n; j++)
{
if (s.charAt(i)
> s.charAt(j))
{
less_than += 1;
}
}
// Count frequency of
// duplicate characters.
long[] d_count = new long[52];
for (int j = i; j < n; j++)
{
// Check whether the
// character is upper
// or lower case and
// then increase the
// specific element of
// the array.
if ((s.charAt(j) >= 'A')
&& s.charAt(j) <= 'Z')
d_count[s.charAt(j) - 'A'] += 1;
else
d_count[s.charAt(j) - 'a' + 26] += 1;
}
// Compute the product of the factorials
// of frequency of characters.
long d_fac = 1;
for (long ele : d_count)
d_fac *= fac(ele);
// add the number of smaller String
// possible from index i to total count.
t_count += (fac(n - i - 1)
* less_than) / d_fac;
}
return (int)t_count;
}
// Driver Code
public static void main(String[] args)
{
// Test case 1
String s1 = "abab";
System.out.print("Rank of " + s1
+ " is: " + lexRank(s1) + "\n");
// Test case 2
String s2 = "settLe";
System.out.print("Rank of " + s2
+ " is: " + lexRank(s2) + "\n");
}
}
// This code is contributed by gauravrajput1
# Python program to find out lexicographic
# rank of a string which may have duplicate
# characters and upper case letters.
# Function to calculate
# factorial of a number.
def fac(n):
if n == 0 or n == 1:
return 1
return n * fac(n - 1)
#Function to calculate
#rank of the String.
def lexRank(s):
n = len(s)
# Initialize total count to 1.
t_count = 1
for i in range(n):
less_than = 0
for j in range(i + 1, n):
if ord(s[i]) > ord(s[j]):
less_than += 1
d_count = [0] * 52
for j in range(i, n):
if ord(s[j]) >= ord('A') and ord(s[j]) <= ord('Z'):
d_count[ord(s[j]) - ord('A')] += 1
else:
d_count[ord(s[j]) - ord('a') + 26] += 1
d_fac = 1
for ele in d_count:
d_fac *= fac(ele)
t_count += (fac(n - i - 1) * less_than) // d_fac
return t_count
# Test case 1
s1 = "abab"
print("Rank of", s1, "is:", lexRank(s1))
# Test case 2
s2 = "settLe"
print("Rank of", s2, "is:", lexRank(s2))
// C# program to find out
// lexicographic rank of a
// String which may have
// duplicate characters and
// upper case letters.
using System;
class GFG {
// Function to calculate
// factorial of a number.
static long fac(long n)
{
if (n == 0 || n == 1)
return 1;
return n * fac(n - 1);
}
// Function to calculate
// rank of the String.
static long lexRank(String s)
{
long n = s.Length;
// Initialize total
// count to 1.
long t_count = 1;
// loop to calculate number
// of smaller Strings.
for (long i = 0; i < n; i++)
{
// Count smaller characters
// than s[i].
long less_than = 0;
for (long j = i + 1; j < n; j++)
{
if (s[i] > s[j])
{
less_than += 1;
}
}
// Count frequency of
// duplicate characters.
long[] d_count = new long[52];
for (int j = i; j < n; j++)
{
// Check whether the character
// is upper or lower case and
// then increase the specific
// element of the array.
if ((s[j] >= 'A') && s[j] <= 'Z')
d_count[s[j] - 'A'] += 1;
else
d_count[s[j] - 'a' + 26] += 1;
}
// Compute the product of the
// factorials of frequency of
// characters.
long d_fac = 1;
foreach(long ele in d_count)
d_fac *= fac(ele);
// add the number of smaller
// String possible from index
// i to total count.
t_count += (fac(n - i - 1)
* less_than) / d_fac;
}
return t_count;
}
// Driver Code
public static void Main(String[] args)
{
// Test case 1
String s1 = "abab";
Console.Write("Rank of " + s1
+ " is: " + lexRank(s1) + "\n");
// Test case 2
String s2 = "settLe";
Console.Write("Rank of " + s2
+ " is: " + lexRank(s2) + "\n");
}
}
// This code is contributed by Rajput-Ji
<script>
// Javascript program to find out lexicographic
// rank of a String which may have duplicate
// characters and upper case letters.
// Function to calculate
// factorial of a number.
function fac(n)
{
if (n == 0 || n == 1)
return 1;
return n * fac(n - 1);
}
// Function to calculate
// rank of the String.
function lexRank(s)
{
n = s.length;
// Initialize total count to 1.
let t_count = 1;
// loop to calculate
// number of smaller Strings.
for (let i = 0; i < n; i++)
{
// Count smaller
// characters than s[i].
let less_than = 0;
for (let j = i + 1; j < n; j++)
{
if (s[i]
> s[j])
{
less_than += 1;
}
}
// Count frequency of
// duplicate characters.
let d_count = new Array(52);
for(let i=0;i<52;i++)
d_count[i]=0;
for (let j = i; j < n; j++)
{
// Check whether the
// character is upper
// or lower case and
// then increase the
// specific element of
// the array.
if ((s[j] >= 'A')
&& s[j] <= 'Z')
d_count[s[j].charCodeAt(0) - 'A'.charCodeAt(0)] += 1;
else
d_count[s[j].charCodeAt(0) - 'a'.charCodeAt(0) + 26] += 1;
}
// Compute the product of the factorials
// of frequency of characters.
let d_fac = 1;
for (let ele=0;ele< d_count.length;ele++)
d_fac *= fac(d_count[ele]);
// add the number of smaller String
// possible from index i to total count.
t_count += (fac(n - i - 1)
* less_than) / d_fac;
}
return t_count;
}
// Driver Code
let s1 = "abab";
document.write("Rank of " + s1
+ " is: " + lexRank(s1) + "<br>");
// Test case 2
let s2 = "settLe";
document.write("Rank of " + s2
+ " is: " + lexRank(s2) + "<br>");
// This code is contributed by patel2127
</script>
OutputRank of abab is: 2
Rank of settLe is: 107
Time complexity: O(n2)
Space complexity: O(1)
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