Lexicographic rank of a string among all its substrings

Given string str, the task is to find the rank of the given string among all its substrings arranged lexicographically.

Examples:

Input: S = “enren”
Output: 7
Explanation:
All the possible substrings in the sorted order are {“e”, “e”, “en”, “en”, “enr”, “enre”, “enren”, “n”, “n”, “nr”, “nre”, “nren”, “r”, “re”, “ren”}.
Therefore, the rank of the given string “enren” is 7.

Input: S = “geeks”
Output: 12
Explanation:
All possible substrings in the sorted order are {“e”, “e”, “ee”, “eek”, “eeks”, “ek”, “eks”, “g”, “ge”, “gee”, “geek”, “geeks”, “k”, “ks”, “s”}.
Therefore, the rank of the given string “geeks” is 12.

Approach: Follow the steps below to solve the problem:



  1. Initialize an array arr[] of vectors of length 26 to store the indices of the characters present in the string and rank as 0.
  2. Store the indices of each character. Indices of a will be stored in arr[0], for b, arr[1] will store all its indices and so on.
  3. Traverse each index stored in the array arr[] up to the characters which are smaller than the first character of the string S.
  4. For each index i, total substrings starting from that index is N – i. Add N – i to rank as all characters with these indices are smaller.
  5. Now, after traversing, store all the substrings starting from the first character of S and sort those substrings in lexicographical order.
  6. Traverse the sorted substrings and compare each substring with the string S and increment the rank until substring equals to S is found.
  7. Print the value of rank + 1 to get the rank of the given string.

Below is the implementation of the above approach:

C++

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// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find lexiographic rank
// of string among all its substring
int lexiographicRank(string s)
{
    // Length of string
    int n = s.length();
 
    vector<int> alphaIndex[26];
 
    // Traverse the given string
    // and store the indices of
    // each character
    for (int i = 0; i < s.length(); i++) {
 
        // Extract the index
        int x = s[i] - 'a';
 
        // Store it in the vector
        alphaIndex[x].push_back(i);
    }
 
    // Traverse the aphaIndex array
    // lesser than the index of first
    // character of given string
    int rank = 0;
 
    for (int i = 0; i < 26
                    && 'a' + i < s[0];
         i++) {
 
        // If alphaIndex[i] size exceeds 0
        if (alphaIndex[i].size() > 0) {
 
            // Traverse over the indices
            for (int j = 0;
                 j < alphaIndex[i].size(); j++) {
 
                // Add count of substring
                // equal to n - alphaIndex[i][j]
                rank = rank
                       + (n
                          - alphaIndex[i][j]);
            }
        }
    }
 
    vector<string> str;
 
    for (int i = 0;
         i < alphaIndex[s[0] - 'a'].size();
         i++) {
 
        // Store all substrings in a vector
        // str starting with the first
        // character of the given string
        string substring;
 
        int j = alphaIndex[s[0] - 'a'][i];
 
        for (; j < n; j++) {
 
            // Insert the current
            // character to substring
            substring.push_back(s[j]);
 
            // Store the substring formed
            str.push_back(substring);
        }
    }
 
    // Sort the substring in the
    // lexicographical order
    sort(str.begin(), str.end());
 
    // Find the rank of given string
    for (int i = 0; i < str.size(); i++) {
 
        if (str[i] != s) {
 
            // increase the rank until
            // the given string is same
            rank++;
        }
 
        // If substring is same as
        // the given string
        else {
            break;
        }
    }
 
    // Add 1 to rank of
    // the given string
    return rank + 1;
}
 
// Driver Code
int main()
{
    // Given string
    string str = "enren";
 
    // Function Call
    cout << lexiographicRank(str);
 
    return 0;
}

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Java

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// Java program for
// the above approach
import java.util.*;
class GFG{
 
// Function to find lexiographic rank
// of String among all its subString
static int lexiographicRank(char []s)
{
  // Length of String
  int n = s.length;
 
  Vector<Integer> []alphaIndex = new Vector[26];
  for (int i = 0; i < alphaIndex.length; i++)
    alphaIndex[i] = new Vector<Integer>();
   
  // Traverse the given String
  // and store the indices of
  // each character
  for (int i = 0; i < s.length; i++)
  {
    // Extract the index
    int x = s[i] - 'a';
 
    // Store it in the vector
    alphaIndex[x].add(i);
  }
 
  // Traverse the aphaIndex array
  // lesser than the index of first
  // character of given String
  int rank = 0;
 
  for (int i = 0; i < 26 &&
           'a' + i < s[0]; i++)
  {
    // If alphaIndex[i] size exceeds 0
    if (alphaIndex[i].size() > 0)
    {
      // Traverse over the indices
      for (int j = 0;
               j < alphaIndex[i].size();
               j++)
      {
        // Add count of subString
        // equal to n - alphaIndex[i][j]
        rank = rank + (n - alphaIndex[i].get(j));
      }
    }
  }
 
  Vector<String> str = new Vector<String>();
 
  for (int i = 0;
           i < alphaIndex[s[0] - 'a'].size();
           i++)
  {
    // Store all subStrings in a vector
    // str starting with the first
    // character of the given String
    String subString = "";
 
    int j = alphaIndex[s[0] - 'a'].get(i);
 
    for (; j < n; j++)
    {
      // Insert the current
      // character to subString
      subString += (s[j]);
 
      // Store the subString formed
      str.add(subString);
    }
  }
 
  // Sort the subString in the
  // lexicographical order
  Collections.sort(str);
 
  // Find the rank of given String
  for (int i = 0; i < str.size(); i++)
  {
    if (!str.get(i).equals(String.valueOf(s)))
    {
      // increase the rank until
      // the given String is same
      rank++;
    }
 
    // If subString is same as
    // the given String
    else
    {
      break;
    }
  }
 
  // Add 1 to rank of
  // the given String
  return rank + 1;
}
 
// Driver Code
public static void main(String[] args)
{
  // Given String
  String str = "enren";
 
  // Function Call
  System.out.print(lexiographicRank(str.toCharArray()));
}
}
 
// This code is contributed by shikhasingrajput

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Python3

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# Python3 program for the above approach
 
# Function to find lexiographic rank
# of among all its substrring
def lexiographicRank(s):
     
    # Length of strring
    n = len(s)
 
    alphaIndex = [[] for i in range(26)]
 
    # Traverse the given strring
    # and store the indices of
    # each character
    for i in range(len(s)):
 
        # Extract the index
        x = ord(s[i]) - ord('a')
 
        # Store it in the vector
        alphaIndex[x].append(i)
 
    # Traverse the aphaIndex array
    # lesser than the index of first
    # character of given strring
    rank = -1
 
    for i in range(26):
        if ord('a') + i >= ord(s[0]):
            break
 
        # If alphaIndex[i] size exceeds 0
        if len(alphaIndex[i]) > 0:
 
            # T raverse over the indices
            for j in range(len(alphaIndex[i])):
 
                # Add count of substrring
                # equal to n - alphaIndex[i][j]
                rank = rank + (n - alphaIndex[i][j])
                 
    # print(rank)
    strr = []
 
    for i in range(len(alphaIndex[ord(s[0]) - ord('a')])):
 
        # Store all substrrings in a vector
        # strr starting with the first
        # character of the given strring
        substrring = []
 
        jj = alphaIndex[ord(s[0]) - ord('a')][i]
 
        for j in range(jj, n):
 
            # Insert the current
            # character to substrring
            substrring.append(s[j])
 
            # Store the subformed
            strr.append(substrring)
 
    # Sort the subin the
    # lexicographical order
    strr = sorted(strr)
 
    # Find the rank of given strring
    for i in range(len(strr)):
        if (strr[i] != s):
 
            # Increase the rank until
            # the given is same
            rank += 1
 
        # If subis same as
        # the given strring
        else:
            break
 
    # Add 1 to rank of
    # the given strring
    return rank + 1
 
# Driver Code
if __name__ == '__main__':
     
    # Given strring
    strr = "enren"
 
    # Function call
    print(lexiographicRank(strr))
 
# This code is contributed by mohit kumar 29

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C#

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// C# program for
// the above approach
using System;
using System.Collections.Generic;
class GFG{
 
// Function to find lexiographic rank
// of String among all its subString
static int lexiographicRank(char []s)
{
  // Length of String
  int n = s.Length;
 
  List<int> []alphaIndex = new List<int>[26];
  for (int i = 0; i < alphaIndex.Length; i++)
    alphaIndex[i] = new List<int>();
 
  // Traverse the given String
  // and store the indices of
  // each character
  for (int i = 0; i < s.Length; i++)
  {
    // Extract the index
    int x = s[i] - 'a';
 
    // Store it in the vector
    alphaIndex[x].Add(i);
  }
 
  // Traverse the aphaIndex array
  // lesser than the index of first
  // character of given String
  int rank = 0;
 
  for (int i = 0; i < 26 &&
           'a' + i < s[0]; i++)
  {
    // If alphaIndex[i] size exceeds 0
    if (alphaIndex[i].Count > 0)
    {
      // Traverse over the indices
      for (int j = 0;
               j < alphaIndex[i].Count; j++)
      {
        // Add count of subString
        // equal to n - alphaIndex[i,j]
        rank = rank + (n - alphaIndex[i][j]);
      }
    }
  }
 
  List<String> str = new List<String>();
 
  for (int i = 0;
           i < alphaIndex[s[0] - 'a'].Count; i++)
  {
    // Store all subStrings in a vector
    // str starting with the first
    // character of the given String
    String subString = "";
 
    int j = alphaIndex[s[0] - 'a'][i];
 
    for (; j < n; j++)
    {
      // Insert the current
      // character to subString
      subString += (s[j]);
 
      // Store the subString formed
      str.Add(subString);
    }
  }
 
  // Sort the subString in the
  // lexicographical order
  str.Sort();
 
  // Find the rank of given String
  for (int i = 0; i < str.Count; i++)
  {
    if (!str[i].Equals(String.Join("", s)))
    {
      // increase the rank until
      // the given String is same
      rank++;
    }
 
    // If subString is same as
    // the given String
    else
    {
      break;
    }
  }
 
  // Add 1 to rank of
  // the given String
  return rank + 1;
}
 
// Driver Code
public static void Main(String[] args)
{
  // Given String
  String str = "enren";
 
  // Function Call
  Console.Write(lexiographicRank(str.ToCharArray()));
}
}
 
// This code is contributed by 29AjayKumar

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Output: 

7







Time Complexity: O(N3)
Auxiliary Space: O(N)

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