Count of same length Strings that exists lexicographically in between two given Strings - GeeksforGeeks

06 May, 2021
Given two string S1 and S2 of length L, the task is to count the number of strings of length L, that exists in between S1 and S2, which are lexicographically greater than S1 but smaller than S2. Examples:  
Input: S1 = “b”, S2 = “f” Output: 3 Explaination: These are 3 strings which come lexicographically in between S1 and S2 i.e. “c”, “d” & “e”Input: S1 = “aby”, S2 = “ace” Output: 5 Explaination: These are 5 strings which come lexicographically in between S1 and S2 i.e. “abz”, “aca”, “acb”, “acc” & “acd”.  
 
Approach: 
First, find out the number of strings lexicographically smaller than the first string S1, as:Let the String S1 of length L 
be represented as c0c1c2...cL-1
where ci is the character in S1 at index i

Therefore, To get the number of strings less than S1,
we will calculate it as 
N(S1) = (number of letters less than c0 * 26L-1)
      + (number of letters less than c1 * 26L-2)
      + (number of letters less than c2 * 26L-3)
      +  ... 
      + (number of letters less than cL-2 * 26)
      + (number of letters less than cL-1)For example:Let S1 = "cbd"

Number of strings less than S1
N(S1) = (number of letters less than 'c' * 262)
      + (number of letters less than 'b' * 26)
      + (number of letters less than 'd')

N(S1) = (2 * 26 * 26) + (1 * 26) + (3) 
      = 1352 + 26 + 3 = 1381.Similarly, find out the number of string lexicographically smaller than S2.Then just find out the difference between the above two values to get the number of string lexicographically greater than S1 but smaller than S2.
First, find out the number of strings lexicographically smaller than the first string S1, as:Let the String S1 of length L 
be represented as c0c1c2...cL-1
where ci is the character in S1 at index i

Therefore, To get the number of strings less than S1,
we will calculate it as 
N(S1) = (number of letters less than c0 * 26L-1)
      + (number of letters less than c1 * 26L-2)
      + (number of letters less than c2 * 26L-3)
      +  ... 
      + (number of letters less than cL-2 * 26)
      + (number of letters less than cL-1)For example:Let S1 = "cbd"

Number of strings less than S1
N(S1) = (number of letters less than 'c' * 262)
      + (number of letters less than 'b' * 26)
      + (number of letters less than 'd')

N(S1) = (2 * 26 * 26) + (1 * 26) + (3) 
      = 1352 + 26 + 3 = 1381.
Let the String S1 of length L 
be represented as c0c1c2...cL-1
where ci is the character in S1 at index i

Therefore, To get the number of strings less than S1,
we will calculate it as 
N(S1) = (number of letters less than c0 * 26L-1)
      + (number of letters less than c1 * 26L-2)
      + (number of letters less than c2 * 26L-3)
      +  ... 
      + (number of letters less than cL-2 * 26)
      + (number of letters less than cL-1)
For example:
Let S1 = "cbd"

Number of strings less than S1
N(S1) = (number of letters less than 'c' * 262)
      + (number of letters less than 'b' * 26)
      + (number of letters less than 'd')

N(S1) = (2 * 26 * 26) + (1 * 26) + (3) 
      = 1352 + 26 + 3 = 1381.
Similarly, find out the number of string lexicographically smaller than S2.
Then just find out the difference between the above two values to get the number of string lexicographically greater than S1 but smaller than S2.
Below is the implementation of the above approach:  
C++
Java
Python3
C#
Javascript
    // C++ program to find the count of// same length Strings that exists lexicographically// in between two given Strings    #include <bits/stdc++.h>using namespace std;  // Function to find the count of strings less// than given string lexicographicallyint LexicoLesserStrings(string s){    int count = 0;    int len;      // Find length of string s    len = s.size();      // Looping over the string characters and    // finding strings less than that character    for (int i = 0; i < len; i++) {        count += (s[i] - 'a')                * pow(26, len - i - 1);    }      return count;}  // Function to find the count of// same length Strings that exists// lexicographically in between two given Stringsint countString(string S1, string S2){    int countS1, countS2, totalString;      // Count string less than S1    countS1 = LexicoLesserStrings(S1);      // Count string less than S2    countS2 = LexicoLesserStrings(S2);      // Total strings between S1 and S2 would    // be difference between the counts - 1    totalString = countS2 - countS1 - 1;      // If S1 is lexicographically greater    // than S2 then return 0, otherwise return    // the value of totalString    return (totalString < 0 ? 0 : totalString);}  // Driver codeint main(){    string S1, S2;    S1 = "cda";    S2 = "cef";      cout << countString(S1, S2);    return 0;}
// Java program to find the count of same length// Strings that exists lexicographically// in between two given Stringsimport java.util.*;  class GFG{  // Function to find the count of strings less// than given string lexicographicallystatic int LexicoLesserStrings(String s){    int count = 0;    int len;      // Find length of string s    len = s.length();      // Looping over the string characters and    // finding strings less than that character    for(int i = 0; i < len; i++)    {        count += (s.charAt(i) - 'a') *                Math.pow(26, len - i - 1);    }    return count;}  // Function to find the count of// same length Strings that exists// lexicographically in between two// given Stringsstatic int countString(String S1, String S2){    int countS1, countS2, totalString;      // Count string less than S1    countS1 = LexicoLesserStrings(S1);      // Count string less than S2    countS2 = LexicoLesserStrings(S2);      // Total strings between S1 and S2 would    // be difference between the counts - 1    totalString = countS2 - countS1 - 1;      // If S1 is lexicographically greater    // than S2 then return 0, otherwise return    // the value of totalString    return (totalString < 0 ? 0 : totalString);}  // Driver codepublic static void main(String args[]){    String S1, S2;    S1 = "cda";    S2 = "cef";      System.out.println(countString(S1, S2));}}  // This code is contributed by apurva raj
# Python3 program to find the count of same# length Strings that exists lexicographically# in between two given Strings  # Function to find the count of strings less# than given string lexicographicallydef LexicoLesserStrings(s):          count = 0      # Find length of string s    length = len(s)      # Looping over the string characters and    # finding strings less than that character    for i in range(length):        count += ((ord(s[i]) - ord('a')) *                pow(26, length - i - 1))                          return count  # Function to find the count of# same length Strings that exists# lexicographically in between two# given Stringsdef countString(S1, S2):      # Count string less than S1    countS1 = LexicoLesserStrings(S1)      # Count string less than S2    countS2 = LexicoLesserStrings(S2)      # Total strings between S1 and S2 would    # be difference between the counts - 1    totalString = countS2 - countS1 - 1;      # If S1 is lexicographically greater    # than S2 then return 0, otherwise return    # the value of totalString    return (0 if totalString < 0 else totalString)  # Driver codeS1 = "cda";S2 = "cef";  print(countString(S1, S2))  # This code is contributed by apurva raj
// C# program to find the count of same length// Strings that exists lexicographically// in between two given Stringsusing System;  class GFG{  // Function to find the count of strings less// than given string lexicographicallystatic int LexicoLesserStrings(String s){    int count = 0;    int len;      // Find length of string s    len = s.Length;      // Looping over the string characters and    // finding strings less than that character    for(int i = 0; i < len; i++)    {        count += ((s[i] - 'a') *                (int)Math.Pow(26, len - i - 1));    }    return count;}  // Function to find the count of// same length Strings that exists// lexicographically in between two// given Stringsstatic int countString(String S1, String S2){    int countS1, countS2, totalString;      // Count string less than S1    countS1 = LexicoLesserStrings(S1);      // Count string less than S2    countS2 = LexicoLesserStrings(S2);      // Total strings between S1 and S2 would    // be difference between the counts - 1    totalString = countS2 - countS1 - 1;      // If S1 is lexicographically greater    // than S2 then return 0, otherwise return    // the value of totalString    return (totalString < 0 ? 0 : totalString);}  // Driver codepublic static void Main(){    String S1, S2;    S1 = "cda";    S2 = "cef";      Console.Write(countString(S1, S2));}}  // This code is contributed by chitranayal
<script>      // JavaScript program to find the count of      // same length Strings that exists lexicographically      // in between two given Strings        // Function to find the count of strings less      // than given string lexicographically      function LexicoLesserStrings(s) {        var count = 0;        var len;          // Find length of string s        len = s.length;          // Looping over the string characters and        // finding strings less than that character        for (var i = 0; i < len; i++) {          count +=            (s[i].charCodeAt(0) - "a".charCodeAt(0)) *            Math.pow(26, len - i - 1);        }          return count;      }        // Function to find the count of      // same length Strings that exists      // lexicographically in between two given Strings      function countString(S1, S2) {        var countS1, countS2, totalString;          // Count string less than S1        countS1 = LexicoLesserStrings(S1);          // Count string less than S2        countS2 = LexicoLesserStrings(S2);          // Total strings between S1 and S2 would        // be difference between the counts - 1        totalString = countS2 - countS1 - 1;          // If S1 is lexicographically greater        // than S2 then return 0, otherwise return        // the value of totalString        return totalString < 0 ? 0 : totalString;      }        // Driver code      var S1, S2;      S1 = "cda";      S2 = "cef";        document.write(countString(S1, S2));</script>     
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Performance Analysis:Time Complexity: In the above approach, we are looping over the two strings of length N, therefore it will take O(N) time where N is the length of each string.Auxiliary Space Complexity: As in the above approach there is no extra space used, therefore the Auxiliary Space complexity will be O(1). 
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