Recuento de strings de la misma longitud que existe lexicográficamente entre dos strings dadas

Dadas dos strings S1 y S2 de longitud L , la tarea es contar el número de strings de longitud L, que existe entre S1 y S2, que son lexicográficamente mayores que S1 pero menores que S2. 
Ejemplos: 
 

Entrada: S1 = «b», S2 = «f» 
Salida: 3 
Explicación: 
Estas son 3 strings que vienen lexicográficamente entre S1 y S2, es decir, «c», «d» y «e»
Entrada: S1 = «aby», S2 = “ace” 
Salida: 5 
Explicación: 
Estas son 5 strings que vienen lexicográficamente entre S1 y S2, es decir, “abz”, “aca”, “acb”, “acc” y “acd”. 
 

Acercarse:
 

1. Primero, encuentre el número de strings lexicográficamente más pequeñas que la primera string S1, como:

   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)

   Por ejemplo:

   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.

2. De manera similar, averigüe el número de strings lexicográficamente más pequeñas que S2.
3. Luego, solo averigüe la diferencia entre los dos valores anteriores para obtener el número de strings lexicográficamente mayor que S1 pero menor que S2.

A continuación se muestra la implementación del enfoque anterior: 
 

    // 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 lexicographically
int 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 Strings
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 code
int 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 Strings
import java.util.*;
  
class GFG{
  
// Function to find the count of strings less
// than given string lexicographically
static 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 Strings
static 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 code
public 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 lexicographically
def 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 Strings
def 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 code
S1 = "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 Strings
using System;
  
class GFG{
  
// Function to find the count of strings less
// than given string lexicographically
static 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 Strings
static 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 code
public 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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Análisis de rendimiento:
Complejidad del tiempo : en el enfoque anterior, estamos recorriendo las dos strings de longitud N, por lo tanto, tomará un tiempo O (N) donde N es la longitud de cada string.
Complejidad del espacio auxiliar : como en el enfoque anterior, no se utiliza espacio adicional, por lo tanto, la complejidad del espacio auxiliar será O(1) .