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:3Explaination:

These are 3 strings which come lexicographically in between S1 and S2 i.e. “c”, “d” & “e”

Input:S1 = “aby”, S2 = “ace”Output:5Explaination:

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**c**where c_{0}c_{1}c_{2}...c_{L-1}_{i}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 c_{0}* 26^{L-1}) + (number of letters less than c_{1}* 26^{L-2}) + (number of letters less than c_{2}* 26^{L-3}) + ... + (number of letters less than c_{L-2}* 26) + (number of letters less than c_{L-1})**For example:**Let S1 = "cbd" Number of strings less than S1 N(S1) = (number of letters less than 'c' * 26

^{2}) + (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++

`// 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

`// 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

`# 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#

`// 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` |

**Output:**

30

**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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