Given a string **S**, the task is to find the length of the longest substring between any pair of occurrences of same character.

**Examples:**

Input:S = “accabbacc”Output:6Explanation:The longest substring satisfying the required conditions is “cabbac”, which lies between S[1](= ‘c’) and s[8](= ‘c’).

Input:S = “aab”Output:0

**Approach: **Follow the steps below to solve the problem:

- Initialize two variables
**res**and**diff**to store the length of the longest substring and the length of the current substring between pair of same characters respectively. - Iterate over characters of the string from left to right.
- Iterate over the remaining string on the right, from right to left up to the current character.
- If two equal characters are obtained, i.e.
**S[i] = S[j],**, store the length of the substring between them in**diff**. - Update the value of
**res**as**max(res, diff).**so that**res**stores the longest substring of required type obtained so far. - After complete traversal of the string, print
**res**as the required answer.

Below is the implementation of the above approach:

## C++

`// C++ program to implement` `// the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to find the longest substring` `// between pair of repetitions of the same character` `int` `longestbetweenequalcharacters(string S)` `{` ` ` `// Length of the string` ` ` `int` `n = S.length();` ` ` `// Stores the maximum length and` ` ` `// length of current substring` ` ` `// satisfying given conditions` ` ` `int` `res = -1, diff = -1;` ` ` `// Traverse the string` ` ` `for` `(` `int` `i = 0; i < n - 1; i++) {` ` ` `// Search for repetition of S[i]` ` ` `for` `(` `int` `j = n - 1; j > i; j--) {` ` ` `// If repetition occurs` ` ` `if` `(S[i] == S[j]) {` ` ` `// Store the length of` ` ` `// the substring` ` ` `diff = j - i - 1;` ` ` `// Update maximum length of` ` ` `// substring obtained` ` ` `res = max(diff, res);` ` ` `}` ` ` `}` ` ` `}` ` ` `// Return obtained maximum length` ` ` `return` `res;` `}` `// Driver Code` `int` `main()` `{` ` ` `string s = ` `"accabbacc"` `;` ` ` `cout << longestbetweenequalcharacters(s);` `}` |

*chevron_right*

*filter_none*

## Java

`// Java program to implement ` `// the above approach ` `class` `GFG{` ` ` `// Function to find the longest substring` `// between pair of repetitions of the ` `// same character` `static` `int` `longestbetweenequalcharacters(String S)` `{` ` ` ` ` `// Length of the string` ` ` `int` `n = S.length();` ` ` ` ` `// Stores the maximum length and` ` ` `// length of current substring` ` ` `// satisfying given conditions` ` ` `int` `res = -` `1` `, diff = -` `1` `;` ` ` ` ` `// Traverse the string` ` ` `for` `(` `int` `i = ` `0` `; i < n - ` `1` `; i++)` ` ` `{` ` ` ` ` `// Search for repetition of S[i]` ` ` `for` `(` `int` `j = n - ` `1` `; j > i; j--)` ` ` `{` ` ` ` ` `// If repetition occurs` ` ` `if` `(S.charAt(i) == S.charAt(j))` ` ` `{` ` ` ` ` `// Store the length of` ` ` `// the substring` ` ` `diff = j - i - ` `1` `;` ` ` ` ` `// Update maximum length of` ` ` `// substring obtained` ` ` `res = Math.max(diff, res);` ` ` `}` ` ` `}` ` ` `}` ` ` ` ` `// Return obtained maximum length` ` ` `return` `res;` `}` ` ` `// Driver code` `public` `static` `void` `main(String[] args)` `{` ` ` ` ` `String s = ` `"accabbacc"` `;` ` ` ` ` `System.out.println(` ` ` `longestbetweenequalcharacters(s));` `}` `}` `// This code is contributed by code_hunt` |

*chevron_right*

*filter_none*

## Python3

`# Python3 program to implement` `# the above approach` `# Function to find the longest` `# substring between pair of ` `# repetitions of the same character` `def` `longestbetweenequalcharacters(S):` ` ` ` ` `# Length of the string` ` ` `n ` `=` `len` `(S)` ` ` `# Stores the maximum length and` ` ` `# length of current substring` ` ` `# satisfying given conditions` ` ` `res ` `=` `-` `1` ` ` `diff ` `=` `-` `1` ` ` `# Traverse the string` ` ` `for` `i ` `in` `range` `(n ` `-` `1` `):` ` ` ` ` `# Search for repetition of S[i]` ` ` `for` `j ` `in` `range` `(n ` `-` `1` `, i, ` `-` `1` `):` ` ` ` ` `# If repetition occurs` ` ` `if` `(S[i] ` `=` `=` `S[j]):` ` ` `# Store the length of` ` ` `# the substring` ` ` `diff ` `=` `j ` `-` `i ` `-` `1` ` ` `# Update maximum length of` ` ` `# substring obtained` ` ` `res ` `=` `max` `(diff, res)` ` ` `# Return obtained maximum length` ` ` `return` `res` `# Driver Code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` ` ` `s ` `=` `"accabbacc"` ` ` ` ` `print` `(longestbetweenequalcharacters(s))` `# This code is contributed by doreamon_` |

*chevron_right*

*filter_none*

## C#

`// C# program to implement` `// the above approach ` `using` `System;` `class` `GFG{` ` ` `// Function to find the longest substring` `// between pair of repetitions of the ` `// same character` `static` `int` `longestbetweenequalcharacters(String S)` `{` ` ` ` ` `// Length of the string` ` ` `int` `n = S.Length;` ` ` ` ` `// Stores the maximum length and` ` ` `// length of current substring` ` ` `// satisfying given conditions` ` ` `int` `res = -1, diff = -1;` ` ` ` ` `// Traverse the string` ` ` `for` `(` `int` `i = 0; i < n - 1; i++)` ` ` `{` ` ` ` ` `// Search for repetition of S[i]` ` ` `for` `(` `int` `j = n - 1; j > i; j--)` ` ` `{` ` ` ` ` `// If repetition occurs` ` ` `if` `(S[i] == S[j])` ` ` `{` ` ` ` ` `// Store the length of` ` ` `// the substring` ` ` `diff = j - i - 1;` ` ` ` ` `// Update maximum length of` ` ` `// substring obtained` ` ` `res = Math.Max(diff, res);` ` ` `}` ` ` `}` ` ` `}` ` ` ` ` `// Return obtained maximum length` ` ` `return` `res;` `}` ` ` `// Driver code` `public` `static` `void` `Main()` `{` ` ` `string` `s = ` `"accabbacc"` `;` ` ` ` ` `Console.WriteLine(` ` ` `longestbetweenequalcharacters(s));` `}` `}` `// This code is contributed by sanjoy_62` |

*chevron_right*

*filter_none*

**Output:**

6

**Time Complexity: **O(N^{2})**Auxiliary Space: **O(1)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready.