# Longest substring such that no three consecutive characters are same

Given string str, the task is to find the length of the longest substring of str such that no three consecutive characters in the substring are same.

Examples:

Input: str = “baaabbabbb”
Output: 7
“aabbabb” is the required substring.

Input: str = “babba”
Output: 5
Given string itself is the longest substring.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The following steps can be followed to solve the problem:

• If the length of the given string is less than 3 then the length of the string will be the answer.
• Initialize temp and ans as 2 initially, since this is the minimum length of the longest substring when the length of the given string is greater than 2.
• Iterate in the string from 2 to N – 1 and increment temp by 1 if str[i] != str[i – 1] or str[i] != str[i – 2].
• Else re-initialize temp = 2 and ans = max(ans, temp).

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the length of the ` `// longest substring such that no three ` `// consecutive characters are same ` `int` `maxLenSubStr(string& s) ` `{ ` `    ``// If the length of the given string ` `    ``// is less than 3 ` `    ``if` `(s.length() < 3) ` `        ``return` `s.length(); ` ` `  `    ``// Initialize temporary and final ans ` `    ``// to 2 as this is the minimum length ` `    ``// of substring when length of the given ` `    ``// string is greater than 2 ` `    ``int` `temp = 2; ` `    ``int` `ans = 2; ` ` `  `    ``// Traverse the string from the ` `    ``// third character to the last ` `    ``for` `(``int` `i = 2; i < s.length(); i++) { ` ` `  `        ``// If no three consecutive characters ` `        ``// are same then increment temporary count ` `        ``if` `(s[i] != s[i - 1] || s[i] != s[i - 2]) ` `            ``temp++; ` ` `  `        ``// Else update the final ans and ` `        ``// reset the temporary count ` `        ``else` `{ ` `            ``ans = max(temp, ans); ` `            ``temp = 2; ` `        ``} ` `    ``} ` ` `  `    ``ans = max(temp, ans); ` ` `  `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``string s = ``"baaabbabbb"``; ` ` `  `    ``cout << maxLenSubStr(s); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `import` `java.util.*; ` ` `  `class` `GFG  ` `{ ` ` `  `// Function to return the length of the ` `// longest substring such that no three ` `// consecutive characters are same ` `static` `int` `maxLenSubStr(String s) ` `{ ` `    ``// If the length of the given string ` `    ``// is less than 3 ` `    ``if` `(s.length() < ``3``) ` `        ``return` `s.length(); ` ` `  `    ``// Initialize temporary and final ans ` `    ``// to 2 as this is the minimum length ` `    ``// of substring when length of the given ` `    ``// string is greater than 2 ` `    ``int` `temp = ``2``; ` `    ``int` `ans = ``2``; ` ` `  `    ``// Traverse the string from the ` `    ``// third character to the last ` `    ``for` `(``int` `i = ``2``; i < s.length(); i++)  ` `    ``{ ` ` `  `        ``// If no three consecutive characters ` `        ``// are same then increment temporary count ` `        ``if` `(s.charAt(i) != s.charAt(i - ``1``) || ` `            ``s.charAt(i) != s.charAt(i - ``2``)) ` `            ``temp++; ` ` `  `        ``// Else update the final ans and ` `        ``// reset the temporary count ` `        ``else` `        ``{ ` `            ``ans = Math.max(temp, ans); ` `            ``temp = ``2``; ` `        ``} ` `    ``} ` `    ``ans = Math.max(temp, ans); ` ` `  `    ``return` `ans; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``String s = ``"baaabbabbb"``; ` ` `  `    ``System.out.println(maxLenSubStr(s)); ` `} ` `}  ` ` `  `// This code is contributed by PrinciRaj1992 `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to return the length of the ` `# longest substring such that no three ` `# consecutive characters are same ` `def` `maxLenSubStr(s): ` `     `  `    ``# If the length of the given string ` `    ``# is less than 3 ` `    ``if` `(``len``(s) < ``3``): ` `        ``return` `len``(s) ` ` `  `    ``# Initialize temporary and final ans ` `    ``# to 2 as this is the minimum length ` `    ``# of substring when length of the given ` `    ``# string is greater than 2 ` `    ``temp ``=` `2` `    ``ans ``=` `2` ` `  `    ``# Traverse the string from the ` `    ``# third character to the last ` `    ``for` `i ``in` `range``(``2``, ``len``(s)): ` ` `  `        ``# If no three consecutive characters ` `        ``# are same then increment temporary count ` `        ``if` `(s[i] !``=` `s[i ``-` `1``] ``or` `s[i] !``=` `s[i ``-` `2``]): ` `            ``temp ``+``=` `1` ` `  `        ``# Else update the final ans and ` `        ``# reset the temporary count ` `        ``else``: ` `            ``ans ``=` `max``(temp, ans) ` `            ``temp ``=` `2` `    ``ans ``=` `max``(temp, ans) ` ` `  `    ``return` `ans ` ` `  `# Driver code ` `s ``=` `"baaabbabbb"` ` `  `print``(maxLenSubStr(s)) ` ` `  `# This code is contributed by Mohit Kumar `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `// Function to return the length of the ` `// longest substring such that no three ` `// consecutive characters are same ` `static` `int` `maxLenSubStr(String s) ` `{ ` `    ``// If the length of the given string ` `    ``// is less than 3 ` `    ``if` `(s.Length < 3) ` `        ``return` `s.Length; ` ` `  `    ``// Initialize temporary and final ans ` `    ``// to 2 as this is the minimum length ` `    ``// of substring when length of the given ` `    ``// string is greater than 2 ` `    ``int` `temp = 2; ` `    ``int` `ans = 2; ` ` `  `    ``// Traverse the string from the ` `    ``// third character to the last ` `    ``for` `(``int` `i = 2; i < s.Length; i++)  ` `    ``{ ` ` `  `        ``// If no three consecutive characters ` `        ``// are same then increment temporary count ` `        ``if` `(s[i] != s[i - 1] || ` `            ``s[i] != s[i - 2]) ` `            ``temp++; ` ` `  `        ``// Else update the final ans and ` `        ``// reset the temporary count ` `        ``else` `        ``{ ` `            ``ans = Math.Max(temp, ans); ` `            ``temp = 2; ` `        ``} ` `    ``} ` `    ``ans = Math.Max(temp, ans); ` ` `  `    ``return` `ans; ` `} ` ` `  `// Driver code ` `static` `public` `void` `Main () ` `{ ` `    ``String s = ``"baaabbabbb"``;  ` `    ``Console.Write(maxLenSubStr(s)); ` `} ` `}  ` ` `  `// This code is contributed by ajit. `

Output:

```7
```

Time Complexity: O(N) where N is the length of the string.
Space Complexity: O(1)

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