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Longest substring such that no three consecutive characters are same

  • Difficulty Level : Medium
  • Last Updated : 11 May, 2021

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

 

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 <bits/stdc++.h>
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.

Javascript




<script>
    // Javascript implementation of the approach
     
    // Function to return the length of the
    // longest substring such that no three
    // consecutive characters are same
    function maxLenSubStr(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
        let temp = 2;
        let ans = 2;
 
        // Traverse the string from the
        // third character to the last
        for (let 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;
    }
     
    let s = "baaabbabbb";
    document.write(maxLenSubStr(s));
 
</script>
Output: 
7

 

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

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