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Length of the longest substring with no consecutive same letters

Last Updated : 23 Mar, 2023
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Given a string str, the task is to find the length of the longest sub-string which does not have any pair of consecutive same characters. 
Examples: 
 

Input: str = “abcdde” 
Output:
“abcd” is the longest
Input: str = “ccccdeededff” 
Output:
“ededf” is the longest 
 

 

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

  • Initialize cnt and maxi as 1 initially, since this is the minimum answer of the length of the longest answer.
  • Iterate in the string from 1 to n – 1 and increment cnt by 1 if str[i] != str[i – 1].
  • If str[i] == str[i – 1], then re-initialize cnt as 1 and maxi to max(maxi, cnt).

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 required sub-string
int longestSubstring(string s)
{
    int cnt = 1;
    int maxi = 1;
 
    // Get the length of the string
    int n = s.length();
 
    // Iterate in the string
    for (int i = 1; i < n; i++) {
 
        // Check for not consecutive
        if (s[i] != s[i - 1])
            cnt++;
        else {
 
            // If cnt greater than maxi
            maxi = max(cnt, maxi);
 
            // Re-initialize
            cnt = 1;
        }
    }
 
    // Check after iteration
    // is complete
    maxi = max(cnt, maxi);
 
    return maxi;
}
 
// Driver code
int main()
{
    string s = "abcdde";
    cout << longestSubstring(s);
 
    return 0;
}


Java




// Java implementation of the approach
import java.lang.Math;
 
class GfG
{
 
    // Function to return the length
    // of the required sub-string
    static int longestSubstring(String s)
    {
        int cnt = 1, maxi = 1;
     
        // Get the length of the string
        int n = s.length();
     
        // Iterate in the string
        for (int i = 1; i < n; i++)
        {
     
            // Check for not consecutive
            if (s.charAt(i) != s.charAt(i-1))
                cnt++;
            else
            {
     
                // If cnt greater than maxi
                maxi = Math.max(cnt, maxi);
     
                // Re-initialize
                cnt = 1;
            }
        }
     
        // Check after iteration is complete
        maxi = Math.max(cnt, maxi);
     
        return maxi;
    }
 
    // Driver code
    public static void main(String []args)
    {
         
        String s = "abcdde";
        System.out.println(longestSubstring(s));
    }
}
 
// This code is contributed by Rituraj Jain


C#




// C# implementation of the approach
using System;
 
class GfG
{
 
    // Function to return the length
    // of the required sub-string
    static int longestSubstring(string s)
    {
        int cnt = 1, maxi = 1;
     
        // Get the length of the string
        int n = s.Length;
     
        // Iterate in the string
        for (int i = 1; i < n; i++)
        {
     
            // Check for not consecutive
            if (s[i] != s[i - 1])
                cnt++;
            else
            {
     
                // If cnt greater than maxi
                maxi = Math.Max(cnt, maxi);
     
                // Re-initialize
                cnt = 1;
            }
        }
     
        // Check after iteration is complete
        maxi = Math.Max(cnt, maxi);
     
        return maxi;
    }
 
    // Driver code
    static void Main()
    {
         
        string s = "abcdde";
        Console.WriteLine(longestSubstring(s));
    }
}
 
// This code is contributed by mits


Python3




# Python3 implementation of the approach
 
# Function to return the length
# of the required sub-string
def longestSubstring(s) :
 
    cnt = 1;
    maxi = 1;
 
    # Get the length of the string
    n = len(s);
 
    # Iterate in the string
    for i in range(1, n) :
 
        # Check for not consecutive
        if (s[i] != s[i - 1]) :
            cnt += 1;
             
        else :
             
            # If cnt greater than maxi
            maxi = max(cnt, maxi);
 
            # Re-initialize
            cnt = 1;
 
    # Check after iteration
    # is complete
    maxi = max(cnt, maxi);
 
    return maxi;
 
# Driver code
if __name__ == "__main__" :
     
    s = "abcdde";
    print(longestSubstring(s));
     
# This code is contributed by Ryuga


PHP




<?php
// PHP implementation of the approach
 
// Function to return the length
// of the required sub-string
function longestSubstring($s)
{
    $cnt = 1;
    $maxi = 1;
 
    // Get the length of the string
    $n = strlen($s);
 
    // Iterate in the string
    for ($i = 1; $i < $n; $i++)
    {
 
        // Check for not consecutive
        if ($s[$i] != $s[$i - 1])
            $cnt++;
        else
        {
 
            // If cnt greater than maxi
            $maxi = max($cnt, $maxi);
 
            // Re-initialize
            $cnt = 1;
        }
    }
 
    // Check after iteration
    // is complete
    $maxi = max($cnt, $maxi);
 
    return $maxi;
}
 
// Driver code
$s = "abcdde";
echo longestSubstring($s);
 
// This code is contributed by Akanksha Rai
?>


Javascript




<script>
// javascript implementation of the approach class GfG
 
// Function to return the length
// of the required sub-string
function longestSubstring(s)
{
    var cnt = 1, maxi = 1;
 
    // Get the length of the string
    var n = s.length;
 
    // Iterate in the string
    for (i = 1; i < n; i++)
    {
 
        // Check for not consecutive
        if (s.charAt(i) != s.charAt(i-1))
            cnt++;
        else
        {
 
            // If cnt greater than maxi
            maxi = Math.max(cnt, maxi);
 
            // Re-initialize
            cnt = 1;
        }
    }
 
    // Check after iteration is complete
    maxi = Math.max(cnt, maxi);
 
    return maxi;
}
 
// Driver code
var s = "abcdde";
document.write(longestSubstring(s));
 
 
// This code contributed by shikhasingrajput
</script>


Output

4

Time Complexity: O(n)
Auxiliary Space: O(1)

Approach 2: Using HashTable:

  • Create a hash table to keep track of the most recent occurrence of each character.
  • Initialize two pointers, left and right, to the beginning of the string.
  • Iterate through the string with the right pointer, updating the hash table and the length of the current substring with each new character encountered.
  • If a repeated character is encountered, move the left pointer to the next character after the previously occurring instance of that character and update the hash table and substring length accordingly.
  • Keep track of the maximum substring length encountered so far.
  • Return the maximum substring length.

Here’s the implementation of this approach in C++:

C++




#include <bits/stdc++.h>
using namespace std;
 
int longestSubstring(string s) {
    unordered_map<char, int> mp;
    int left = 0, right = 0, maxLen = 0;
    while (right < s.length()) {
        char c = s[right];
        if (mp.find(c) != mp.end() && mp >= left) {
            left = mp + 1;
        }
        mp = right;
        maxLen = max(maxLen, right - left + 1);
        right++;
    }
    return maxLen;
}
 
int main() {
    string s = "abcdde";
    cout << longestSubstring(s) << endl;
    return 0;
}


Java




import java.util.*;
 
public class Main {
 
    public static int longestSubstring(String s) {
        Map<Character, Integer> mp = new HashMap<>();
        int left = 0, right = 0, maxLen = 0;
        while (right < s.length()) {
            char c = s.charAt(right);
            if (mp.containsKey(c) && mp.get(c) >= left) {
                left = mp.get(c) + 1;
            }
            mp.put(c, right);
            maxLen = Math.max(maxLen, right - left + 1);
            right++;
        }
        return maxLen;
    }
 
    public static void main(String[] args) {
        String s = "abcdde";
        System.out.println(longestSubstring(s));
    }
}


Python3




def longestSubstring(s):
    mp = {}
    left = 0
    maxLen = 0
    for right in range(len(s)):
        if s[right] in mp and mp[s[right]] >= left:
            left = mp[s[right]] + 1
        mp[s[right]] = right
        maxLen = max(maxLen, right - left + 1)
    return maxLen
 
if __name__ == "__main__":
    s = "abcdde"
    maxLen = longestSubstring(s)
    print(maxLen)


C#




// Added C# code for the above approach
using System;
using System.Collections.Generic;
 
class Program {
    static int LongestSubstring(string s)
    {
        Dictionary<char, int> mp
            = new Dictionary<char, int>();
        int left = 0, right = 0, maxLen = 0;
        while (right < s.Length) {
            char c = s[right];
            if (mp.ContainsKey(c) && mp >= left) {
                left = mp + 1;
            }
            mp = right;
            maxLen = Math.Max(maxLen, right - left + 1);
            right++;
        }
        return maxLen;
    }
 
    static void Main(string[] args)
    {
        string s = "abcdde";
        Console.WriteLine(LongestSubstring(s));
    }
}
 
// Contributed by adityasha4x71


Javascript




function longestSubstring(s) {
  const mp = {};
  let left = 0;
  let maxLen = 0;
  for (let right = 0; right < s.length; right++) {
    if (s[right] in mp && mp[s[right]] >= left) {
      left = mp[s[right]] + 1;
    }
    mp[s[right]] = right;
    maxLen = Math.max(maxLen, right - left + 1);
  }
  return maxLen;
}
 
const s = "abcdde";
const maxLen = longestSubstring(s);
console.log(maxLen);


Output

4

Time Complexity: O(n)
Auxiliary Space: O(min(m, n))



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