Length of the longest substring with no consecutive same letters

Namachila November 19, 2025

0 0 6 minutes read

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: 4
“abcd” is the longest
Input: str = “ccccdeededff”
Output: 5
“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

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

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

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!