Length of the longest substring with no consecutive same letters
Last Updated :
23 Mar, 2023
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 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# 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 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 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
?>
<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>
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[c] >= left) {
left = mp[c] + 1;
}
mp[c] = right;
maxLen = max(maxLen, right - left + 1);
right++;
}
return maxLen;
}
int main() {
string s = "abcdde";
cout << longestSubstring(s) << endl;
return 0;
}
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));
}
}
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)
// 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[c] >= left) {
left = mp[c] + 1;
}
mp[c] = 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
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);
Time Complexity: O(n)
Auxiliary Space: O(min(m, n))
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