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-stringint 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 codeint 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-stringfunction 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 codevar 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 maxLenif __name__ == "__main__": s = "abcdde" maxLen = longestSubstring(s) print(maxLen) |
C#
// Added C# code for the above approachusing 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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