CSHARP

longest substring without repeating characters

// JavaScript Solution - Credit to @albertchanged on LC
// Sliding window implementation
var lengthOfLongestSubstring = function(s) {
  if (!s.length) return 0;
  
  let left = 0, right = 0;
  let maxLength = -Infinity;
  const set = new Set();

  while (right < s.length) {
    // If s[right] has not been seen yet
    if (!set.has(s[right])) {
      // Add it to the set
      set.add(s[right]);
      // Increase size of window to right
      right++;
      // Update maxLength; set size represents length of unique substring
      maxLength = Math.max(maxLength, set.size);
    } else {
      // We've seen s[right] so we need to shrink the window
      // Delete s[left] from set
      set.delete(s[left]);
      // Shrink window from left
      left++;
    }
  }

  return maxLength;
}

Longest Substring Without Repeating Characters

/*
    Javascript Solution - With Better Performance
    Methods used in this algorithm are:
    1. Two Pointer Technique
    2. Sliding Window Technique
    3. Hash Map

    Time complexity: O(N)
    Space complexity: O(N),
    If the string does not contain repeated characters, 
    then the hash map will contain N characters.
*/

const s = "sdtvtsgh"

const lengthOfLongestSubstring = function(s) {
    let max_length = 0
    let map = {}
    let l = 0

    for (let r = 0; r < s.length; r++) {
        if (map.hasOwnProperty(s[r])) {
            const collision_index = map[s[r]]

            if (collision_index >= l) {
                l = collision_index + 1
            }

            delete map[s[collision_index]]
        }

        if (!map.hasOwnProperty(s[r])) {
            map[s[r]] = r
        }

        if (max_length < (r + 1) - l) {
            max_length = (r + 1) - l
        }

        console.log(map, ':   (', r + 1, '-', l, ')')
    }

    return max_length
}


console.log(lengthOfLongestSubstring(s))

// [Log]:
{ s: 0 } :   ( 1 - 0 )
{ s: 0, d: 1 } :   ( 2 - 0 )
{ s: 0, d: 1, t: 2 } :   ( 3 - 0 )
{ s: 0, d: 1, t: 2, v: 3 } :   ( 4 - 0 )
{ s: 0, d: 1, v: 3, t: 4 } :   ( 5 - 3 )
{ d: 1, v: 3, t: 4, s: 5 } :   ( 6 - 3 )
{ d: 1, v: 3, t: 4, s: 5, g: 6 } :   ( 7 - 3 )
{ d: 1, v: 3, t: 4, s: 5, g: 6, h: 7 } :   ( 8 - 3 )

max_length = 5

longest substring without repeating characters

class Solution {
    public int lengthOfLongestSubstring(String s) {int max=0;
        HashMap<Character,Integer>hm=new HashMap<>();
        for(int i=0,j=0;i<s.length();i++){
            if(hm.containsKey(s.charAt(i))){
                j=Math.max(j,hm.get(s.charAt(i))+1);
            }
            hm.put(s.charAt(i),i);
            max=Math.max(max,i-j+1);
        }
        return max;
        
    }
}

longest substring without repeating characters

public class Solution 
{
    public int LengthOfLongestSubstring(string str) 
    {
        var dict = new Dictionary<char, int>();
        var max = 0;
        int start = 0;
        for (int i = 0; i < str.Length; i++)
        {
            var x = str[i];
            if (!dict.ContainsKey(x))
            {
                dict.Add(x, 1);
            }
            else
            {
                dict[x] += 1;
                while (start <= i && dict.ContainsKey(str[start]) && dict[x] > 1)
                {
                    dict[str[start]]--;
                    if (dict[str[start]] == 0)
                        dict.Remove(str[start]);
                    start++;
                }
            }
            max = Math.Max(max, i - start + 1);
        }
        return max;
    }
}

3. Longest Substring Without Repeating Characters

import java.io.*;
 
class GFG {
    public static int longestUniqueSubsttr(String str)
    {
        String test = "";
 
        // Result
        int maxLength = -1;
 
        // Return zero if string is empty
        if (str.isEmpty()) {
            return 0;
        }
        // Return one if string length is one
        else if (str.length() == 1) {
            return 1;
        }
        for (char c : str.toCharArray()) {
            String current = String.valueOf(c);
 
            // If string already contains the character
            // Then substring after repeating character
            if (test.contains(current)) {
                test = test.substring(test.indexOf(current)
                                      + 1);
            }
            test = test + String.valueOf(c);
            maxLength = Math.max(test.length(), maxLength);
        }
 
        return maxLength;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        String str = "geeksforgeeks";
        System.out.println("The input string is " + str);
 
        int len = longestUniqueSubsttr(str);
        System.out.println("The length of the longest "
                           + "non-repeating character "
                           + "substring is " + len);
    }
}
 
// This code is contributed by Alex Bennet

Longest Substring Without Repeating Characters

class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        
    }
};

Longest Substring Without Repeating Characters

class Solution {
    public int lengthOfLongestSubstring(String s) {
        
    }
}

Longest Substring Without Repeating Characters

int lengthOfLongestSubstring(char * s){

}

Longest Substring Without Repeating Characters

/**
 * @param {string} s
 * @return {number}
 */
var lengthOfLongestSubstring = function(s) {
    
};

Longest Substring Without Repeating Characters

# @param {String} s
# @return {Integer}
def length_of_longest_substring(s)
    
end

Longest Substring Without Repeating Characters

class Solution {

    /**
     * @param String $s
     * @return Integer
     */
    function lengthOfLongestSubstring($s) {
        
    }
}

Longest Substring Without Repeating Characters

function lengthOfLongestSubstring(s: string): number {

};

find longest substring without repeating characters

One thing needs to be mentioned is that when asked to find maximum substring, we should update maximum after the inner while loop to guarantee that the substring is valid. On the other hand, when asked to find minimum substring, we should update minimum inside the inner while loop.

The code of solving Longest Substring with At Most Two Distinct Characters is below:

int lengthOfLongestSubstringTwoDistinct(string s) {
        vector<int> map(128, 0);
        int counter=0, begin=0, end=0, d=0; 
        while(end<s.size()){
            if(map[s[end++]]++==0) counter++;
            while(counter>2) if(map[s[begin++]]--==1) counter--;
            d=max(d, end-begin);
        }
        return d;
    }
The code of solving Longest Substring Without Repeating Characters is below:

Update 01.04.2016, thanks @weiyi3 for advise.

int lengthOfLongestSubstring(string s) {
        vector<int> map(128,0);
        int counter=0, begin=0, end=0, d=0; 
        while(end<s.size()){
            if(map[s[end++]]++>0) counter++; 
            while(counter>0) if(map[s[begin++]]-->1) counter--;
            d=max(d, end-begin); //while valid, update d
        }
        return d;
    }
I think this post deserves some upvotes! : )

Longest Substring Without Repeating Characters

public class Solution {
    public int LengthOfLongestSubstring(string s) {
        
    }
}

Longest Substring Without Repeating Characters

class Solution {
    func lengthOfLongestSubstring(_ s: String) -> Int {
        
    }
}