Create Longest_Substring_Without_Repeating_Characters.cpp

others/Longest_Substring_Without_Repeating_Characters.cpp

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+/**
+* 3.Longest_Substring_Without_Repeating_Characters
+* Problem link :- https://leetcode.com/problems/longest-substring-without-repeating-characters/description/
+* Intiuation :
+  Intuition is strightforward and simple, we track the frequencey and as we know we can't use string to track longest substring without repeating characters, as poping a char from front of string is not in O(1) which could be optimized by deque approch.
+* Approch :
+    1) At first we initialize a unordered_map to track the frequncy and then.
+    2) We initialize deque for pushing the characters and if temp > res we update our res deque to temp as we need longest substring here.
+    3) And while loop is used to reduce the frequency from front doing i++ and removing charecters from temp deque as we no longer need them.
+    4) return res.size() as we only need size not string.
+* Time Complexity :
+  O(N)
+
+* Space Complexity :
+  O(N)
+* Examples:-
+  testcase1:- s = "abcabcbb", o/p = 3.
+  testcase2:- s = "bbbbb", o/p = 1.
+  testcase3:- s = "pwwkew", o/p = 3.
+* I hope this helps to understand.
+* Thank you!!
+**/
+
+// ----------------- Header files ----------------------------------
+#include <iostream> // for input and output read/write.
+#include <unordered_map> // to use it for character frequency.
+#include <deque> // for push and pop operations at O(1) time.
+#include <string> // for taking string as input
+
+using namespace std; // using the namspace standard to reduce the redundent usage of std:: when calling functions (basically reduicng an extra overhead).
+
+class Solution {
+public:
+    int lengthOfLongestSubstring(string s) {
+        // if size of string is 1 then it will be the answer.
+        if(s.size()==1) return 1;
+
+        // map used to store the characters frequency.
+        unordered_map<char,int>m;
+        int n = s.length();
+
+        // dequeue to remove from back if repeating charcters are present.
+        deque<char>temp;
+        deque<char>res;
+        int i,j;
+        for(i=0,j=0;i<n && j<n;) {
+            m[s[j]]++;
+
+            if(m[s[j]]>1) {
+                if(temp.size()>res.size()) {
+                    res = temp;
+                }
+
+                while(m[s[j]]>1) {
+                    // as repeating charecter is present pop_front() is called.
+                    temp.pop_front();
+                    m[s[i]]--;
+                    i++;
+                }
+            }
+
+            temp.push_back(s[j]);
+            j++;
+        }
+
+        if(temp.size()>res.size()) {
+            res = temp;
+        }
+
+        return res.size();
+    }
+};
+
+//-------------------- Main function -------------------------------
+int main() {
+  // object of class Solution created for function call.
+  Solution soln;
+
+
+  // user inputted string. 
+  string s;
+  cout<<"Enter the string : "<<endl;
+  cin>>s;
+
+  // function call to get the ans of length of longest substring without repeating characters.
+  cout<<soln.lengthOfLongestSubstring(s)<<endl;
+
+  return 0;
+}