Create Longest_Substring_Without_Repeating_Characters.cpp

others/longest_substring_without_repeating_characters.cpp

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+/**
+ * @file
+ * @brief Solution for Longest Substring Without Repeating Characters problem.
+ * @details
+ * Problem link: https://leetcode.com/problems/longest-substring-without-repeating-characters/description/
+ * 
+ * Intuition:
+ * The intuition is straightforward and simple. We track the frequency of characters.
+ * Since we can't use a string to track the longest substring without repeating characters
+ * efficiently (as removing a character from the front of a string isn't O(1)),
+ * we optimize the solution using a deque approach.
+ *
+ * Approach:
+ * 1) Initialize an unordered_map to track the frequency of characters.
+ * 2) Use a deque for pushing characters, and update the result deque (`res`) with the current deque (`temp`) 
+ *    whenever we find a longer substring.
+ * 3) Use a while loop to reduce the frequency from the front, incrementing `i`, 
+ *    and removing characters from the `temp` deque as we no longer need them.
+ * 4) Return `res.size()` as we are interested in the length of the longest substring.
+ * 
+ * Time Complexity: O(N)
+ * Space Complexity: O(N)
+ * 
+ * Examples:
+ * - testcase1: s = "abcabcbb", output = 3.
+ * - testcase2: s = "bbbbb", output = 1.
+ * - testcase3: s = "pwwkew", output = 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 namespace standard to reduce redundant usage of std::
+
+//------------------ Longest_Substring Class -------------------------------
+/**
+ * @class Solution
+ * @brief Class that solves the Longest Substring Without Repeating Characters problem.
+ */
+class Longest_Substring {
+public:
+    /**
+     * @brief Function to find the length of the longest substring without repeating characters.
+     * @param s Input string.
+     * @returns Length of the longest substring.
+     */
+    int lengthOfLongestSubstring(string s) {
+        // If the size of string is 1, then it will be the answer.
+        if (s.size() == 1) return 1;
+
+        // Map used to store the character frequency.
+        unordered_map<char, int> m;
+        int n = s.length();
+
+        // Deque to remove from back if repeating characters are present.
+        deque<char> temp;
+        deque<char> res;
+        int i, j;
+
+        // Sliding window approach using two pointers.
+        for (i = 0, j = 0; i < n && j < n;) {
+            m[s[j]]++;
+
+            // If repeating character found, update result and remove from the front.
+            if (m[s[j]] > 1) {
+                if (temp.size() > res.size()) {
+                    res = temp;
+                }
+
+                while (m[s[j]] > 1) {
+                    temp.pop_front();
+                    m[s[i]]--;
+                    i++;
+                }
+            }
+
+            // Add the current character to the deque.
+            temp.push_back(s[j]);
+            j++;
+        }
+
+        // Final check to update result.
+        if (temp.size() > res.size()) {
+            res = temp;
+        }
+
+        return res.size();  // Return the length of the longest substring.
+    }
+};
+
+//-------------------- Main function -------------------------------
+/**
+ * @brief Main function.
+ * @returns 0 on successful execution.
+ */
+int main() {
+    // Create an object of the Solution class to call the function.
+    Longest_Substring s;
+
+    // User inputted string.
+    string str;
+    cout << "Enter the string: " << endl;
+    cin >> str;
+
+    // Function call to get the length of the longest substring without repeating characters.
+    cout << soln.lengthOfLongestSubstring(str) << endl;
+
+    return 0;
+}