fineanmol · J-B-Mugundh · Sep 30, 2024
diff --git a/...s/LeetCode-Top-300-Liked-Problems-Solutions/0002-add-two-numbers/0002-add-two-numbers.cpp b/...s/LeetCode-Top-300-Liked-Problems-Solutions/0002-add-two-numbers/0002-add-two-numbers.cpp
@@ -0,0 +1,36 @@
+/**
+ * Definition for singly-linked list.
+ * struct ListNode {
+ *     int val;
+ *     ListNode *next;
+ *     ListNode() : val(0), next(nullptr) {}
+ *     ListNode(int x) : val(x), next(nullptr) {}
+ *     ListNode(int x, ListNode *next) : val(x), next(next) {}
+ * };
+ */
+class Solution {
+public:
+    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
+        ListNode *ans = new ListNode(0);
+        ListNode *temp = ans;
+        int carry = 0;
+        while(l1 || l2 || carry){
+            int sum = carry;
+            if(l1){
+                sum += l1->val;
+                l1 = l1->next;
+            }
+
+            if(l2){
+                sum += l2->val;
+                l2 = l2->next;
+            }
+
+            carry = sum / 10;
+            ListNode* node = new ListNode(sum % 10);
+            temp->next = node;
+            temp = temp->next;
+        }
+        return ans->next;
+    }
+};
diff --git a/...butors/LeetCode-Top-300-Liked-Problems-Solutions/0002-add-two-numbers/README.md b/...butors/LeetCode-Top-300-Liked-Problems-Solutions/0002-add-two-numbers/README.md
@@ -0,0 +1,35 @@
+<h2><a href="https://leetcode.com/problems/add-two-numbers">2. Add Two Numbers</a></h2><h3>Medium</h3><hr><p>You are given two <strong>non-empty</strong> linked lists representing two non-negative integers. The digits are stored in <strong>reverse order</strong>, and each of their nodes contains a single digit. Add the two numbers and return the sum&nbsp;as a linked list.</p>
+
+<p>You may assume the two numbers do not contain any leading zero, except the number 0 itself.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2020/10/02/addtwonumber1.jpg" style="width: 483px; height: 342px;" />
+<pre>
+<strong>Input:</strong> l1 = [2,4,3], l2 = [5,6,4]
+<strong>Output:</strong> [7,0,8]
+<strong>Explanation:</strong> 342 + 465 = 807.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> l1 = [0], l2 = [0]
+<strong>Output:</strong> [0]
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
+<strong>Output:</strong> [8,9,9,9,0,0,0,1]
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li>The number of nodes in each linked list is in the range <code>[1, 100]</code>.</li>
+	<li><code>0 &lt;= Node.val &lt;= 9</code></li>
+	<li>It is guaranteed that the list represents a number that does not have leading zeros.</li>
+</ul>
diff --git a/...ring-without-repeating-characters/0003-longest-substring-without-repeating-characters.cpp b/...ring-without-repeating-characters/0003-longest-substring-without-repeating-characters.cpp
@@ -0,0 +1,20 @@
+class Solution {
+public:
+    int lengthOfLongestSubstring(string s) {
+        int n = s.length(), maxLen = 0;
+        int l = 0, r = 0;
+        unordered_map<char, int> mp;
+
+        while (r < n) {
+            if (mp.find(s[r]) != mp.end()) {
+                l = max(mp[s[r]] + 1, l); // moving left ptr only if the current left value was less than the previous occurance of current character
+            }
+            maxLen = max(maxLen, r - l + 1); 
+            mp[s[r]] = r;
+            r++;
+            if(n - l - 1 < maxLen) break;
+        }
+
+        return maxLen;
+    }
+};
diff --git a/...roblems-Solutions/0003-longest-substring-without-repeating-characters/README.md b/...roblems-Solutions/0003-longest-substring-without-repeating-characters/README.md
@@ -0,0 +1,35 @@
+<h2><a href="https://leetcode.com/problems/longest-substring-without-repeating-characters">3. Longest Substring Without Repeating Characters</a></h2><h3>Medium</h3><hr><p>Given a string <code>s</code>, find the length of the <strong>longest</strong> <span data-keyword="substring-nonempty"><strong>substring</strong></span> without repeating characters.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;abcabcbb&quot;
+<strong>Output:</strong> 3
+<strong>Explanation:</strong> The answer is &quot;abc&quot;, with the length of 3.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;bbbbb&quot;
+<strong>Output:</strong> 1
+<strong>Explanation:</strong> The answer is &quot;b&quot;, with the length of 1.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;pwwkew&quot;
+<strong>Output:</strong> 3
+<strong>Explanation:</strong> The answer is &quot;wke&quot;, with the length of 3.
+Notice that the answer must be a substring, &quot;pwke&quot; is a subsequence and not a substring.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>0 &lt;= s.length &lt;= 5 * 10<sup>4</sup></code></li>
+	<li><code>s</code> consists of English letters, digits, symbols and spaces.</li>
+</ul>
diff --git a/...-Problems-Solutions/0004-median-of-two-sorted-arrays/0004-median-of-two-sorted-arrays.cpp b/...-Problems-Solutions/0004-median-of-two-sorted-arrays/0004-median-of-two-sorted-arrays.cpp
@@ -0,0 +1,44 @@
+class Solution {
+public:
+    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
+        // T.C = O(log (m+n)), S.C = O(1)
+        // Try to figure out the minimum size array and do binary search from that to reduce T.C
+        if(nums2.size() < nums1.size()) return findMedianSortedArrays(nums2, nums1);
+        int n1 = nums1.size();
+        int n2 = nums2.size();
+
+        // Edge cases
+        if(n1 == 0)
+            return n2 % 2 == 1 ? nums2[n2 / 2] : (nums2[n2 / 2] + nums2[n2 / 2 - 1]) / 2.0;
+
+        if(n2 == 0)
+            return n1 % 2 == 1 ? nums1[n1 / 2] : (nums1[n1 / 2] + nums1[n1 / 2 - 1]) / 2.0;
+
+        int low = 0, high = n1;
+
+        while(low <= high)
+        {
+            int cut1 = (low + high) >> 1;
+            int cut2 = (n1 + n2 + 1) / 2 - cut1;
+
+            int left1 = cut1 == 0 ? INT_MIN : nums1[cut1 - 1];
+            int left2 = cut2 == 0 ? INT_MIN : nums2[cut2 - 1];
+
+            int right1 = cut1 >= n1 ? INT_MAX : nums1[cut1];
+            int right2 = cut2 >= n2 ? INT_MAX : nums2[cut2];
+
+            if(left1 <= right2 && left2 <= right1)
+            {
+                if((n1 + n2) % 2 == 0)
+                    return (max(left1, left2) + min(right1, right2)) / 2.0;
+                else
+                    return max(left1, left2);
+            }
+            else if(left1 > right2)
+                high = cut1 - 1;
+            else
+                low = cut1 + 1;
+        }
+        return 0.0;
+    }
+};
diff --git a/...ode-Top-300-Liked-Problems-Solutions/0004-median-of-two-sorted-arrays/README.md b/...ode-Top-300-Liked-Problems-Solutions/0004-median-of-two-sorted-arrays/README.md
@@ -0,0 +1,32 @@
+<h2><a href="https://leetcode.com/problems/median-of-two-sorted-arrays">4. Median of Two Sorted Arrays</a></h2><h3>Hard</h3><hr><p>Given two sorted arrays <code>nums1</code> and <code>nums2</code> of size <code>m</code> and <code>n</code> respectively, return <strong>the median</strong> of the two sorted arrays.</p>
+
+<p>The overall run time complexity should be <code>O(log (m+n))</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums1 = [1,3], nums2 = [2]
+<strong>Output:</strong> 2.00000
+<strong>Explanation:</strong> merged array = [1,2,3] and median is 2.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> nums1 = [1,2], nums2 = [3,4]
+<strong>Output:</strong> 2.50000
+<strong>Explanation:</strong> merged array = [1,2,3,4] and median is (2 + 3) / 2 = 2.5.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>nums1.length == m</code></li>
+	<li><code>nums2.length == n</code></li>
+	<li><code>0 &lt;= m &lt;= 1000</code></li>
+	<li><code>0 &lt;= n &lt;= 1000</code></li>
+	<li><code>1 &lt;= m + n &lt;= 2000</code></li>
+	<li><code>-10<sup>6</sup> &lt;= nums1[i], nums2[i] &lt;= 10<sup>6</sup></code></li>
+</ul>
diff --git a/...blems-Solutions/0005-longest-palindromic-substring/0005-longest-palindromic-substring.cpp b/...blems-Solutions/0005-longest-palindromic-substring/0005-longest-palindromic-substring.cpp
@@ -0,0 +1,31 @@
+class Solution {
+public:
+
+string expandAroundCenter(string s, int left, int right) {
+    while (left >= 0 && right < s.length() && s[left] == s[right]) {
+        left--;
+        right++;
+    }
+    return s.substr(left + 1, (right - 1) - (left + 1) + 1);
+}
+
+string longestPalindrome(string s) {
+        int n = s.length();
+        string longest = "";
+
+        for(int i = 0; i < n; i++){
+            string oddPal = expandAroundCenter(s, i, i);
+            string evenPal = expandAroundCenter(s, i, i + 1);
+
+            if(oddPal.length() > longest.length())
+                longest = oddPal;
+
+            if(evenPal.length() > longest.length())
+                longest = evenPal;
+        }
+
+        return longest;
+    }
+
+
+};
diff --git a/...e-Top-300-Liked-Problems-Solutions/0005-longest-palindromic-substring/README.md b/...e-Top-300-Liked-Problems-Solutions/0005-longest-palindromic-substring/README.md
@@ -0,0 +1,25 @@
+<h2><a href="https://leetcode.com/problems/longest-palindromic-substring">5. Longest Palindromic Substring</a></h2><h3>Medium</h3><hr><p>Given a string <code>s</code>, return <em>the longest</em> <span data-keyword="palindromic-string"><em>palindromic</em></span> <span data-keyword="substring-nonempty"><em>substring</em></span> in <code>s</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;babad&quot;
+<strong>Output:</strong> &quot;bab&quot;
+<strong>Explanation:</strong> &quot;aba&quot; is also a valid answer.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;cbbd&quot;
+<strong>Output:</strong> &quot;bb&quot;
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s.length &lt;= 1000</code></li>
+	<li><code>s</code> consist of only digits and English letters.</li>
+</ul>
diff --git a/...-Problems-Solutions/0010-regular-expression-matching/0010-regular-expression-matching.cpp b/...-Problems-Solutions/0010-regular-expression-matching/0010-regular-expression-matching.cpp
@@ -0,0 +1,26 @@
+class Solution {
+public:
+    bool isMatch(string s, string p) {
+        int m = s.length(), n = p.length();
+        vector<vector<bool>> dp(m + 1, vector<bool>(n + 1, false));
+        dp[0][0] = true;
+
+        for (int j = 1; j <= n; ++j) {
+            if (p[j - 1] == '*') {
+                dp[0][j] = dp[0][j - 2];
+            }
+        }
+
+        for (int i = 1; i <= m; ++i) {
+            for (int j = 1; j <= n; ++j) {
+                if (s[i - 1] == p[j - 1] || p[j - 1] == '.') {
+                    dp[i][j] = dp[i - 1][j - 1];
+                } else if (p[j - 1] == '*') {
+                    dp[i][j] = dp[i][j - 2] || (dp[i - 1][j] && (s[i - 1] == p[j - 2] || p[j - 2] == '.'));
+                }
+            }
+        }
+
+        return dp[m][n];
+    }
+};
diff --git a/...ode-Top-300-Liked-Problems-Solutions/0010-regular-expression-matching/README.md b/...ode-Top-300-Liked-Problems-Solutions/0010-regular-expression-matching/README.md
@@ -0,0 +1,44 @@
+<h2><a href="https://leetcode.com/problems/regular-expression-matching">10. Regular Expression Matching</a></h2><h3>Hard</h3><hr><p>Given an input string <code>s</code>&nbsp;and a pattern <code>p</code>, implement regular expression matching with support for <code>&#39;.&#39;</code> and <code>&#39;*&#39;</code> where:</p>
+
+<ul>
+	<li><code>&#39;.&#39;</code> Matches any single character.</li>
+	<li><code>&#39;*&#39;</code> Matches zero or more of the preceding element.</li>
+</ul>
+
+<p>The matching should cover the <strong>entire</strong> input string (not partial).</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;aa&quot;, p = &quot;a&quot;
+<strong>Output:</strong> false
+<strong>Explanation:</strong> &quot;a&quot; does not match the entire string &quot;aa&quot;.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;aa&quot;, p = &quot;a*&quot;
+<strong>Output:</strong> true
+<strong>Explanation:</strong> &#39;*&#39; means zero or more of the preceding element, &#39;a&#39;. Therefore, by repeating &#39;a&#39; once, it becomes &quot;aa&quot;.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;ab&quot;, p = &quot;.*&quot;
+<strong>Output:</strong> true
+<strong>Explanation:</strong> &quot;.*&quot; means &quot;zero or more (*) of any character (.)&quot;.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s.length&nbsp;&lt;= 20</code></li>
+	<li><code>1 &lt;= p.length&nbsp;&lt;= 20</code></li>
+	<li><code>s</code> contains only lowercase English letters.</li>
+	<li><code>p</code> contains only lowercase English letters, <code>&#39;.&#39;</code>, and&nbsp;<code>&#39;*&#39;</code>.</li>
+	<li>It is guaranteed for each appearance of the character <code>&#39;*&#39;</code>, there will be a previous valid character to match.</li>
+</ul>
diff --git a/...LeetCode-Top-300-Liked-Problems-Solutions/0012-integer-to-roman/0012-integer-to-roman.cpp b/...LeetCode-Top-300-Liked-Problems-Solutions/0012-integer-to-roman/0012-integer-to-roman.cpp
@@ -0,0 +1,20 @@
+class Solution {
+public:
+    string intToRoman(int num) {
+        string roman[] = {"M","CM","D","CD","C","XC","L","XL","X","IX","V","IV","I"},result = "";
+        int values[] = {1000,900,500,400,100,90,50,40,10,9,5,4,1};
+        int len = sizeof(roman)/sizeof(roman[0]);
+        int i = 0;
+
+        while(num && i < len){
+            if(num - values[i] >= 0)
+            {
+                result += roman[i];
+                num -= values[i];
+            }
+            else
+                i++;
+        }
+        return result;
+    }
+};