Added Cpp

src/main/cpp/g0001_0100/s0001_two_sum/Solution.cpp

@@ -0,0 +1,21 @@
+// #Easy #Top_100_Liked_Questions #Top_Interview_Questions #Array #Hash_Table
+// #Data_Structure_I_Day_2_Array #Level_1_Day_13_Hashmap #Udemy_Arrays #Big_O_Time_O(n)_Space_O(n)
+// #2024_05_12_Time_4_ms_(94.42%)_Space_14.1_MB_(17.14%)
+
+#include <vector>
+#include <unordered_map>
+
+class Solution {
+public:
+    std::vector<int> twoSum(std::vector<int>& numbers, int target) {
+        std::unordered_map<int, int> indexMap;
+        for (int i = 0; i < numbers.size(); i++) {
+            int requiredNum = target - numbers[i];
+            if (indexMap.find(requiredNum) != indexMap.end()) {
+                return {indexMap[requiredNum], i};
+            }
+            indexMap[numbers[i]] = i;
+        }
+        return {-1, -1};
+    }
+};

src/main/cpp/g0001_0100/s0001_two_sum/readme.md

@@ -0,0 +1,38 @@
+1\. Two Sum
+
+Easy
+
+Given an array of integers `nums` and an integer `target`, return _indices of the two numbers such that they add up to `target`_.
+
+You may assume that each input would have **_exactly_ one solution**, and you may not use the _same_ element twice.
+
+You can return the answer in any order.
+
+**Example 1:**
+
+**Input:** nums = [2,7,11,15], target = 9
+
+**Output:** [0,1]
+
+**Explanation:** Because nums[0] + nums[1] == 9, we return [0, 1]. 
+
+**Example 2:**
+
+**Input:** nums = [3,2,4], target = 6
+
+**Output:** [1,2] 
+
+**Example 3:**
+
+**Input:** nums = [3,3], target = 6
+
+**Output:** [0,1] 
+
+**Constraints:**
+
+*   <code>2 <= nums.length <= 10<sup>4</sup></code>
+*   <code>-10<sup>9</sup> <= nums[i] <= 10<sup>9</sup></code>
+*   <code>-10<sup>9</sup> <= target <= 10<sup>9</sup></code>
+*   **Only one valid answer exists.**
+
+**Follow-up:** Can you come up with an algorithm that is less than <code>O(n<sup>2</sup>)</code> time complexity?

src/main/cpp/g0001_0100/s0002_add_two_numbers/Solution.cpp

@@ -0,0 +1,42 @@
+// #Medium #Top_100_Liked_Questions #Top_Interview_Questions #Math #Linked_List #Recursion
+// #Data_Structure_II_Day_10_Linked_List #Programming_Skills_II_Day_15
+// #Big_O_Time_O(max(N,M))_Space_O(max(N,M)) #2024_05_12_Time_16_ms_(82.17%)_Space_76.1_MB_(52.68%)
+
+/**
+ * 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* dummyHead = new ListNode(0);
+        ListNode* p = l1;
+        ListNode* q = l2;
+        ListNode* curr = dummyHead;
+        int carry = 0;
+        while (p != nullptr || q != nullptr) {
+            int x = (p != nullptr) ? p->val : 0;
+            int y = (q != nullptr) ? q->val : 0;
+            int sum = carry + x + y;
+            carry = sum / 10;
+            curr->next = new ListNode(sum % 10);
+            curr = curr->next;
+            if (p != nullptr) {
+                p = p->next;
+            }
+            if (q != nullptr) {
+                q = q->next;
+            }
+        }
+        if (carry > 0) {
+            curr->next = new ListNode(carry);
+        }
+        return dummyHead->next;
+    }
+};

src/main/cpp/g0001_0100/s0002_add_two_numbers/readme.md

@@ -0,0 +1,35 @@
+2\. Add Two Numbers
+
+Medium
+
+You are given two **non-empty** linked lists representing two non-negative integers. The digits are stored in **reverse order**, and each of their nodes contains a single digit. Add the two numbers and return the sum as a linked list.
+
+You may assume the two numbers do not contain any leading zero, except the number 0 itself.
+
+**Example 1:**
+
+![](https://assets.leetcode.com/uploads/2020/10/02/addtwonumber1.jpg)
+
+**Input:** l1 = [2,4,3], l2 = [5,6,4]
+
+**Output:** [7,0,8]
+
+**Explanation:** 342 + 465 = 807. 
+
+**Example 2:**
+
+**Input:** l1 = [0], l2 = [0]
+
+**Output:** [0] 
+
+**Example 3:**
+
+**Input:** l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
+
+**Output:** [8,9,9,9,0,0,0,1] 
+
+**Constraints:**
+
+*   The number of nodes in each linked list is in the range `[1, 100]`.
+*   `0 <= Node.val <= 9`
+*   It is guaranteed that the list represents a number that does not have leading zeros.

src/main/cpp/g0001_0100/s0003_longest_substring_without_repeating_characters/Solution.cpp

@@ -0,0 +1,31 @@
+// #Medium #Top_100_Liked_Questions #Top_Interview_Questions #String #Hash_Table #Sliding_Window
+// #Algorithm_I_Day_6_Sliding_Window #Level_2_Day_14_Sliding_Window/Two_Pointer #Udemy_Strings
+// #Big_O_Time_O(n)_Space_O(1) #2024_05_12_Time_5_ms_(86.87%)_Space_10.4_MB_(74.49%)
+
+#include <string>
+#include <vector>
+#include <algorithm>
+
+class Solution {
+public:
+    int lengthOfLongestSubstring(std::string s) {
+        std::vector<int> lastIndices(256, -1);
+        int maxLen = 0;
+        int curLen = 0;
+        int start = 0;
+        for (int i = 0; i < s.length(); i++) {
+            char cur = s[i];
+            if (lastIndices[cur] < start) {
+                lastIndices[cur] = i;
+                curLen++;
+            } else {
+                int lastIndex = lastIndices[cur];
+                start = lastIndex + 1;
+                curLen = i - start + 1;
+                lastIndices[cur] = i;
+            }
+            maxLen = std::max(maxLen, curLen);
+        }
+        return maxLen;
+    }
+};

src/main/cpp/g0001_0100/s0003_longest_substring_without_repeating_characters/readme.md

@@ -0,0 +1,40 @@
+3\. Longest Substring Without Repeating Characters
+
+Medium
+
+Given a string `s`, find the length of the **longest substring** without repeating characters.
+
+**Example 1:**
+
+**Input:** s = "abcabcbb"
+
+**Output:** 3
+
+**Explanation:** The answer is "abc", with the length of 3. 
+
+**Example 2:**
+
+**Input:** s = "bbbbb"
+
+**Output:** 1
+
+**Explanation:** The answer is "b", with the length of 1. 
+
+**Example 3:**
+
+**Input:** s = "pwwkew"
+
+**Output:** 3
+
+**Explanation:** The answer is "wke", with the length of 3. Notice that the answer must be a substring, "pwke" is a subsequence and not a substring. 
+
+**Example 4:**
+
+**Input:** s = ""
+
+**Output:** 0 
+
+**Constraints:**
+
+*   <code>0 <= s.length <= 5 * 10<sup>4</sup></code>
+*   `s` consists of English letters, digits, symbols and spaces.

src/main/cpp/g0001_0100/s0004_median_of_two_sorted_arrays/Solution.cpp

@@ -0,0 +1,43 @@
+// #Hard #Top_100_Liked_Questions #Top_Interview_Questions #Array #Binary_Search #Divide_and_Conquer
+// #Big_O_Time_O(log(min(N,M)))_Space_O(1) #2024_05_22_Time_19_ms_(85.75%)_Space_94.4_MB_(74.87%)
+
+#include <vector>
+#include <algorithm>
+#include <climits>
+
+class Solution {
+public:
+    double findMedianSortedArrays(std::vector<int>& nums1, std::vector<int>& nums2) {
+        if (nums2.size() < nums1.size()) {
+            return findMedianSortedArrays(nums2, nums1);
+        }
+
+        int n1 = nums1.size();
+        int n2 = nums2.size();
+        int low = 0;
+        int high = n1;
+
+        while (low <= high) {
+            int cut1 = (low + high) / 2;
+            int cut2 = (n1 + n2 + 1) / 2 - cut1;
+
+            int l1 = (cut1 == 0) ? INT_MIN : nums1[cut1 - 1];
+            int l2 = (cut2 == 0) ? INT_MIN : nums2[cut2 - 1];
+            int r1 = (cut1 == n1) ? INT_MAX : nums1[cut1];
+            int r2 = (cut2 == n2) ? INT_MAX : nums2[cut2];
+
+            if (l1 <= r2 && l2 <= r1) {
+                if ((n1 + n2) % 2 == 0) {
+                    return (std::max(l1, l2) + std::min(r1, r2)) / 2.0;
+                }
+                return std::max(l1, l2);
+            } else if (l1 > r2) {
+                high = cut1 - 1;
+            } else {
+                low = cut1 + 1;
+            }
+        }
+
+        return 0.0;
+    }
+};

src/main/cpp/g0001_0100/s0004_median_of_two_sorted_arrays/readme.md

@@ -0,0 +1,118 @@
+4\. Median of Two Sorted Arrays
+
+Hard
+
+Given two sorted arrays `nums1` and `nums2` of size `m` and `n` respectively, return **the median** of the two sorted arrays.
+
+The overall run time complexity should be `O(log (m+n))`.
+
+**Example 1:**
+
+**Input:** nums1 = [1,3], nums2 = [2]
+
+**Output:** 2.00000
+
+**Explanation:** merged array = [1,2,3] and median is 2. 
+
+**Example 2:**
+
+**Input:** nums1 = [1,2], nums2 = [3,4]
+
+**Output:** 2.50000
+
+**Explanation:** merged array = [1,2,3,4] and median is (2 + 3) / 2 = 2.5. 
+
+**Example 3:**
+
+**Input:** nums1 = [0,0], nums2 = [0,0]
+
+**Output:** 0.00000 
+
+**Example 4:**
+
+**Input:** nums1 = [], nums2 = [1]
+
+**Output:** 1.00000 
+
+**Example 5:**
+
+**Input:** nums1 = [2], nums2 = []
+
+**Output:** 2.00000 
+
+**Constraints:**
+
+*   `nums1.length == m`
+*   `nums2.length == n`
+*   `0 <= m <= 1000`
+*   `0 <= n <= 1000`
+*   `1 <= m + n <= 2000`
+*   <code>-10<sup>6</sup> <= nums1[i], nums2[i] <= 10<sup>6</sup></code>
+
+To solve the Median of Two Sorted Arrays problem in Java using a `Solution` class, we'll follow these steps:
+
+1. Define a `Solution` class with a method named `findMedianSortedArrays`.
+2. Calculate the total length of the combined array (m + n).
+3. Determine the middle index or indices of the combined array based on its length (for both even and odd lengths).
+4. Implement a binary search algorithm to find the correct position for partitioning the two arrays such that elements to the left are less than or equal to elements on the right.
+5. Calculate the median based on the partitioned arrays.
+6. Handle edge cases where one or both arrays are empty or where the combined length is odd or even.
+7. Return the calculated median.
+
+Here's the implementation:
+
+```java
+public class Solution {
+
+    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
+        int m = nums1.length;
+        int n = nums2.length;
+        int totalLength = m + n;
+        int left = (totalLength + 1) / 2;
+        int right = (totalLength + 2) / 2;
+        return (findKth(nums1, 0, nums2, 0, left) + findKth(nums1, 0, nums2, 0, right)) / 2.0;
+    }
+
+    private int findKth(int[] nums1, int i, int[] nums2, int j, int k) {
+        if (i >= nums1.length) return nums2[j + k - 1];
+        if (j >= nums2.length) return nums1[i + k - 1];
+        if (k == 1) return Math.min(nums1[i], nums2[j]);
+
+        int midVal1 = (i + k / 2 - 1 < nums1.length) ? nums1[i + k / 2 - 1] : Integer.MAX_VALUE;
+        int midVal2 = (j + k / 2 - 1 < nums2.length) ? nums2[j + k / 2 - 1] : Integer.MAX_VALUE;
+
+        if (midVal1 < midVal2) {
+            return findKth(nums1, i + k / 2, nums2, j, k - k / 2);
+        } else {
+            return findKth(nums1, i, nums2, j + k / 2, k - k / 2);
+        }
+    }
+
+    public static void main(String[] args) {
+        Solution solution = new Solution();
+
+        // Test cases
+        int[] nums1_1 = {1, 3};
+        int[] nums2_1 = {2};
+        System.out.println("Example 1 Output: " + solution.findMedianSortedArrays(nums1_1, nums2_1));
+
+        int[] nums1_2 = {1, 2};
+        int[] nums2_2 = {3, 4};
+        System.out.println("Example 2 Output: " + solution.findMedianSortedArrays(nums1_2, nums2_2));
+
+        int[] nums1_3 = {0, 0};
+        int[] nums2_3 = {0, 0};
+        System.out.println("Example 3 Output: " + solution.findMedianSortedArrays(nums1_3, nums2_3));
+
+        int[] nums1_4 = {};
+        int[] nums2_4 = {1};
+        System.out.println("Example 4 Output: " + solution.findMedianSortedArrays(nums1_4, nums2_4));
+
+        int[] nums1_5 = {2};
+        int[] nums2_5 = {};
+        System.out.println("Example 5 Output: " + solution.findMedianSortedArrays(nums1_5, nums2_5));
+    }
+}
+```
+
+This implementation provides a solution to the Median of Two Sorted Arrays problem in Java with a runtime complexity of O(log(min(m, n))).