Add a Solution for Longest Increasing Subsequence (LeetCode Problem 300)

dsa-solutions/lc-solutions/0300-0399/0300-longest-increasing-subsequence.md

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+---
+id: longest-increasing-subsequence
+title: Longest Increasing Subsequence (LeetCode)
+sidebar_label: 0300-LongestIncreasingSubsequence
+---
+
+## Problem Description
+
+| Problem Statement | Solution Link | LeetCode Profile |
+| :---------------- | :------------ | :--------------- |
+| [Merge Two Sorted Lists](https://leetcode.com/problems/longest-increasing-subsequence/) | [Merge Two Sorted Lists Solution on LeetCode](https://leetcode.com/problems/longest-increasing-subsequence/solutions/) |  [VijayShankerSharma](https://leetcode.com/u/darkknight648/) |
+
+## Problem Description
+
+Given an integer array nums, return the length of the longest strictly increasing subsequence.
+
+### Examples
+
+#### Example 1:
+
+- **Input:** nums = [10,9,2,5,3,7,101,18]
+- **Output:** 4
+- **Explanation:** The longest increasing subsequence is [2,3,7,101], therefore the length is 4.
+
+#### Example 2:
+
+- **Input:** nums = [0,1,0,3,2,3]
+- **Output:** 4
+
+#### Example 3:
+
+- **Input:** nums = [7,7,7,7,7,7,7]
+- **Output:** 1
+
+### Constraints:
+
+- $1 <= nums.length <= 2500$
+- $-10^4 <= nums[i] <= 10^4$
+
+## Approach
+
+We can solve this problem using dynamic programming or binary search. Here, we will discuss the binary search approach which achieves $O(n log(n))$ time complexity.
+
+1. Initialize an empty list tails to keep track of the smallest ending element for all increasing subsequences.
+2. Iterate through each element num in the input nums.
+3. If num is greater than the last element in tails, append num to tails.
+4. Otherwise, perform a binary search on tails to find the smallest element greater than or equal to num and update that element with num.
+5. After iterating through all elements in nums, return the length of tails.
+
+## Solution Code
+
+#### Python
+
+```py
+class Solution:
+    def lengthOfLIS(self, nums):
+        tails = []
+        for num in nums:
+            left, right = 0, len(tails) - 1
+            while left <= right:
+                mid = (left + right) // 2
+                if tails[mid] < num:
+                    left = mid + 1
+                else:
+                    right = mid - 1
+            if left < len(tails):
+                tails[left] = num
+            else:
+                tails.append(num)
+        return len(tails)
+```
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    int lengthOfLIS(vector<int>& nums) {
+        vector<int> tails;
+        for (int num : nums) {
+            auto it = lower_bound(tails.begin(), tails.end(), num);
+            if (it == tails.end())
+                tails.push_back(num);
+            else
+                *it = num;
+        }
+        return tails.size();
+    }
+};
+```
+
+#### Java
+
+```java
+class Solution {
+    public int lengthOfLIS(int[] nums) {
+        List<Integer> tails = new ArrayList<>();
+        for (int num : nums) {
+            int left = 0, right = tails.size() - 1;
+            while (left <= right) {
+                int mid = left + (right - left) / 2;
+                if (tails.get(mid) < num)
+                    left = mid + 1;
+                else
+                    right = mid - 1;
+            }
+            if (left < tails.size())
+                tails.set(left, num);
+            else
+                tails.add(num);
+        }
+        return tails.size();
+    }
+}
+```
+
+## Conclusion
+
+The "Longest Increasing Subsequence" problem can be efficiently solved using the binary search approach, achieving $O(n log(n))$ time complexity. The provided solution code implements this approach in Python, C++, and Java, providing an optimal solution to the problem.