Longest Increasing subsequence using binary search most optimal approach (Modified) (#2776)

* Longest Increasing subsequence using binary search most optimal approach for this problem

* Longest Increasing subsequence using binary search most optimal approach for this problem

* Longest Increasing subsequence using binary search most optimal approach for this problem (Modified)

* Longest Increasing subsequence using binary search most optimal approach for this problem

* Longest Increasing subsequence using binary search most optimal approach for this problem(done)

* Longest Increasing subsequence using binary search most optimal approach for this problem

* Floyd warshall

* Longest Increasing subsequence using binary search most optimal approach for this problem

* Longest Increasing subsequence using binary search most optimal approach for this problem

* Longest Increasing subsequence using binary search most optimal approach for this problem

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Co-authored-by: realstealthninja <68815218+realstealthninja@users.noreply.github.com>

search/Longest_Increasing_Subsequence_using_binary_search.cpp

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+/**
+ * @file
+ * @brief find the length of the Longest Increasing Subsequence (LIS)
+ * using [Binary Search](https://en.wikipedia.org/wiki/Longest_increasing_subsequence)
+ * @details
+ * Given an integer array nums, return the length of the longest strictly
+ * increasing subsequence.
+ * The longest increasing subsequence is described as a subsequence of an array
+ * where: All elements of the subsequence are in increasing order. This subsequence
+ * itself is of the longest length possible.
+
+ * For solving this problem we have Three Approaches :-
+
+ * Approach 1 :- Using Brute Force
+ * The first approach that came to your mind is the Brute Force approach where we
+ * generate all subsequences and then manually filter the subsequences whose
+ * elements come in increasing order and then return the longest such subsequence.
+ * Time Complexity :- O(2^n)
+ * It's time complexity is exponential. Therefore we will try some other
+ * approaches.
+
+ * Approach 2 :- Using Dynamic Programming
+ * To generate all subsequences we will use recursion and in the recursive logic we
+ * will figure out a way to solve this problem. Recursive Logic to solve this
+ * problem:-
+ * 1. We only consider the element in the subsequence if the element is grater then
+ * the last element present in the subsequence
+ * 2. When we consider the element we will increase the length of subsequence by 1
+ * Time Complexity: O(N*N)
+ * Space Complexity: O(N*N) + O(N)
+
+ * This approach is better then the previous Brute Force approach so, we can
+ * consider this approach.
+
+ * But when the Constraints for the problem is very larger then this approach fails
+
+ * Approach 3 :- Using Binary Search
+ * Other approaches use additional space to create a new subsequence Array.
+ * Instead, this solution uses the existing nums Array to build the subsequence
+ * array. We can do this because the length of the subsequence array will never be
+ * longer than the current index.
+
+ * Time complexity: O(n∗log(n))
+ * Space complexity: O(1)
+
+ * This approach consider Most optimal Approach for solving this problem
+
+ * @author [Naman Jain](https://github.com/namanmodi65)
+ */
+
+#include <cassert>   /// for std::assert
+#include <iostream>  /// for IO operations
+#include <vector>    /// for std::vector
+#include <algorithm> /// for std::lower_bound
+#include <cstdint>   /// for std::uint32_t
+
+/**
+ * @brief Function to find the length of the Longest Increasing Subsequence (LIS)
+ * using Binary Search
+ * @tparam T The type of the elements in the input vector
+ * @param nums The input vector of elements of type T
+ * @return The length of the longest increasing subsequence
+ */
+template <typename T>
+std::uint32_t longest_increasing_subsequence_using_binary_search(std::vector<T>& nums) {
+    if (nums.empty()) return 0;
+
+    std::vector<T> ans;
+    ans.push_back(nums[0]);
+    for (std::size_t i = 1; i < nums.size(); i++) {
+        if (nums[i] > ans.back()) {
+            ans.push_back(nums[i]);
+        } else {
+            auto idx = std::lower_bound(ans.begin(), ans.end(), nums[i]) - ans.begin();
+            ans[idx] = nums[i];
+        }
+    }
+    return static_cast<std::uint32_t>(ans.size());
+}
+
+/**
+ * @brief Test cases for Longest Increasing Subsequence function
+ * @returns void
+ */
+static void tests() {
+    std::vector<int> arr = {10, 9, 2, 5, 3, 7, 101, 18};
+    assert(longest_increasing_subsequence_using_binary_search(arr) == 4);
+
+    std::vector<int> arr2 = {0, 1, 0, 3, 2, 3};
+    assert(longest_increasing_subsequence_using_binary_search(arr2) == 4);
+
+    std::vector<int> arr3 = {7, 7, 7, 7, 7, 7, 7};
+    assert(longest_increasing_subsequence_using_binary_search(arr3) == 1);
+
+    std::vector<int> arr4 = {-10, -1, -5, 0, 5, 1, 2};
+    assert(longest_increasing_subsequence_using_binary_search(arr4) == 5);
+
+    std::vector<double> arr5 = {3.5, 1.2, 2.8, 3.1, 4.0};
+    assert(longest_increasing_subsequence_using_binary_search(arr5) == 4);
+
+    std::vector<char> arr6 = {'a', 'b', 'c', 'a', 'd'};
+    assert(longest_increasing_subsequence_using_binary_search(arr6) == 4);
+
+    std::vector<int> arr7 = {};
+    assert(longest_increasing_subsequence_using_binary_search(arr7) == 0);
+
+    std::cout << "All tests have successfully passed!\n";
+}
+
+/**
+ * @brief Main function to run tests
+ * @returns 0 on exit
+ */
+int main() {
+    tests();  // run self test implementation
+    return 0;
+}