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

Author: namanmodi65

search/Longest_Increasing_Subsequence_using_binary_search.cpp

@@ -1,47 +1,66 @@
 
-// Given an integer array nums, return the length of the longest strictly 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.
+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 :-
+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 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)
+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.
+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
+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.
+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)
+Time complexity: O(n∗log(n))
+Space complexity: O(1)
 
-// This approach consider Most optimal Approach for solving this problem
+This approach consider Most optimal Approach for solving this problem
 
-// Implementaion:-
+ *******************************************************************************/
 
-#include <iostream>
-#include <vector>
-using namespace std;
+#include <cassert>  /// for std::assert
+#include <iostream> /// for IO operations
+#include <vector>   /// for std::vector
 
-int Longest_Increasing_Subsequence_using_binary_search(vector<int>& nums){
+/**
+ * @brief Function to find the length of Longest Increasing Subsequence (LIS)
+ * using Binary Search
+ * @param nums Input integer array
+ * @return Length of the longest increasing subsequence
+ */
+int Longest_Increasing_Subsequence_using_binary_search(std::vector<int>& nums){
+    
     if(nums.size() == 0) return 0;
 
-    vector<int> ans;
+    std::vector<int> ans;
     ans.push_back(nums[0]);
     for(int i=1;i<nums.size();i++){
         if(nums[i] > ans.back()){
@@ -55,8 +74,28 @@ int Longest_Increasing_Subsequence_using_binary_search(vector<int>& nums){
     return ans.size();
 }
 
-int main(){
-    vector<int> arr = {10,9,2,5,3,7,101,18};
-    cout<<"longest increasing subsequence : "<<Longest_Increasing_Subsequence_using_binary_search(arr);
-   return 0;
+/**
+ * @brief test implementations
+ * @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::cout << "All tests have successfully passed!\n";
+}
+
+/**
+ * @brief Main function
+ * @returns 0 on exit
+ */
+int main() {
+    tests();
+    return 0;
 }