Longest Increasing Subsequence file added

algorithms/CPlusPlus/Dynamic-Programming/longest-increasing-subsequence.cpp

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+/*Longest Increasing Subsequence
+
+ Description:
+ Given an integer array nums, return the length of the longest strictly increasing subsequence.
+ A subsequence is a sequence that can be derived from an array by deleting some or no elements
+ without changing the order of the remaining elements.
+
+ For example, [3,6,2,7] is a subsequence of the array [0,3,1,6,2,2,7].
+
+
+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.
+
+*/
+
+#include <bits/stdc++.h> //importing all necessary header files
+using namespace std;
+
+int lengthOfLIS(vector<int> &nums) // function returning the length of LIS.
+{
+    int n = nums.size();
+    vector<int> dp(n, 1);
+
+    for (int i = 0; i < n; ++i) // iterating i for all the index from 0 to the n-1.
+    {
+        for (int j = 0; j < i; ++j) // iterating j to all values less than i
+        {
+            if (nums[i] > nums[j]) // if the value at j is less than value at i
+            {
+                dp[i] = max(dp[i], 1 + dp[j]); // then store the max of value of dp[i] and 1+dp[j]
+            }
+        }
+    }
+    return *max_element(dp.begin(), dp.end()); // using max_element stl function to return max value
+                                               //  in the dp vector and finallly returning it.
+}
+
+int main()
+{
+    int n;
+    cout << "Enter the total number of elements : " << endl;
+    cin >> n; // Taking the input for size of array
+    cout << endl;
+
+    vector<int> nums(n); // Created a vector of size n
+
+    cout << "Enter the elements : " << endl;
+    for (int i = 0; i < n; i++)
+        cin >> nums[i]; // Taking the input for vector;
+    
+    cout<<endl;
+    cout << "The Length of Longest Increasing Subsequence in the given array is :" << endl;
+    cout << lengthOfLIS(nums) << endl; // Calling the function lengthOfLIS and passing nums as arguement.
+
+    return 0;
+}
+
+
+
+//Time Complexity : O(n^2)