Longest increasing subsequence in dp

Code/C++/longest_increasing_subsequence_in_dp.cpp

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+#include <iostream>
+#include <bits/stdc++.h>
+using namespace std;
+
+int lis(int* input, int n) {
+	int* dp = new int[n];
+	dp[0] = 1;
+	for (int i = 1; i < n; i++) {
+		dp[i] = 1;
+		for (int j = i - 1; j >= 0; j--) {
+			if (input[j] > input[i]) {
+				continue;
+			}
+			int possibleAns = dp[j] + 1;
+			if (possibleAns > dp[i]) {
+				dp[i] = possibleAns;
+			}
+		}
+	}
+	int best = 0;
+	for (int i = 0; i < n; i++) {
+		if (best < dp[i]) {
+			best = dp[i];
+		}
+	}
+	delete [] dp;
+	return best;
+}
+
+int main() {
+	int n;
+	cin >> n;
+	int * input = new int[n];
+	for (int i = 0; i < n; i++) {
+		cin >> input[i];
+	}
+	int ans = lis(input, n);
+	cout << ans << endl;
+	delete [] input;
+}
+
+
+//    n=5
+//    input=[50,3,10,20,30]
+//    ans=4(3,10,20,30)
+
+//    test case:
+//    sample input:
+//    6
+//    50 3 10 7 40 80
+//    Sample output:
+//    4
+
+//    Explanation:
+//    {3,7,40,80}
+//    dp=[1,1,2,2,3,4]
+//    As first niumber is 50 so increasing number is only 1 so:
+//    dp[0]=1,then after that the number is 3 which is not increasing so 
+//    dp[1]=1,then 10 is greater than 3 so ,
+//    dp[2]=(1+1=2),now then 7 is lesser than 10 so:
+//    dp[3]=dp[2]as (dp[2]>1) ,now next no is 40 which is greater than 7 so:
+//    dp[4]=dp[3]+1=3,so now the last number is 80 which is greater than 40 so:
+//    dp[5]=dp[4]+1=4,so the greatest value is 4 ,
+//    and the answer is 4.
+
+// Time Complexity:O(N^2)
+// Space Complexity:O(N)