From 730142fa68109003e160f2a2671342e6deeadebe Mon Sep 17 00:00:00 2001
From: hussein-saad <hussein.saad2021@gmail.com>
Date: Thu, 29 Dec 2022 17:32:04 +0200
Subject: [PATCH] add LIS dp algorithm

---
 .../longest-increasing-subsequence.cpp        | 72 +++++++++++++++++++
 algorithms/CPlusPlus/README.md                |  1 +
 2 files changed, 73 insertions(+)
 create mode 100644 algorithms/CPlusPlus/Dynamic-Programming/longest-increasing-subsequence.cpp

diff --git a/algorithms/CPlusPlus/Dynamic-Programming/longest-increasing-subsequence.cpp b/algorithms/CPlusPlus/Dynamic-Programming/longest-increasing-subsequence.cpp
new file mode 100644
index 00000000..64a1719b
--- /dev/null
+++ b/algorithms/CPlusPlus/Dynamic-Programming/longest-increasing-subsequence.cpp
@@ -0,0 +1,72 @@
+/* ---problem Defintation---
+ *
+ * find the length of the longest subsequence of a given sequence
+ * such that all elements of the subsequence are sorted in increasing order
+ *
+ * ---approche---
+ * this problem has a lot of similarities with the 0/1 knapsack problem
+ * that we can pick an element that make the sequence in increasing order or leave it
+ * So we need to save our prviouse choice to help us predict our future
+ *
+ */
+
+#include <bits/stdc++.h>
+
+using namespace std;
+
+const int MAX = 1e3; // max length of the array given
+
+int dp[MAX][MAX]; // dp arrary to memeraize our choices
+
+int length; // length of the array given
+vector<int> given; // the given array
+
+int lis(int idx,int prv_idx){
+    // base case if the idx out of bounds
+    if (idx >= length)
+        return 0;
+
+    // insted of dealing with (dp[idx][prv_idx]) we take a reverence from it for short
+    auto &ret = dp[idx][prv_idx];
+    // we know in binary representation (-1) all of it's bits is (1)
+    // and this (~) is a bitwise not in c++ whice convert each bit one to zero and vice versa
+    // so if we have calculated this answer we don't need to recalculate it again so we return it
+    if (~ret)
+        return ret;
+
+    int choice1 = lis(idx+1,prv_idx); // leave
+    int choice2 = 0;
+
+    // if it's the first element or it can be in the sequence we take it
+    if (prv_idx == length || given[idx] > given[prv_idx])
+        choice2 = 1 + lis(idx+1,idx); // take
+
+
+    return ret = max(choice1,choice2);
+}
+
+int main() {
+
+    cin >> length;
+    given.resize(length);
+    for (int i = 0; i < length; i++)
+        cin >> given[i];
+    // initialize dp table with -1
+    memset(dp,-1, sizeof(dp));
+    // make prv_idx == length to indcaite this is our first element in the sequence
+    cout << lis(0,length);
+}
+
+/*
+ * ---complexity analysis---
+ *  time:  O(n^2)
+ *  memeory:  O(n^2)
+ *
+ *
+ *  sample
+ *  input:
+ *  5
+ *  2 1 5 3 4
+ *
+ *  output: 3
+*/
\ No newline at end of file
diff --git a/algorithms/CPlusPlus/README.md b/algorithms/CPlusPlus/README.md
index 064a599b..2a01bb99 100644
--- a/algorithms/CPlusPlus/README.md
+++ b/algorithms/CPlusPlus/README.md
@@ -44,6 +44,7 @@
 - [Edit Distance](Dynamic-Programming/edit-distance.cpp)
 - [Fibonacci](Dynamic-Programming/fibonacci.cpp)
 - [Rod Cutting](Dynamic-Programming/rod-cutting.cpp)
+- [Longest Increasing Subsequence](Dynamic-Programming/longest-increasing-subsequence.cpp)
 
 ## Graphs