diff --git a/README.md b/README.md
index 2aeaa5d77..70f1285aa 100644
--- a/README.md
+++ b/README.md
@@ -26,6 +26,7 @@ Compiled pages are published at [https://cp-algorithms.com/](https://cp-algorith
 
 ### New articles
 
+- (28 January 2024) [Longest Common Subsequence](https://cp-algorithms.com/sequences/longest-common-subsequence.html)
 - (28 January 2024) [Introduction to Dynamic Programming](https://cp-algorithms.com/dynamic_programming/intro-to-dp.html)
 - (8 December 2023) [Hungarian Algorithm](https://cp-algorithms.com/graph/hungarian-algorithm.html)
 - (10 September 2023) [Tortoise and Hare Algorithm](https://cp-algorithms.com/others/tortoise_and_hare.html)
diff --git a/src/navigation.md b/src/navigation.md
index de682c536..fd38f44e0 100644
--- a/src/navigation.md
+++ b/src/navigation.md
@@ -202,6 +202,7 @@ search:
     - Sequences
         - [RMQ task (Range Minimum Query - the smallest element in an interval)](sequences/rmq.md)
         - [Longest increasing subsequence](sequences/longest_increasing_subsequence.md)
+        - [Longest common subsequence](sequences/longest-common-subsequence.md)
         - [Search the subsegment with the maximum/minimum sum](others/maximum_average_segment.md)
         - [K-th order statistic in O(N)](sequences/k-th.md)
         - [MEX task (Minimal Excluded element in an array)](sequences/mex.md)
diff --git a/src/sequences/longest-common-subsequence.md b/src/sequences/longest-common-subsequence.md
new file mode 100644
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@@ -0,0 +1,208 @@
+---
+tags:
+    - Original
+---
+
+# Longest Common Subsequence
+
+Given two strings with of length $a$ and $b$ respectively, where $0 \leq a, b \leq n$. The task is to find the _length_ of **longest subsequence** which exists in both the strings.
+
+**Subsequence** of a string is considered collection of characters from that string where some chars are deleted and relative ordering of characters are not changed.
+
+## Solution using Recursion
+
+Intution is, for the position $x$ and $y$ representing current index respectively for $text1$ and $text2$, we have two options:
+
+- If $text1[x]$ matches $text2[y]$, we'll increase both x and y and will increment the _lcs_ length by 1.
+
+- If it doesn't match, we'll take max result of one time incrementing current index for $text1$ and recursively calling the function, and one time incrementing current index for $text2$ and recursively calling the function.
+
+### Implementation
+
+```cpp title="lcs_recursion" linenums="1"
+int longest_common_subsequence_recursion(string text1, string text2) {
+	int a = text1.size(), b = text2.size();
+
+	function<int(int x, int y)> get_lcs = [&](int x, int y) -> int {
+		if (x >= a || y >= b) return 0;
+
+		int length = 0;
+		if (text1[x] == text1[y]) {
+			length = 1 + get_lcs(x + 1, y + 1);
+		} else length = max(get_lcs(x + 1, y), get_lcs(x, y + 1));
+
+		return length;
+	};
+
+	return lcs(0, 0);
+}
+```
+
+## Solution using Caching
+
+Solution using caching, or memoized solution is when additional space is used to store the result of the algorithm/program at a particular state, so that it can be used repeatedly without the need of doing calculation for that particular state again.
+
+The recursive solution from above, the state is depended upon two numbers, index $x$ and $y$. Thus, we need 2-dimensional space to store the states.
+
+For each, $x$ and $y$, we'll check if solution already exists or not. If it does, we'll use that solution. Also, we'll keep storing the solutions for new $x$ and $y$. Implementation goes as follow:
+
+### Implementation
+
+```cpp title="lcs_caching" linenums="1"
+int longest_common_subsequence_caching(string text1, string text2) {
+	int a = text1.size(), b = text2.size();
+	vector<vector<int>> space(a, vector<int>(b, -1));
+
+	function<int(int x, int y)> lcs = [&](int x, int y) -> int {
+		if (x >= a || y >= b) return 0;
+
+		if (space[x][y] != -1) return space[x][y];
+
+		int length = 0;
+		if (text1[x] == text1[y]) {
+			length = 1 + lcs(x + 1, y + 1);
+		} else length = max(lcs(x + 1, y), lcs(x, y + 1));
+
+		return space[x][y] = length;
+	};
+
+	return lcs(0, 0);
+}
+```
+
+## Solution using Dynamic Programming
+
+The problem can also be solved using **Dynamic Programming**. Well, the caching solution from above is also considered as DP but is not pure DP.
+
+This approach, which is also known as **Tabulated DP** is better than caching solution in some senses as follow:
+
+* It is iterative and thus when you have large set of input (ie., length of strings provided is quite large), this approach would work faster than caching solution from above, which is based on recursive solution.
+
+* This approach is also useful when you need the solution for given input on every state depending on $x$ and $y$, instead of just some subsets of state.
+
+Based on how you write the _transition_ part of the solution, this DP approach can be of following two types:
+
+- Pull DP
+- Push DP
+
+When using tabulated solution, we deal with base cases before starting to iterate for all the states ie., just before when loop starts. For this particular problem, base case is already set to zero, as we are initializing (resizing the empty space) the space with 0. The base case is when either when $text1$ is empty and thus
+
+```cpp
+for (int i = 0; i < text2.size(); i++) {
+	space[0][i] = 0;
+}
+```
+
+Or when $text2$ is empty and thus
+
+```cpp
+for (int i = 0; i < text1.size(); i++) {
+	space[i][0] = 0;
+}
+```
+
+or both $text1$ and $text2$ both are empty, ie., $space[0][0] = 0$
+
+Lets see the implementation in both ways of tabulated solution.
+
+## Pull DP
+
+Pull DP is when, the solution for current state relies on the previous calculated state/s.
+
+
+```cpp title="lcs_pull_dp" linenums="1"
+int longest_common_subsequence_pulldp(string text1, string text2) {
+	int a = text1.size(), b = text2.size();
+	vector<vector<int>> space(a + 1, vector<int>(b + 1, 0));
+
+	for (int i = 1; i <= a; i++) {
+		for (int j = 1; j <= b; j++) {
+			if (text1[i - 1] == text2[j - 1]) space[i][j] = space[i - 1][j - 1] + 1;
+			else space[i][j] = max(space[i - 1][j], space[i][j - 1]);
+		}
+	}
+
+	return space[a][b];
+}
+```
+
+Here, lets focus on one part of the transition from above solution
+
+$$space[i][j] = space[i - 1][j - 1] + 1;$$
+
+We can see that the current state represented by $i$ and $j$ as current index for both the strings as $space[i][j]$, is dependent upon the previous already calculated state as $space[i - 1][j - 1]$
+
+## Push DP
+
+Push DP is when the solution for current state contributes to the future state/s of the solution.
+
+```cpp title="lcs_push_dp" linenums="1"
+int longest_common_subsequence_pushdp(string text1, string text2) {
+	int a = text1.size(), b = text2.size();
+	vector<vector<int>> space(a + 1, vector<int>(b + 1, 0));
+
+	for (ll i = 0; i < a; i++) {
+		for (ll j = 0; j < b; j++) {
+			if (text1[i] == text2[j]) space[i + 1][j + 1] = space[i][j] + 1;
+			else space[i + 1][j + 1] = max(space[i + 1][j], space[i][j + 1]);
+		}
+	}
+
+	return space[a][b];
+}
+```
+
+Here, lets focus on one part of the transition from above solution
+
+$$space[i + 1][j + 1] = space[i][j] + 1;$$
+
+We can see that the current state represented by $i$ and $j$ as current index for both the strings as $space[i][j]$, is contribu.ting for the solution of upcoming state for $space[i + 1][j + 1]$
+
+Push DP and Pull DP are not that much different from each other, you can decide for yourself which idea to solve tabulated DP clicks you.
+
+## Retrieve Longest Common Subsequence from DP solution
+
+Once we have filled the $space$ (the 2-dimensional vector used in above tabulated solutions), we can retrieve the _longest common subsequence_ string itself from that 2-dimensional space itself.
+
+Intuition is simple. Let's initialize a empty string $res$, which will store the longest common subsequence. We know that the very end cell of the 2D space is the result containing the length of longest common subsequence. We will start from there.
+
+For $i$ and $j$, we'll start with very end cell ie., $i = text1.size()$ and $j = text2.size()$. We'll move as follow:
+
+- If $text1[i - 1] == $text2[j - 1]$ matches then it means that the character is common in both the strings, and we'll add that character to $res$.
+- If it doesn't match, then we'll move to the next adjacent cell (either above or left side) whichever contains max value, which means that we're following the path of common subsequence which is longest and in which direction we move, we decrement either $i$ or $j$.
+
+Once the $res$ is filled properly, that longest common substring itself is in reverse, so to get answer properly we'll have to reverse the string $res$ itself. C++ stl provides `reverse()` method, but I'm providing this small algorithm here:
+
+```cpp title="reverse string" linenums="1"
+int sz = res.size();
+for (int i = 0; i < sz / 2; i++) {
+	char temp = res[i];
+	res[i] = res[sz - i - 1];
+	res[sz - i - 1] = temp;
+}
+```
+
+Here's the implementation, which will make the idea clear:
+
+### Implementation
+
+```cpp title="retrieve lcs string" linenums="1"
+string retreive_longest_common_subsequence(string text1, string text2, vector<vector<int>>& space) {
+	string res{};
+
+	int i = text1.size(), j = text2.size();
+	while (i > 0 && j > 0) {
+		if (text1[i - 1] == text2[j - 1]) {
+			res += text1[i - 1];
+			i--, j--;
+		} else {
+			if (space[i - 1][j] > space[i][j - 1]) i--;
+			else j--;
+		}
+	}
+
+	reverse(res.begin(), res.end());
+
+	return res;
+}
+```