From 92571f619700a05019853aff0230015fe4b85a2d Mon Sep 17 00:00:00 2001
From: saiumasankar <narsipuramsaiumashankar@gmail.com>
Date: Wed, 5 Jun 2024 11:29:23 +0530
Subject: [PATCH] update dynamic programming

Added LIS problem and Tabulation code of LCS and 0-1 Knapsack
---
 contrib/ds-algorithms/dynamic-programming.md | 124 ++++++++++++++++++-
 1 file changed, 121 insertions(+), 3 deletions(-)

diff --git a/contrib/ds-algorithms/dynamic-programming.md b/contrib/ds-algorithms/dynamic-programming.md
index 43149f8..54c6fcb 100644
--- a/contrib/ds-algorithms/dynamic-programming.md
+++ b/contrib/ds-algorithms/dynamic-programming.md
@@ -84,8 +84,32 @@ Y = "GXTXAYB"
 print("Length of Longest Common Subsequence:", longest_common_subsequence(X, Y, len(X), len(Y)))
 ```
 
+## Longest Common Subsequence Code in Python (Bottom-Up Approach)
+
+```python
+
+def longestCommonSubsequence(X, Y, m, n):
+    L = [[None]*(n+1) for i in range(m+1)]
+    for i in range(m+1):
+        for j in range(n+1):
+            if i == 0 or j == 0:
+                L[i][j] = 0
+            elif X[i-1] == Y[j-1]:
+                L[i][j] = L[i-1][j-1]+1
+            else:
+                L[i][j] = max(L[i-1][j], L[i][j-1])
+    return L[m][n]
+
+
+S1 = "AGGTAB"
+S2 = "GXTXAYB"
+m = len(S1)
+n = len(S2)
+print("Length of LCS is", longestCommonSubsequence(S1, S2, m, n))
+```
+
 ## Complexity Analysis
-- **Time Complexity**: O(m * n) for the top-down approach, where m and n are the lengths of the input sequences
+- **Time Complexity**: O(m * n) for both approaches, where m and n are the lengths of the input sequences
 - **Space Complexity**: O(m * n) for the memoization table
 
 </br>
@@ -122,11 +146,105 @@ capacity = 50
 n = len(weights)
 print("Maximum value that can be obtained:", knapsack(weights, values, capacity, n))
 ```
+## 0-1 Knapsack Problem Code in Python (Bottom-up Approach)
 
+```python
+def knapSack(capacity, weights, values, n):
+    K = [[0 for x in range(capacity + 1)] for x in range(n + 1)]
+    for i in range(n + 1):
+        for w in range(capacity + 1):
+            if i == 0 or w == 0:
+                K[i][w] = 0
+            elif weights[i-1] <= w:
+                K[i][w] = max(values[i-1]
+                              + K[i-1][w-weights[i-1]],
+                              K[i-1][w])
+            else:
+                K[i][w] = K[i-1][w]
+
+    return K[n][capacity]
+
+values = [60, 100, 120]
+weights = [10, 20, 30]
+capacity = 50
+n = len(weights)
+print(knapSack(capacity, weights, values, n))
+```
 ## Complexity Analysis
-- **Time Complexity**: O(n * W) for the top-down approach, where n is the number of items and W is the capacity of the knapsack
+- **Time Complexity**: O(n * W) for both approaches, where n is the number of items and W is the capacity of the knapsack
 - **Space Complexity**: O(n * W) for the memoization table
 
 </br>
 <hr>
-</br>
\ No newline at end of file
+</br>
+
+# 4. Longest Increasing Subsequence
+
+The Longest Increasing Subsequence (LIS) is a task is to find the longest subsequence that is strictly increasing, meaning each element in the subsequence is greater than the one before it. This subsequence must maintain the order of elements as they appear in the original sequence but does not need to be contiguous. The goal is to identify the subsequence with the maximum possible length.
+
+**Algorithm Overview:**
+- **Base cases:** If the sequence is empty, the LIS length is 0.
+- **Memoization:** Store the results of previously computed subproblems to avoid redundant computations.
+- **Recurrence relation:** Compute the LIS length by comparing characters of the sequences and making decisions based on their values.
+
+## Longest Increasing Subsequence Code in Python (Top-Down Approach using Memoization)
+
+```python
+
+import sys
+
+def f(idx, prev_idx, n, a, dp):
+	if (idx == n):
+		return 0
+
+	if (dp[idx][prev_idx + 1] != -1):
+		return dp[idx][prev_idx + 1]
+
+	notTake = 0 + f(idx + 1, prev_idx, n, a, dp)
+	take = -sys.maxsize - 1
+	if (prev_idx == -1 or a[idx] > a[prev_idx]):
+		take = 1 + f(idx + 1, idx, n, a, dp)
+
+	dp[idx][prev_idx + 1] = max(take, notTake)
+	return dp[idx][prev_idx + 1]
+
+def longestSubsequence(n, a):
+
+	dp = [[-1 for i in range(n + 1)]for j in range(n + 1)]
+	return f(0, -1, n, a, dp)
+
+a = [3, 10, 2, 1, 20]
+n = len(a)
+
+print("Length of lis is", longestSubsequence(n, a))
+
+```
+
+## Longest Increasing Subsequence Code in Python (Bottom-Up Approach)
+
+```python
+def lis(arr):
+	n = len(arr)
+	lis = [1]*n
+
+	for i in range(1, n):
+		for j in range(0, i):
+			if arr[i] > arr[j] and lis[i] < lis[j] + 1:
+				lis[i] = lis[j]+1
+
+	maximum = 0
+	for i in range(n):
+		maximum = max(maximum, lis[i])
+
+	return maximum
+
+arr = [10, 22, 9, 33, 21, 50, 41, 60]
+print("Length of lis is", lis(arr))
+```
+## Complexity Analysis
+- **Time Complexity**: O(n * n) for both approaches, where n is the length of the array.
+- **Space Complexity**: O(n * n) for the memoization table in Top-Down Approach, O(n) in Bottom-Up Approach.
+
+</br>
+<hr>
+</br>