added tabulation method in nth_fibo problem and memoization method in…

Algorithms_and_Data_Structures/Design_and_Analysis_of_Algorithms/Dynammic_Programming/lcs.py

@@ -18,6 +18,36 @@ def get_lcs_length(S1, S2):
     # The last cell contains the length of the longest common subsequence
     return dp[m][n]
 
+""" memoization method """
+def lcs_memo(S1:str ,S2:str ,i:int ,j:int ,dp:list[list[int]]):
+    """
+    Compute the length of the longest common subsequence (LCS) between two strings using memoization.
+
+    Parameters:
+    S1 (str): The first string.
+    S2 (str): The second string.
+    i (int): Current index of string S1 (starts from len(S1)-1).
+    j (int): Current index of string S2 (starts from len(S2)-1).
+    dp (list[list[int]]): A memoization table initialized with -1 to store intermediate results.
+
+    Returns:
+    int: Length of the longest common subsequence between S1 and S2
+    """
+    #base case : if either string is ended return 0
+    if i<0 or j<0:   
+        return 0;
+
+    #if result of subproblem is already calculated then no need to calculate again
+    if dp[i][j] != -1: 
+        return dp[i][j]
+
+    if S1[i] == S2[j]:
+        dp[i][j] = 1 + lcs_memo(S1,S2,i-1,j-1,dp)
+    else:   
+        dp[i][j] = max(lcs_memo(S1,S2,i-1,j,dp),lcs_memo(S1,S2,i,j-1,dp))
+
+    return dp[i][j]
+
 if __name__ == "__main__":
     # Take input strings from the user
     S1 = input("Enter the first string: ")

Algorithms_and_Data_Structures/Design_and_Analysis_of_Algorithms/Dynammic_Programming/matrix_multiplication.py

@@ -22,6 +22,37 @@ def minMult(arr):
     # Return the minimum cost to multiply the entire chain of matrices
     return dp[0][n - 1]
 
+""" memoization method """
+def minMult_memo(arr:list ,i:int ,j:int ,dp:list[list[int]]):
+    """
+    Compute the minimum number of scalar multiplications required to multiply 
+    a chain of matrices using memoization (Matrix Chain Multiplication problem).
+
+    Parameters:
+    arr (list): List of integers where the i-th matrix has dimensions arr[i-1] x arr[i].
+    i (int): Starting index of the matrix chain.
+    j (int): Ending index of the matrix chain.
+    dp (list[list[int]]): A memoization table initialized with -1 to store intermediate results.
+
+    Returns:
+    int: Minimum number of scalar multiplications needed to multiply matrices from index i to j.
+    if i==j:
+        return 0
+    """
+    if dp[i][j] != -1:
+        return dp[i][j]
+
+    ans = sys.maxsize
+
+    for k in range(i,j):
+
+        res = minMult_memo(arr ,i ,k ,dp) + minMult_memo(arr ,k+1 ,j ,dp) + arr[i-1]* arr[j] * arr[k]
+
+        ans = min(ans,res)
+
+    dp[i][j] = ans
+    return dp[i][j]
+
 if __name__ == "__main__":
     # Input the matrix dimensions as space-separated integers
     arr = list(map(int, input("Enter the dimensions of matrices separated by spaces: ").split()))

Algorithms_and_Data_Structures/Design_and_Analysis_of_Algorithms/Dynammic_Programming/nth_fibonacci.py

@@ -11,11 +11,29 @@ def nth_fibonacci(n, memo={}):
     memo[n] = nth_fibonacci(n - 1, memo) + nth_fibonacci(n - 2, memo)
     return memo[n]
 
+""" tabulation method for solving nth_fibonnaci """
+def nth_fibonacci_tab(n:int) -> int:
+    """ 
+    Calculate n_th fibonnaci number in O(n)
+    params:
+    n : term value
+
+    returns:
+    int: nth term of fibonnaci sequence
+    """
+    dp = [0] * (n+1)
+    dp[1] = 1
+    for i in range(2,n+1):
+
+        dp[i] = dp[i-1] + dp[i-2]
+
+    return dp[n]
+
 # Get input from the user for the Fibonacci number position
 n = int(input("Enter the position of the Fibonacci number to find: "))
 
 # Calculate the nth Fibonacci number
-result = nth_fibonacci(n)
+result = nth_fibonacci_tab(n)
 
 # Print the result
 print(f"The {n}th Fibonacci number is: {result}")