Name: 17. Linear Algebra | Cayley Hamilton Theorem
Uploaded: 2021-12-06T00:00:00+00:00
Duration: T11M12S
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Cayley Hamilton Theorem

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17. Linear Algebra | Cayley Hamilton Theorem

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17. Linear Algebra  Cayley Hamilton Theorem

In this video, we will discuss three methods to check if a singly linked list is a palindrome. The first approach uses a stack, where we push each node’s data onto the stack and then compare the nodes with the popped elements in the second traversal. This method runs in O(n) time and uses O(n) space due to the stack. The second approach employs recursion to compare nodes from the left and right, moving start forward and recursively checking the rest. This also runs in O(n) time but uses O(n) space for the recursion stack.The third approach is the most efficient, utilizing an iterative method with two pointers: slow and fast. The fast pointer moves twice as fast as the slow pointer, allowing the slow pointer to reach the middle of the list. We then reverse the second half of the list and compare it with the first half. This method runs in O(n) time and uses O(1) space, making it optimal. Watch the video to learn the details of each approach and see how they work to check if a linked list is a palindrome. For more details, please go through - <a href="https://www.geeksforgeeks.org/function-to-check-if-a-singly-linked-list-is-palindrome/">Palindrome Linked List</a>

Palindrome Linked Lists

Insert In BST (Python)

For more details, please go through - <a href="https://www.geeksforgeeks.org/dsa/deletion-in-binary-search-tree/">Deletion in Binary Search Tree (BST)</a>

Deletion in BST.

AAAIn this video, we explore the process of serializing and deserializing a binary tree, which allows us to store and later restore a tree structure. Serialization is the process of converting the tree into an array, while deserialization reconstructs the tree from that array. We cover two approaches for serialization: using preorder traversal and using level order traversal. Both methods ensure that the tree structure is preserved, with the latter being more suitable for complete trees. We also explain the time and space complexities of each approach.The first approach uses preorder traversal to serialize the tree, where each node’s value is recorded, and a marker (-1) is used for null nodes. The second approach employs level order traversal, which processes nodes level by level, making it suitable for complete binary trees. The deserialization process for both methods involves reconstructing the tree from the array, creating nodes where necessary, and linking them back together. Finally, we discuss potential optimizations for more efficient storage, such as using a single bit for leaf nodes to minimize markers. For more details, please go through - <a href="https://www.geeksforgeeks.org/serialize-deserialize-binary-tree/">Serialize and Deserialize a Binary Tree</a>

Serialize and Deserialize Binary Tree

In this video, we explore how to find the Lowest Common Ancestor (LCA) in a Binary Search Tree (BST). The LCA of two nodes is the lowest node that has both nodes as descendants, with the added condition that the node itself can be a descendant of itself. We present two approaches: the recursive method and the iterative method. In the recursive approach, we check if both nodes are smaller or greater than the current node, recursively traversing left or right subtrees until the LCA is found. If one node is smaller and the other is larger, the current node is the LCA.The iterative method improves upon the recursive approach by eliminating the need for recursion, optimizing space complexity to O(1). We traverse the BST iteratively, checking the values of n1 and n2 at each node. If both n1 and n2 are smaller than the current node, we move left; if both are greater, we move right. Once we find a node where one value is smaller and the other is larger, that node is the LCA. This method maintains the efficiency of the recursive approach while using constant space. For more details, lease go through -<a href="https://www.geeksforgeeks.org/lowest-common-ancestor-in-a-binary-search-tree/">LCA in BST – Lowest Common Ancestor in Binary Search Tree</a>

LCA in BST – Lowest Common Ancestor in Binary Search Tree

In this video, we explore how to check if a given binary tree is a Binary Search Tree (BST) or not. A Binary Search Tree (BST) is a binary tree where the left subtree's nodes are smaller than the root, and the right subtree's nodes are greater, with the added condition that both subtrees must also be BSTs. We discuss three different approaches to solving this problem: using a specified range of minimum and maximum values, using inorder traversal, and using Morris Traversal. Each method ensures that the binary tree satisfies the properties of a BST, allowing you to verify if the tree is a BST or not.The first approach uses a recursive helper function, isBSTUtil(), which checks whether a subtree is a BST within a given range. The second approach leverages inorder traversal, checking if the values of the nodes are sorted in ascending order. Finally, the third approach uses Morris Traversal, which eliminates the need for extra space while performing the check. This method temporarily alters the tree structure to create threads, helping maintain the BST properties while traversing the tree efficiently. For more details, please go through - <a href="https://www.geeksforgeeks.org/a-program-to-check-if-a-binary-tree-is-bst-or-not/">Check if a Binary Tree is BST or not</a>

Check if a Binary Tree is BST or not

Morris Traversal is an efficient method for performing an inorder traversal of a binary tree without using recursion or a stack. The key idea behind Morris Traversal is to use threaded binary trees, which create temporary links (or "threads") to the inorder successors of nodes, allowing traversal without additional space. The process involves traversing the tree and making changes to the tree's structure temporarily, which are reverted after the traversal is complete.The algorithm works by setting the current node as the root and iterating through the tree. If a node does not have a left child, the node's value is printed and the traversal moves to the right child. If a node has a left child, the inorder predecessor is found (the rightmost node in the left subtree). If the right child of the predecessor points back to the current node, the changes made to the tree structure are undone, and the current node’s data is printed before moving to the right child. This process ensures that no extra space is used, and the tree is restored to its original form at the end. The time complexity of this approach is O(n), where n is the number of nodes in the tree, and the space complexity is O(1), as no extra data structures are used. For more details, please go through - <a href="https://www.geeksforgeeks.org/inorder-tree-traversal-without-recursion-and-without-stack/">Morris traversal for Inorder</a>

Morris traversal for Inorder

The diameter of a binary tree is the length of the longest path between any two nodes in the tree, measured in terms of the number of edges. There are two main approaches to calculate the diameter of a binary tree: the naive approach using top-down recursion and the more efficient bottom-up approach.In the naive approach, the diameter is calculated by recursively determining the height of the left and right subtrees for each node. For each node, the sum of the heights of its left and right subtrees is calculated, and the maximum of these sums is compared to the current maximum diameter. This process results in O(n^2) time complexity as the height is computed separately for each node. On the other hand, the bottom-up recursive approach optimizes this by calculating the height and diameter in the same recursive call. For each node, it computes the height of its subtrees and updates the diameter by checking if the sum of the left and right subtree heights exceeds the current maximum. This approach improves the time complexity to O(n), where n is the number of nodes in the tree, with auxiliary space complexity of O(h), where h is the height of the tree. For more details, please go through - <a href="http://geeksforgeeks.org/diameter-of-a-binary-tree/">Diameter of a Binary Tree</a>

Diameter of a Binary Tree.

Level Order Traversal, also known as Breadth First Search (BFS), is a technique to traverse all nodes in a binary tree level by level. This approach ensures that nodes at the same level are visited before those at the next level. The traversal can be implemented in two main ways: using recursion with a stack or iteratively with a queue.In the recursive approach, a helper function is used to traverse the tree, passing the current node and its level. Nodes at the same level are grouped together in a result array, where the index corresponds to the level of the node. On the other hand, the iterative approach uses a queue to process the nodes level by level. Starting from the root, the left and right children of each node are enqueued for processing. This method maintains a First-In-First-Out (FIFO) order, ensuring that all nodes at the current level are processed before moving to the next level. Both approaches have a time and space complexity of O(n), where n is the number of nodes in the tree. For more details, please go through - <a href="https://www.geeksforgeeks.org/level-order-tree-traversal/">Level Order Traversal (Breadth First Search or BFS) of Binary Tree</a>

Level Order Traversal.

In this video, we explore multiple approaches to flattening a linked list of linked lists, where each sub-linked list is sorted. The first method is a simple yet inefficient approach, where we collect all the node values into an array, sort the array, and then create a new linked list by iterating through the sorted array. This method has a time complexity of O(n * m * log(n * m)) and uses O(n * m) space, making it less efficient for large lists.The second approach uses recursion to merge two sorted linked lists at a time, similar to how we merge lists in merge sort. This results in a time complexity of O(n * n * m), where n is the number of main linked list nodes and m is the size of a single sub-linked list. The third approach optimizes the process by using a Min Heap (priority queue). By always extracting the minimum element from the heap, it ensures that nodes are flattened in sorted order efficiently. This method has a time complexity of O(n * m * log(n)) and uses O(n) space, making it the most efficient. Watch the video to dive deeper into these methods and see them in action! For more details, please go through - <a href="https://www.geeksforgeeks.org/flattening-a-linked-list/">Flattening a Linked List</a>

Flattening a Linked List.

In this video, we will discuss several approaches to merge K sorted linked lists into a single sorted list. The first method is a simple approach where we merge the lists one by one. Starting with an empty result list, we merge each linked list into the result using the standard merge algorithm for two sorted lists. This approach has a time complexity of O(n * k * k), where n is the length of each list and k is the number of lists, and uses O(1) space.The second approach improves this by selecting the smallest node from all the current head nodes and adding it to the result list. By doing this iteratively, we eventually merge all the lists. This method also has a time complexity of O(n * k * k) and uses O(1) space. The third approach, using a min heap, is more efficient for unevenly sized lists. It reduces the time complexity to O(n * k * log k) by efficiently finding the minimum element across all lists. Lastly, the divide and conquer approach splits the lists into two halves recursively, merging them pair by pair, which is similar to merge sort. This method works best when the lists are of equal size and has a time complexity of O(n * k * log k). Watch the video to learn the details of these approaches and how to merge K sorted linked lists efficiently. For more details, please go through - <a href="https://www.geeksforgeeks.org/merge-k-sorted-linked-lists/">Merge K sorted linked lists</a>

Merge K sorted linked list

In this video, we will discuss three approaches to merge two sorted linked lists into one sorted linked list. The first approach uses an array. We start by extracting all the elements from both linked lists and storing them in an array. After sorting the array, we create a new linked list by sequentially inserting the sorted elements. This approach has a time complexity of O((n+m) * log(n+m)) due to the sorting operation, where n and m are the lengths of the two linked lists, and uses O(n+m) space for storing the elements in the array.The second approach uses recursion to merge the two lists. In the base cases, if one of the lists is empty, we return the other list. For the recursive step, we compare the current nodes of both lists, attach the smaller node to the merged list, and recursively merge the remaining nodes. This approach efficiently merges the lists with a time complexity of O(n+m) and space complexity of O(n+m) due to the recursion stack.The third and most efficient approach uses iteration to merge the lists. It eliminates the need for extra space used by recursion or arrays. By iteratively comparing the nodes of both lists, we can construct the merged list in a single pass. This approach also runs in O(n+m) time, but uses O(1) space as it does not require extra storage. Watch the video to understand the step-by-step process for each approach and how they efficiently merge two sorted linked lists. For more details, please g through - <a href="https://www.geeksforgeeks.org/merge-two-sorted-linked-lists/">Merge two sorted linked lists</a>