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LeetCards Data Structures Fundamentals Pack (Linked List, Trees, Graphs, Arrays)
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Two Sum
Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution, and you may not use the same element twice.
Two Sum
Given an array of integers, return indices of the two numbers such that they add up to a spe...
To solve this problem, we can use a brute force approach where we iterate through the array and check for every possible pair of elements that could sum up to the target. This would take O(n^2) time. A better solution would be to use a hash map to store the elements as we iterate through the array. For each element, we can check if the complement (target - element) exists in the map. If it does, we can return the indices. This would take O(n) time.
Maximum Depth of Binary Tree
Given a binary tree, find its maximum depth. The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node. Note: A leaf is a node with no children.
Maximum Depth of Binary Tree
Given a binary tree, find its maximum depth. The maximum depth is the n...
The idea is to use depth-first search. We can start from the root and explore the depth of each subtree. The maximum depth is the maximum of the depths of the subtrees.
Time complexity: O(n) where n is the number of nodes in the tree. Datastructure: Binary Tree
Kth Smallest Element in a BST
Given a binary search tree, write a function kthSmallest to find the kth smallest element in it. Note: You may assume k is always valid, 1 ≤ k ≤ BST's total elements.
Kth Smallest Element in a BST
Given a binary search tree, write a function kthSmallest to find the k...
We can in-order traverse the tree and keep track of the number of nodes seen so far. As soon as the number of nodes seen so far equals k, we know that the current node is the k-th smallest node in the tree. The time complexity of this approach is O(N), where N is the number of nodes in the tree, since we need to visit every node in the tree at least once. We can also make use of the fact that this is a binary search tree. We can do a modified in-order traversal of the tree, where we first recurse on the left subtree, then visit the root, and finally recurse on the right subtree.
Time complexity: O(H + k), where H is the height of the tree and k is the desired element.
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