Leetcode 2534

Study Plan.

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Leetcode 2534

There is a long and thin painting that can be represented by a number line. Painting the same area multiple times will create an uneven painting so you only want to paint each area of the painting at most once. Create a boolean array to store whether each block has been painted. We can check each of these blocks, painting the unpainted ones we also keep count of how many blocks we paint because that's what the question asks for. In the worst case, we have to check every block on every day. Let n n n be the number of days up to and let m m m be the largest number that appears in the input up to This is not fast enough. Instead of using a boolean array, we can use a BBST balanced binary search tree to store the indices of the unpainted blocks. When we paint a node, we delete its node from the BBST. In our time complexity analysis, it will become clear why we chose to use a BBST. Otherwise, in future days, we'd have to keep checking whether each block has been painted. A BBST can do what we need: find and delete single items quickly. Instead of storing the unpainted blocks, we can store the painted segments. We store them as left, right pairs in a BBST, where no segments intersect. This solution is trickier to implement—code will not be presented here.

Reverse Words in a String II

There are 5 states involved. State can be transformed by folds, etc, elegantly although perhaps not with great performance. Have you tried solving the problem you referenced? Thanks to Algebraic Data Types and pattern-matching, implementing algorithms on trees and lists is a breeze e. However, some algorithms do require state and mutability for efficiency dynamic programming, doubly-linked lists, trees with links to parents and iterators over trees, many algorithms in computational geometry, etc.

In this problem, there are n persons who want to enter or exit through a door. Each person can enter or exit through the door once, taking one second. We are given a non-decreasing integer array arrival of size n , where arrival[i] is the arrival time of the i th person at the door. Also, there is an array state of size n , where state[i] is 0 if person i wants to enter through the door or 1 if they want to exit through the door. The main goal is to find an array answer of size n where answer[i] is the second at which the i th person crosses the door, considering various rules regarding entering and exiting the door. We can use two pointers approach for solving this problem. First, we can create two lists, one for the entering persons and one for the exiting persons, and initialize pointers at the beginning of those lists.

Leetcode 2534

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Crassula bitkisi faydaları

Array Nesting Logger Rate Limiter Linked List Cycle II Closest Leaf in a Binary Tree Binary Watch Spiral Matrix III Subtree of Another Tree Open the Lock Number of Atoms Majority Element II Equal Tree Partition Android Unlock Patterns Unique Substrings in Wraparound String Escape a Large Maze

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N-ary Tree Preorder Traversal Replace Words String Compression Poor Pigs Palindrome Partitioning II Paint House II Short Encoding of Words Maximum Sum Circular Subarray Advanced Lowest Common Ancestor. Number of Lines To Write String Design Circular Deque Target Sum Super Pow