Range Update Maximum Query

Range Update Maximum Query

Range Update Maximum Query

Problem Description

You are given an integer array nums and a list of queries queries, where each query queries[i] is of the form [left, right, M]. Each query represents an update operation that modifies elements in nums between indices left and right (inclusive). The update sets nums[i] = max(nums[i], M) for left <= i <= right. Your task is to return the array nums after applying all the update queries.

Problem Analysis

The given problem involves multiple range updates and then point queries. It is a combination of several programming concepts, including small-to-large merging, offline queries, and sweep line. This problem is not commonly found and requires a novel approach to solve.

The time complexity of the solution is not immediately obvious and requires a proof to understand. Therefore, this problem can be classified as hard, as the alternative solutions also involve implementing not-so-common and difficult-to-implement data structures.

Solution Approach

The problem can be solved without resorting to complex data structures such as lazy segment trees. Although lazy segment trees are commonly used to solve range update problems, our problem can be solved using a simpler technique.

If the problem involved adding a value X to a range, it could be easily implemented using a difference array. However, applying the same technique to handle range maximum updates is not obvious but possible.

The solution approach involves maintaining two arrays: adds and subs. These arrays are of length N + 1 and store ordered sets (or priority queues with lazy deletion). For each query [l, r, M], we add M to adds[l] and subs[r + 1].

To find the active updates that apply to each index while traversing the array from left to right, we merge all the sets in adds up to the current index and remove all the elements present in subs. This can be achieved efficiently using a technique called small-to-large merging.

Small-to-large merging states that when merging two sets, it is optimal to push all elements of the smaller set to the larger set. The optimality of this approach is proven here.

By using small-to-large merging, we can solve the problem in O(N log^2 N) in the worst case.

Playground Link

You can access the playground link that includes a random test case generator and a checker function to validate the optimal solution.

Example Test Cases

Test Case 1:

Input:
nums = [7, 2, 8, 5, 9]
queries = [[1, 3, 41], [2, 4, 39], [0, 0, 10], [3, 4, 7]]

Output:
[10, 41, 41, 41, 39]

Test Case 2:

Input:
nums = [5, 4, 8, 7, 1]
queries = [[1, 2, 20], [0, 4, 16], [4, 4, 48], [2, 4, 14], [0, 1, 10]]

Output:
[16, 20, 20, 16, 48]

Tags

Heap (Priority Queue), Ordered Set, Segment Tree

Comments

Post a Comment

Popular posts from this blog

Exploring a Number Conversion Algorithm

Exploring a Number Conversion Algorithm Exploring a Number Conversion Algorithm Category: algorithms Description In this blog post, we will explore an interesting number conversion algorithm that utilizes the Counter data structure from the collections module. We will walk through the provided code snippet and understand its functionality. Code Snippet from collections import Counter def convert(n): if n <= 0: return Counter() x = 1 while pow(x + 1, x + 1) <= n: x += 1 return convert(n - pow(x, x)) + Counter({x: 1}) print(convert(38748358348)) Explanation The code begins by importing the Counter class from the collections module. The Counter class is a powerful tool for counting elements in an iterable. It allows us to easily keep track of the frequency of elements in a collection. The convert function takes an integer n as input and returns a Counter object. Let's dive into the implementation of the convert function: def conv...
Read more

Toeplitz Matrix

Image
Given an m x n matrix , return  true  if the matrix is Toeplitz. Otherwise, return false . A matrix is Toeplitz if every diagonal from top-left to bottom-right has the same elements.   Example 1: Input: matrix = [[1,2,3,4],[5,1,2,3],[9,5,1,2]] Output: true Explanation: In the above grid, the diagonals are: "[9]", "[5, 5]", "[1, 1, 1]", "[2, 2, 2]", "[3, 3]", "[4]". In each diagonal all elements are the same, so the answer is True. Example 2: Input: matrix = [[1,2],[2,2]] Output: false Explanation: The diagonal "[1, 2]" has different elements.   Constraints: m == matrix.length n == matrix[i].length 1 <= m, n <= 20 0 <= matrix[i][j] <= 99   Follow up: What if the matrix is stored on disk, and the memory is limited such that you can only load at most one row of the matrix into the memory at once? What if the matrix is so large ...
Read more