wagi | Write HTTP handlers in WebAssembly with a minimal amount
kandi X-RAY | wagi Summary
kandi X-RAY | wagi Summary
WARNING: This is experimental code. It is not considered production-grade by its developers, neither is it "supported" software. DeisLabs is experimenting with many WASM technologies right now. This is one of a multitude of projects (including Krustlet) designed to test the limits of WebAssembly as a cloud-based runtime.
Support
Quality
Security
License
Reuse
Top functions reviewed by kandi - BETA
Currently covering the most popular Java, JavaScript and Python libraries. See a Sample of wagi
wagi Key Features
wagi Examples and Code Snippets
Community Discussions
Trending Discussions on wagi
QUESTION
We have list A that after sorting needs to look like list B and we have effort or "weight" of each number so when we are swapping in order effort will swap also ,they are connected.
Knowing how list should look like at the end find what is the lowest effort needed to sort list A to look like lis B
I've found answear to my question but it's in c++ code is at the bottom
...ANSWER
Answered 2021-Feb-24 at 10:53I decided to sit down a bit and try tackling this problem. If I understand correctly, we are performing swaps on list in order to get them sorted like in the destination list.
There are two types of operations we can do. Swapping two numbers to destination, which is swapping two numbers and they both land where they belong. And swapping one number to destination, which makes one of them land where it belongs and puts other in the incorrect location.
Perfect swap should always be prioritized over one-element to destination. After writing it down on paper, I also concluded that to minimize the sum when doing one-element to desination swaps, the more you move the smallest element, the smaller the sum.
So, the algorithm I came up with is: Find smallest weight element in target, find what should be in its place, switch them. Then remove all elements that are on their correct places from both target and origin lists(in order to find new smallest weight if previous one is already in destination), loop until the lists are empty.
Program will use one-to-destination swaps to move the lowest weight as long as it can, and when its done, chooses next smallest element. two-way perfect swaps will solve themselves when program picks one of those elements as the smallest weight.
I am not sure if this algorithm is perfectly correct, especially on the corner cases, but it's the best I could come up with, with the little time I have.
Community Discussions, Code Snippets contain sources that include Stack Exchange Network
Vulnerabilities
No vulnerabilities reported
Install wagi
Get the latest binary release
Unpack it tar -zxf wagi-VERSION-OS.tar.gz
Run the wagi --help command
Support
Reuse Trending Solutions
Find, review, and download reusable Libraries, Code Snippets, Cloud APIs from over 650 million Knowledge Items
Find more librariesStay Updated
Subscribe to our newsletter for trending solutions and developer bootcamps
Share this Page