voracious | chrome extension that displays definitions | Browser Plugin library

 by   sanchitnevgi JavaScript Version: Current License: MIT

kandi X-RAY | voracious Summary

kandi X-RAY | voracious Summary

voracious is a JavaScript library typically used in Plugin, Browser Plugin applications. voracious has no bugs, it has no vulnerabilities, it has a Permissive License and it has low support. You can download it from GitHub.

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            kandi-support Support

              voracious has a low active ecosystem.
              It has 21 star(s) with 3 fork(s). There are 4 watchers for this library.
              OutlinedDot
              It had no major release in the last 6 months.
              voracious has no issues reported. There are no pull requests.
              It has a neutral sentiment in the developer community.
              The latest version of voracious is current.

            kandi-Quality Quality

              voracious has no bugs reported.

            kandi-Security Security

              voracious has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.

            kandi-License License

              voracious is licensed under the MIT License. This license is Permissive.
              Permissive licenses have the least restrictions, and you can use them in most projects.

            kandi-Reuse Reuse

              voracious releases are not available. You will need to build from source code and install.

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            voracious Key Features

            No Key Features are available at this moment for voracious.

            voracious Examples and Code Snippets

            No Code Snippets are available at this moment for voracious.

            Community Discussions

            QUESTION

            Why doesn't an image show up in WhatsApp link sharing?
            Asked 2020-Apr-14 at 10:01

            I'm trying to enrich a web site with Open Graph data, to let the page author decide which image and texts show up in a preview on social media. However, I can't get WhatsApp to show a small image in the sharing card.

            My current markup looks like this:

            ...

            ANSWER

            Answered 2019-Sep-14 at 17:24

            Your og implementation only works on some versions of whats app. And that is because the url (of the entire page, not just the image) contains non ascii characters.

            I have copy/paste your html code on one of my web-sites and it worked ok, as long as the file was named index.html or ogtest.html.

            After I renamed the file with Greek characters, og-protocol stopped working.

            Source https://stackoverflow.com/questions/57833714

            QUESTION

            Why does my C++ solution to the "Fish" challange by codility fail?
            Asked 2019-Sep-18 at 18:51

            I'm attempting Codility fish challange which is described as follows:

            You are given two non-empty arrays A and B consisting of N integers. Arrays A and B represent N voracious fish in a river, ordered downstream along the flow of the river.

            The fish are numbered from 0 to N − 1. If P and Q are two fish and P < Q, then fish P is initially upstream of fish Q. Initially, each fish has a unique position.

            Fish number P is represented by A[P] and B[P]. Array A contains the sizes of the fish. All its elements are unique. Array B contains the directions of the fish. It contains only 0s and/or 1s, where:

            0 represents a fish flowing upstream, 1 represents a fish flowing downstream. If two fish move in opposite directions and there are no other (living) fish between them, they will eventually meet each other. Then only one fish can stay alive − the larger fish eats the smaller one. More precisely, we say that two fish P and Q meet each other when P < Q, B[P] = 1 and B[Q] = 0, and there are no living fish between them. After they meet:

            If A[P] > A[Q] then P eats Q, and P will still be flowing downstream, If A[Q] > A[P] then Q eats P, and Q will still be flowing upstream. We assume that all the fish are flowing at the same speed. That is, fish moving in the same direction never meet. The goal is to calculate the number of fish that will stay alive.

            For example, consider arrays A and B such that:

            A[0] = 4 B[0] = 0 A1 = 3 B1 = 1 A2 = 2 B2 = 0
            A[3] = 1 B[3] = 0 A[4] = 5 B[4] = 0 Initially all the fish are alive and all except fish number 1 are moving upstream. Fish number 1 meets fish number 2 and eats it, then it meets fish number 3 and eats it too. Finally, it meets fish number 4 and is eaten by it. The remaining two fish, number 0 and 4, never meet and therefore stay alive.

            Write a function:

            int solution(vector &A, vector &B);

            that, given two non-empty arrays A and B consisting of N integers, returns the number of fish that will stay alive.

            For example, given the arrays shown above, the function should return 2, as explained above.

            Write an efficient algorithm for the following assumptions:

            N is an integer within the range [1..100,000]; each element of array A is an integer within the range [0..1,000,000,000]; each element of array B is an integer that can have one of the following values: 0, 1; the elements of A are all distinct.

            My solution is as follows:

            ...

            ANSWER

            Answered 2019-Sep-18 at 18:32

            I'm going to post some algorithm thoughts and ideas for you. Hopefully this will help you visual the problem.

            Some initial observations - all fish swimming upstream at the beginning of the array can never eat or be eaten. Same for fish at the bottom of the array swimming downstream. The action occur when scanning from front to back when a downstream meets an up.

            First, walk your array front to back until you find a down swimming fish.

            Next continue walking your down/up array until it swaps from down to up.

            Resolve combat and increment a combat counter.

            If down wins, keep walking down. If up wins, save the index of your location in the array(s), then walk it up the array until it is eaten or you hit a fish swimming the same direction.

            If it finds a fish swimming its direction, it will live, jump back to where it started scan downwards again.

            If it is eaten, take this fish and start comparing where you left off in the down array.

            Repeat until you reach the end of the list. Return the (starting number of fish - number of combats)

            The neat part is this is in-place. No moving of memory, no need to mark dead fish, etc. It will run linear time - at most nearly 2x linear.

            This the the pseudo code I would use to resolve this.

            Source https://stackoverflow.com/questions/57998346

            QUESTION

            Time and Space complexity for calculating many fish are alive?
            Asked 2019-Aug-03 at 16:16

            I was working on a Codility problem:

            You are given two non-empty zero-indexed arrays A and B consisting of N integers. Arrays A and B represent N voracious fish in a river, ordered downstream along the flow of the river.

            The fish are numbered from 0 to N − 1. If P and Q are two fish and P < Q, then fish P is initially upstream of fish Q. Initially, each fish has a unique position.

            Fish number P is represented by A[P] and B[P]. Array A contains the sizes of the fish. All its elements are unique. Array B contains the directions of the fish. It contains only 0s and/or 1s, where:

            0 represents a fish flowing upstream, 1 represents a fish flowing downstream. If two fish move in opposite directions and there are no other (living) fish between them, they will eventually meet each other. Then only one fish can stay alive − the larger fish eats the smaller one. More precisely, we say that two fish P and Q meet each other when P < Q, B[P] = 1 and B[Q] = 0, and there are no living fish between them. After they meet:

            If A[P] > A[Q] then P eats Q, and P will still be flowing downstream, If A[Q] > A[P] then Q eats P, and Q will still be flowing upstream. We assume that all the fish are flowing at the same speed. That is, fish moving in the same direction never meet. The goal is to calculate the number of fish that will stay alive.

            ...

            ANSWER

            Answered 2018-Mar-14 at 23:57

            It's hard to follow this code with the strange data structures and lack of variable names, but I think I have the needed understanding ...

            Space complexity: The only dimensioned space you have is myQ, which is bounded by the total quantity of fish. Thus, this is O(n).

            Time Complexity: Given your strange logic, this was harder to follow. The paired decrements of remFish and the abuse of while -- break confused me for a couple of minutes. However, the simpler analysis is ...

            The while -- break turns that loop into a simple if statement, since you always break the loop on first iteration. Therefore, your only true iteration is the for loop, bounded by the quantity of fish. Thus, this is also O(n).

            Among other properties, note that you decrement numFish on each iteration, and it never drops as far as 0.

            Why do you gauge one iteration on a.length and another on b.length? Those must be the same, the starting quantity of fish.

            Source https://stackoverflow.com/questions/49288218

            Community Discussions, Code Snippets contain sources that include Stack Exchange Network

            Vulnerabilities

            No vulnerabilities reported

            Install voracious

            You can download it from GitHub.

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            CLONE
          • HTTPS

            https://github.com/sanchitnevgi/voracious.git

          • CLI

            gh repo clone sanchitnevgi/voracious

          • sshUrl

            git@github.com:sanchitnevgi/voracious.git

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