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const levenshteinDistance = (s, t) => {
if (!s.length) return t.length;
if (!t.length) return s.length;
const arr = [];
for (let i = 0; i <= t.length; i++) {
arr[i] = [i];
for (let j = 1; j <= s.length; j++) {
arr[i][j]
def levenshtein_distance(first_word: str, second_word: str) -> int:
"""Implementation of the levenshtein distance in Python.
:param first_word: the first word to measure the difference.
:param second_word: the second word to measure th
function minkowskiDistance (x, lx, y, ly, p) {
var d;
var i;
if (lx !== ly) {
throw 'Both vectors should have same dimension';
}
if (isNaN(p)) {
throw 'The order "p" must be a number';
}
if
function levenshteinDistance (s, ls, t, lt) {
var memo = [];
var currRowMemo;
var i;
var k;
for (k = 0; k <= lt; k += 1) {
memo[k] = k;
}
for (i = 1; i <= ls; i += 1) {
currRowMemo = [
Community Discussions
Trending Discussions on coding-test
QUESTION
I tried to split an input
...ANSWER
Answered 2019-Jan-01 at 13:03You could try trimming your value before you split the string:
QUESTION
In this problem , I have been given an array , and a binary matrix ( matrix consisting of 0 and 1 only ) , i and j value can be thought of index of array elment and if matrix[i][j]==1 then we can swap a[i] and a[j] , now what I have to do is to get the minimum possible permutation using all those one in matrix in any order, this is initial array suppose n = 5 size array is there
4 2 1 5 3
now this is given matrix, which is nXn
00100
00011
10010
01101
01010
using all those one,we can get minimum possible permutation like this (using 1-based indexing to explain)
4 2 1 5 3 initial
we do (p1<->p3)
we get,1 2 4 5 3
now we do (p4<->p5)
we get, 1 2 4 3 5
and now we do (p3<->p4)
we get, 1 2 3 4 5
this is minimum possible we can get using one's
I can think of brute force but that would of course give TIME LIMIT EXCEEDED, so I am wondering how to approach this problem in a better way.
for more detail, here is the link to the problem, https://www.hackerrank.com/contests/pre-placement-coding-test/challenges/smallest-permutation/problem .
...ANSWER
Answered 2018-Dec-27 at 17:13If you would interpret "possible swaps" matrix as graph then you could figure out, that in every connected component you could rearrange numbers in any order you want.
So the solution is to find all components and sort numbers independently in every one.
QUESTION
I am trying to parse JSON using URLSession and without using Alamofire or anything else.
I just want to take the JSON and put it into a UITableView.
I am trying to piece together what I learned from learning how to Parse JSON using Alamofire with what I can find on google. Many of the answers on youtube or Stack etc use NS for everything..NSURL, NSDictionary, etc etc..Or are just typing code without explaining what/why.
I THINK I am almost there, but I need help understanding what I have left to do.
SO.
I Allowed arbitrary loads in the plst
In a Swift File I have the following
...ANSWER
Answered 2017-Apr-17 at 10:48The method downloadJSON()
should be implemented in the ViewController
since it is returning the array of Potter
data. Then in the URLSession
response you should create one array which will be act as the tableview datasource. (i.e self.arrTableData = try JSONSerialization.jsonObject(with: content, options: JSONSerialization.ReadingOptions.mutableContainers) as? [[String : AnyObject]]
)
Then in for the tableView
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