# Swift Hackerrank Dynamic Array

I've solved this question. Wasn't sure if I was suppose to use a dictionary in this case, as the question was to use a dynamic array, but I didnt know how else to do it. Was wondering if I can get feedback on my code or if there's a better way to approach it:

import Foundation

// reads input for N(number of sequences) and Q(number of queries)
let input = readLine()!.componentsSeparatedByString(" ").map { Int($0)! } let numOfSequences = input[0] let numOfQueries = input[1] var lastAns = 0 var seqDictionary = [Int: [Int]]() // loop through numOfQueries for _ in 0..<numOfQueries { // each query is given in format [1, x, y] or [2, x, y] let query = readLine()!.componentsSeparatedByString(" ").map { Int($0)! }
// check to see if query[0] is 1 or 2
switch query[0] {
case 1:
// formula for seqDictionaryIndex
// ^ is the XOR operator
let seqDictionaryIndex = ((query[1] ^ lastAns) % numOfSequences)
// since we initalized an empty dictionary, we gotta set the first key to a value to start the array so we can start appending
guard seqDictionary[seqDictionaryIndex] != nil else { seqDictionary[seqDictionaryIndex] = [query[2]] ; continue }
seqDictionary[seqDictionaryIndex]?.append(query[2])
case 2:
let seqDictionaryIndex = ((query[1] ^ lastAns) % numOfSequences)
// calculate the size of the particular sequence in seqDictionary
let size = seqDictionary[seqDictionaryIndex]?.count
// formula for finding index in particular sequence in seqDictionary
let index = query[2] % size!
// set last answer to that element at the index calculated above
lastAns = seqDictionary[seqDictionaryIndex]![index]
print(lastAns)
default: break
}
}


## 1 Answer

The task is about dynamic arrays, and using an array (instead of a dictionary, as you did) simplifies things considerably.

If you start with an array of numOfSequences empty arrays:

var seqList = [[Int]](count: numOfSequences, repeatedValue: [])


then the 1 x y query simplifies to

let seqIndex = ((x ^ lastAns) % numOfSequences)
seqList[seqIndex].append(y)


without the need to check for an empty dictionary value.

Some more suggestions:

Reading a list of integers occurs at two places in the program, that justifies a separate function:

func readIntegers() -> [Int] {
return readLine()!.componentsSeparatedByString(" ").map { Int($0)! } }  (Remark: Usually, forced unwrapping with ! should be avoided, but here we have a programming challenge with well-defined input data.) To increase the legibility of the code, I would assign the three parameters of a query to variables type, x, y. The code then becomes almost self-explaining. The entire code then looks like this: func readIntegers() -> [Int] { return readLine()!.componentsSeparatedByString(" ").map { Int($0)! }
}

// Read  N(number of sequences) and Q(number of queries):
let input = readIntegers()
let (numOfSequences, numOfQueries) = (input[0], input[1])

var lastAns = 0
// Array of numOfSequences empty arrays:
var seqList = [[Int]](count: numOfSequences, repeatedValue: [])

for _ in 0..<numOfQueries {
let input = readIntegers()
let (type, x, y) = (input[0], input[1], input[2])

switch type {
case 1:
let seqIndex = ((x ^ lastAns) % numOfSequences)
seqList[seqIndex].append(y)
case 2:
let seqIndex = ((x ^ lastAns) % numOfSequences)
let index = y % seqList[seqIndex].count
lastAns = seqList[seqIndex][index]
print(lastAns)
default:
fatalError("Unexpected query")
}
}

• Thanks Martin! Your solution is amazing, well explained, and clear to understand Jul 22, 2016 at 7:02