HackerRank Dynamic Array Problem Solution

In this post, we will solve the Dynamic Array Problem in HackerRank. Declare a 2-dimensional array, arr, with n empty arrays, all zero-indexed.

Declare an integer, last Answer, and initialize it to 0.
You need to process two types of queries:

Query: 1 x y

Compute idx = (xlast Answer).
Append the integer y to arr[idx].

Query: 2 x y

Compute idx = (xlast Answer).
Set last Answer = arr[idx][y%size(arr[idx])].
Store the new value of the last Answer in an answers array.

Function Description
Complete the dynamicArray function with the following parameters:

int n: the number of empty arrays to initialize in arr
int queries [q][3]: 2-D array of integers

Returns
int: the results of each type 2 query in the order they are presented

Input Format
The first line contains two space-separated integers, n, the size of arr to create, and q. the number of queries, respectively. Each of the q subsequent lines contains a query string. queries[I].

Constraints

1≤n, q≤105
0 < x, y ≤ 109
It is guaranteed that query type 2 will never query an empty array or index.

Sample Input

STDIN    Function
-----    --------
2 5      size of arr[] n = 2, size of queries[] q = 5
1 0 5    queries = [[1,0,5],[1,1,7],[1,0,3],[2,1,0],[2,1,1]]
1 1 7
1 0 3
2 1 0
2 1 1

Sample Output

7
3

HackerRank Dynamic Array Solution in C

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

struct sequence {
    int *data;
    int len;
    int cap;
};
struct sequence *seqList;

void push_list(int item, struct sequence *n) {
   if (n->len >= n->cap) {
       int new_size = n->cap == 0 ? 1 : 2*n->cap;
       n->data=(int*)realloc(n->data,new_size);
       n->cap=new_size;
   }
   n->data[n->len++]=item;
}

void print_list(struct sequence *n) {
   int i;
    
    for (i=0; i<n->len; i++)
        printf("%d ",n->data[i]);
    putchar('\n');
}

int main() {
    int N, Q, i;
    int query, x, y;
    int lastAns =0;
    struct sequence *chosen;
        
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */   
    scanf("%d %d", &N, &Q);
    seqList = (struct sequence *) malloc(N * sizeof(struct sequence));
    for (i=0; i<N; i++) {
        seqList[i].data = NULL;
        seqList[i].len  = 0;
        seqList[i].cap  = 0;
    }
    
    for (i=0; i<Q; i++) {
        scanf("%d %d %d", &query, &x, &y);
        chosen = &seqList[(x^lastAns)%N];
        if (query==1) {
            push_list(y, chosen);
        } else if (query == 2) {
            lastAns = chosen->data[y%chosen->len];
            printf("%d\n",lastAns);
        }
    }
    
    return 0;
}

HackerRank Dynamic Array Solution in C++

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;


int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */
    int N = 0;
    int Q = 0;
    cin >> N >> Q;
    
    vector<int>* sequences = new vector<int>[N];
    int lastans = 0;

    for (int i=0; i<Q; i++) {
        int command_type = 0;
        int x = 0;
        int y = 0;
        cin >> command_type >> x >> y;
        
        if (command_type == 1) {
            sequences[(x^lastans) % N].push_back(y);
        }
        else if (command_type == 2) {
            lastans = sequences[(x^lastans) % N][y % sequences[(x^lastans) % N].size()];
            cout << lastans << "\n";
        }
    }
    return 0;
}

HackerRank Dynamic Array Solution in Java

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int N = in.nextInt();
        int Q = in.nextInt();
        
        ArrayList<ArrayList<Integer>> arrays = new ArrayList<ArrayList<Integer>>(N);
        for (int i = 0; i < N; ++i) {
            arrays.add(new ArrayList<Integer>());
        }
        
        int lastans = 0;
        
        for (int i = 0; i < Q; ++i) {
            int operation = in.nextInt();
            int x = in.nextInt();
            int y = in.nextInt();
            int idx = (x ^ lastans) % N;
            ArrayList<Integer> array = arrays.get(idx);
            
            if (operation == 1) {
                // "1 x y" - Insert y at the end of the ((x⊕lastans) mod N)th sequence.
                array.add(y);
            } else {
                // "2 x y" - Print the value of the (y mod size)th element of the ((x⊕lastans) mod N)th sequence. Here, size denotes the size of the related sequence. Then, assign this integer to lastans.
                lastans = array.get(y % array.size());
                System.out.println(lastans);
            }
        }
    }
}

HackerRank Dynamic Array Solution in JavaScript

function processData(input) {
    
    let queries = input.split("\n");
    let [n, q] = queries.shift().split(" ").map(Number);
    
    let lastAnswer = 0;
    let arrOfArr = [];
    for(let i = 0; i < q; i++) {
        let [t, x, y] = queries[i].split(" ").map(Number);
        let index = ((x ^ lastAnswer) % n);
        
        if(t === 1) {
            if(typeof arrOfArr[index] === "undefined") {
                arrOfArr[index] = [];
            }
            arrOfArr[index].push(y);
        }
        else if(t === 2) {
            lastAnswer = arrOfArr[index][y % arrOfArr[index].length];
            console.log(lastAnswer);
        }
    }
    
} 

process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
    _input += input;
});

process.stdin.on("end", function () {
   processData(_input);
});

HackerRank Dynamic Array Solution in Python

# Enter your code here. Read input from STDIN. Print output to STDOUT
seqList = []
lastAns = 0

N,Q = map(int,raw_input().split())

for i in range(N):
    seqList.append([])

for i in range(Q):
    query = map(int,raw_input().split())
    cmd   = query[0]
    if cmd == 1:
        seqList[(query[1]^lastAns) % N].append(query[2])    
    elif cmd == 2:
        seq = seqList[ (query[1]^lastAns) % N]
        lastAns = seq[query[2] % len(seq)]  
        print lastAns

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