Dynamic Array
Problem Statement :
Create a list, seqList, of n empty sequences, where each sequence is indexed from 0 to n-1. The elements within each of the n sequences also use 0-indexing.
Create an integer, lastAnswer, and initialize it to 0.
There are 2 types of queries that can be performed on the list of sequences:
1. Query: 1 x y
a. Find the sequence, seq, at index ((x xor lastAnswer)%n) in seqList.
b. Append integer y to sequence seq.
2. Query: 2 x y
a. Find the sequence, seq, at index ((x xor lastAnswer)%n) in seqList.
b. Find the value of element y%size in seq (where size is the size of seq) and assign it to
lastAnswer.
c. Print the new value of lastAnswer on a new line
Note: xor is the bitwise XOR operation, which corresponds to the ^ operator in most languages. Learn more about it on Wikipedia. is the modulo operator.
Function Description:
Complete the dynamicArray function below.
dynamicArray has the following parameters:
- int n: the number of empty sequences to initialize in seqList.
- string queries[q]: an array of query strings
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 number of sequences) and q (the number of queries), respectively.
Each of the subsequent lines contains a query in the format defined above, queries[i].
Constraints:
1. 1<=n,q<=10^5
2. 0<=x<=10^9
3. 0<=y<=10^9
4. It is guaranteed that query type will never query an empty sequence or index.
Solution :
Solution in C :
In C:
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#define MAX_N_Q 100000
#define MAX_X_Y 1000000000
typedef struct linklist {
int data;
struct linklist *next;
}linknode, *linklistp;
typedef struct {
linklistp head;
int size;
}SEQUENCE;
linklistp insert_tail(linklistp head,linklistp newnode)
{
if (head == NULL) {
head = newnode;
} else {
linklistp temp = head;
while(temp->next != NULL)
temp = temp->next;
newnode->next = NULL;
temp->next = newnode;
}
return head;
}
void outputlinklist(linklistp head) {
linklistp temp = head;
while(temp) {
printf("%d",temp->data);
temp = temp->next;
if (temp!=NULL)
printf("\n");
}
}
int main()
{
int N, Q;
int i = 0;
int choice, x, y, index;
int lastans = 0;
SEQUENCE *s = NULL;
linklistp node, temp, output = NULL, outnode;
scanf("%d", &N);
if (N < 1 || N > MAX_N_Q) {
return 1;
}
scanf("%d", &Q);
if (Q < 1 || Q > MAX_N_Q) {
return 1;
}
s = malloc(sizeof(SEQUENCE)*N);
if(s == NULL) {
return 1;
}
memset(s, 0, sizeof(SEQUENCE)*N);
do {
scanf("%d", &choice);
scanf("%d", &x);
scanf("%d", &y);
if (choice != 1 && choice != 2) {
return 1;
}
if (x < 0 || x > MAX_X_Y) {
return 1;
}
if (y < 0 || y > MAX_X_Y) {
return 1;
}
switch(choice){
case 1:
index = (x^lastans)%N;
node = malloc(sizeof(linknode));
if (node == NULL)
return 1;
node->data = y;
node->next = NULL;
s[index].size ++;
s[index].head = insert_tail(s[index].head, node);
break;
case 2:
index = (x^lastans)%N;
y = y % (s[index].size);
temp = s[index].head;
while ( y > 0) {
temp = temp->next;
y--;
}
lastans = temp->data;
outnode = malloc(sizeof(linknode));
outnode ->data = temp->data;
output = insert_tail(output, outnode);
break;
}
i++;
}while (i < Q);
if (output != NULL)
outputlinklist(output);
return 0;
}
In C++:
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int main()
{
int n, q;
int lastans = 0;
cin >> n >> q;
vector<vector<int>>sqces(n);
while (q--)
{
int a;
long long x, y;
cin >> a >> x >> y;
long long t = (x^lastans) % n;
if (a == 1)
{
sqces[t].push_back(y);
}
else
{
long long size = sqces[t].size();
long long b;
if (size != 0)
b = y%size;
else
continue;
cout << sqces[t][b] << endl;
lastans =sqces[t][b];
}
}
return 0;
}
In Java:
import java.io.*;
import java.util.*;
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int q = in.nextInt();
List<Integer>[] sequences = new List[n];
for (int n_i = 0; n_i < n; n_i++) {
sequences[n_i] = new ArrayList<>();
}
int lastans = 0;
for (int i = 0; i < q; i++) {
int t = in.nextInt();
int x = in.nextInt();
int y = in.nextInt();
List<Integer> sequence = sequences[(x^lastans)%n];
switch (t) {
case 1:
sequence.add(y);
break;
case 2:
lastans = sequence.get(y%sequence.size());
System.out.println(lastans);
break;
}
}
}
}
In Python 3:
n, q = map(int, input().split())
l = [[] for _ in range(n)]
latsans = 0
for _ in range(q):
a, x, y = map(int, input().split())
if a == 1:
l[(x^latsans)%n].append(y)
else:
t = (x^latsans)%n
latsans = l[t][y%len(l[t])]
print(latsans)
#print(a, x, y, l)
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