# 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
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

int data;

typedef struct {
int size;
}SEQUENCE;

{

} else {
while(temp->next != NULL)
temp = temp->next;

newnode->next = NULL;
temp->next = newnode;
}

}

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;
if (node == NULL)
return 1;
node->data = y;
node->next = NULL;

s[index].size ++;
break;

case 2:
index = (x^lastans)%N;
y = y % (s[index].size);

while ( y > 0) {
temp = temp->next;
y--;
}

lastans = temp->data;
outnode ->data = temp->data;
output = insert_tail(output, outnode);
break;

}

i++;

}while (i < Q);

if (output != NULL)

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:
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)```
```

## Inserting a Node Into a Sorted Doubly Linked List

Given a reference to the head of a doubly-linked list and an integer ,data , create a new DoublyLinkedListNode object having data value data and insert it at the proper location to maintain the sort. Example head refers to the list 1 <-> 2 <-> 4 - > NULL. data = 3 Return a reference to the new list: 1 <-> 2 <-> 4 - > NULL , Function Description Complete the sortedInsert function

## Reverse a doubly linked list

This challenge is part of a tutorial track by MyCodeSchool Given the pointer to the head node of a doubly linked list, reverse the order of the nodes in place. That is, change the next and prev pointers of the nodes so that the direction of the list is reversed. Return a reference to the head node of the reversed list. Note: The head node might be NULL to indicate that the list is empty.

## Tree: Preorder Traversal

Complete the preorder function in the editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's preorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the preOrder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the tree's

## Tree: Postorder Traversal

Complete the postorder function in the editor below. It received 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's postorder traversal as a single line of space-separated values. Input Format Our test code passes the root node of a binary tree to the postorder function. Constraints 1 <= Nodes in the tree <= 500 Output Format Print the

## Tree: Inorder Traversal

In this challenge, you are required to implement inorder traversal of a tree. Complete the inorder function in your editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's inorder traversal as a single line of space-separated values. Input Format Our hidden tester code passes the root node of a binary tree to your \$inOrder* func

## Tree: Height of a Binary Tree

The height of a binary tree is the number of edges between the tree's root and its furthest leaf. For example, the following binary tree is of height : image Function Description Complete the getHeight or height function in the editor. It must return the height of a binary tree as an integer. getHeight or height has the following parameter(s): root: a reference to the root of a binary