Array and simple queries


Problem Statement :


Given two numbers N  and M. N  indicates the number of elements in the array  A[](1-indexed) and  M indicates number of queries. You need to perform two types of queries on the array A[] .

You are given  queries. Queries can be of two types, type 1 and type 2.

Type 1 queries are represented as 1 i j : Modify the given array by removing elements from i to j and adding them to the front.

Type 2 queries are represented as 2 i j : Modify the given array by removing elements from i to j and adding them to the back.

Your task is to simply print | A[1] -  A[N] | of the resulting array after the execution of  queries followed by the resulting array.

Note While adding at back or front the order of elements is preserved.

Input Format


First line consists of two space-separated integers, N and M .
Second line contains N integers,  which represent the elements of the array.
M queries follow. Each line contains a query of either type 1 or type 2 in the form type i j


Output Format

Print the absolute value i.e. abs(A[1] - A[N] )  in the first line.
Print elements of the resulting array in the second line. Each element should be seperated by a single space.


Solution :



title-img 


                            Solution in C :

In C ++ :




#include <iostream>
#include <cstdlib>
#include <vector>
#include <ctime>

struct ctree {
	long value, size;
	int prio;
	ctree *left, *right;
};

static const long MAX_N = 100000l;
static ctree nodes[MAX_N];
static long values[MAX_N];

static long csize(ctree *t)
{ return t ? t->size : 0; }

static ctree *cmerge(ctree *l, ctree *r)
{
	if (!l) return r;
	if (!r) return l;

	if (l->prio > r->prio) {
		l->right = cmerge(l->right, r);
		l->size = csize(l->left) + csize(l->right) + 1;
		return l;
	}
	r->left = cmerge(l, r->left);
	r->size = csize(r->left) + csize(r->right) + 1;
	return r;
}

static void csplit(ctree *t, long idx, ctree *&l, ctree *&r)
{
	if (!t) {
		l = r = 0;
		return;
	}

	const long cur = csize(t->left) + 1;
	if (cur <= idx) {
		csplit(t->right, idx - cur, t->right, r);
		l = t;
	} else {
		csplit(t->left, idx, l, t->left);
		r = t;
	}
	t->size = csize(t->left) + csize(t->right) + 1;
}

static ctree *cextract(ctree *&t, long from, long to)
{
	ctree *l, *m, *r;

	csplit(t, from, l, m);
	csplit(m, to - from, m, r);
	t = cmerge(l, r);
	return m;
}

static long traverse(ctree *tree, long i)
{
	if (!tree)
		return i;

	i = traverse(tree->left, i);
	values[i++] = tree->value;
	return traverse(tree->right, i);
}

int main()
{
	srand(time(NULL));
	for (long i = 0; i != MAX_N; ++i) {
		nodes[i].prio = rand();
		nodes[i].size = 1;
	}

	long n, m;
	std::cin >> n >> m;

	ctree *tree = 0;
	for (long i = 0; i != n; ++i) {
		std::cin >> nodes[i].value;
		tree = cmerge(tree, nodes + i);
	}

	for (long i = 0; i != m; ++i) {
		long l, r;
		int type;

		std::cin >> type >> l >> r;

		ctree *sub = cextract(tree, l - 1, r);
		if (type == 1)
			tree = cmerge(sub, tree);
		else
			tree = cmerge(tree, sub);
	}

	traverse(tree, 0);
	std::cout << std::abs(values[0] - values[n - 1]) << "\n";
	for (long i = 0; i != n; ++i)
		std::cout << values[i] << " ";
	std::cout << "\n";

	return 0;
}








In C :




#include <stdio.h>
#include <stdlib.h>
typedef struct _ct_node{
  int size;
  int priority;
  int value;
  struct _ct_node *left,*right;
} ct_node;
void tar(ct_node *root);
int get_size(ct_node *root);
ct_node* merge(ct_node *L,ct_node *R);
int sizeOf(ct_node *root);
void recalc(ct_node *root);
void split(int x,ct_node **L,ct_node **R,ct_node *root);
void computeTree();
int P[100000],T[100000],st[100000],N;
ct_node pool[100000];

int main(){
  int M,x,y,z,i;
  ct_node *root,*l,*m,*r,*t;
  scanf("%d%d",&N,&M);
  for(i=0;i<N;i++){
    scanf("%d",&pool[i].value);
    P[i]=pool[i].priority=rand();
    pool[i].left=pool[i].right=NULL;
  }
  computeTree();
  for(i=0;i<N;i++)
    if(T[i]==-1)
      root=&pool[i];
    else
      if(i<T[i])
        pool[T[i]].left=&pool[i];
      else
        pool[T[i]].right=&pool[i];
  get_size(root);
  for(i=0;i<M;i++){
    scanf("%d%d%d",&x,&y,&z);
    switch(x){
      case 1:
        split(y-2,&l,&t,root);
        split(z-y,&m,&r,t);
        root=merge(merge(m,l),r);
        break;
      default:
        split(y-2,&l,&t,root);
        split(z-y,&m,&r,t);
        root=merge(merge(l,r),m);
    }
  }
  N=0;
  tar(root);
  printf("%d\n",(T[0]>T[N-1])?T[0]-T[N-1]:T[N-1]-T[0]);
  for(i=0;i<N;i++)
    printf("%d ",T[i]);
  return 0;
}
void tar(ct_node *root){
  if(!root)
    return;
  tar(root->left);
  T[N++]=root->value;
  tar(root->right);
  return;
}
int get_size(ct_node *root){
  if(!root)
    return 0;
  root->size=get_size(root->left)+get_size(root->right)+1;
  return root->size;
}
ct_node* merge(ct_node *L,ct_node *R){
  if(!L)
    return R;
  if(!R)
    return L;
  if(L->priority>R->priority){
    L->right=merge(L->right,R);
    recalc(L);
    return L;
  }
  R->left=merge(L,R->left);
  recalc(R);
  return R;
}
int sizeOf(ct_node *root){
  return (root)?root->size:0;
}
void recalc(ct_node *root){
  root->size=sizeOf(root->left)+sizeOf(root->right)+1;
  return;
}
void split(int x,ct_node **L,ct_node **R,ct_node *root){
  if(!root){
    *L=*R=NULL;
    return;
  }
  int curIndex=sizeOf(root->left);
  ct_node *t;
  if(curIndex<=x){
    split(x-curIndex-1,&t,R,root->right);
    root->right=t;
    recalc(root);
    *L=root;
  }
  else{
    split(x,L,&t,root->left);
    root->left=t;
    recalc(root);
    *R=root;
  }
  return;
}
void computeTree(){
  int i,k,top=-1;
  for(i=0;i<N;i++){
    k=top;
    while(k>=0 && P[st[k]]<P[i])
      k--;
    if(k!=-1)
      T[i]=st[k];
    if(k<top)
      T[st[k+1]]=i;
    st[++k]=i;
    top=k;
  }
  T[st[0]]=-1;
  return;
}







In Java :




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

public class Solution {
    
    Node root;
    
    public Solution() {
        root = null;
    }
    
    public int size(Node root) {
        if (root == null)
            return 0;
        return root.size;
    }
    
    private void update(Node root) {
        if (root != null)
            root.size = 1 + size(root.left) + size(root.right);
    }
    
    public DNode split(Node root, int key) {
    	DNode pair = new DNode();
        if (root == null) {
            return pair;
        }
        if (size(root.left) >= key) {
            pair = split(root.left, key);
            root.left = pair.right;
            update(root);
            pair.right = root;
        } else {
            pair = split(root.right, key - size(root.left) - 1);
            root.right = pair.left;
            update(root);
            pair.left = root;
        }
        return pair;
    }
    
    public Node merge(Node left, Node right) {
        if (left == null) return right;
        if (right == null) return left;
        
        if (left.pri > right.pri) {
            left.right = merge(left.right, right);
            update(left);
            return left;
        } else {
            right.left = merge(left, right.left);
            update(right);
            return right;
        }
    }
    
    public Node getFirst() {
    	if (root == null)
    		return null;
    	return getFirst(root);
    }
    
    private Node getFirst(Node root) {
    	if (root.left != null)
    		return getFirst(root.left);
    	return root;
    }
    
    public Node getLast() {
    	if (root == null)
    		return null;
    	return getLast(root);
    }
    
    private Node getLast(Node root) {
    	if (root.right != null)
    		return getLast(root.right);
    	return root;
    }
    
    public void query1(int left, int right) {
    	DNode pair1 = split(root, right);
    	DNode pair2 = split(pair1.left, left-1);
    	
    	root = merge(pair2.left, pair1.right);
    	root = merge(pair2.right, root);
    }
    
    public void query2(int left, int right) {
    	DNode pair1 = split(root, right);
    	DNode pair2 = split(pair1.left, left-1);
    	
    	root = merge(pair2.left, pair1.right);
    	root = merge(root, pair2.right);
    }
    
    public void add(int val, int pri) {
        Node n = new Node(val, pri);
        root = merge(root, n);
    }
    
    public void inorder() {
        inorder(root);
        System.out.println();
    }
    
    private void inorder(Node n) {
        if (n == null)
            return;
        inorder(n.left);
        System.out.print(n + " ");
        inorder(n.right);
    }
    
    private static class DNode {
        Node left;
        Node right;
        
        DNode() {
            left = null;
            right = null;
        }
        
        public String toString() {
            return "L:" + left + " R:" + right;
        }
    }
    
    private static class Node {
        int val;
        int pri;
        int size;
        
        Node left;
        Node right;
        
        Node(int val, int pri) {
            this.val = val;
            this.pri = pri;
            size = 1;
            left = null;
            right = null;
        }
        
        public String toString() {
        	return String.valueOf(val);
        }
    }
    
    public static void main(String[] args) {
        Solution t = new Solution();
        Scanner sc = new Scanner(System.in);
        Random rd = new Random();
        int n = sc.nextInt();
        int m = sc.nextInt();
        for (int i = 0;  i < n; i++) {
            int v = sc.nextInt();
            t.add(v, rd.nextInt(n * 10));
        }
        for (int i = 0; i < m; i++) {
        	int op = sc.nextInt();
            int low = sc.nextInt();
            int high = sc.nextInt();
            if (op == 1)
            	t.query1(low, high);
            else
            	t.query2(low, high);
        }
        System.out.println(Math.abs(t.getFirst().val - t.getLast().val));
        t.inorder();
        sc.close();
    }
}








In python3 :




from array import array

n, n_queries = map(int, input().split())
data = array('L', map(int, input().split()))
assert len(data) == n
for m in range(n_queries):
    t, i, j = map(int, input().split())
    if t == 1:
        aux1 = i-1
        aux2 = j - aux1
        data[:aux2], data[aux2:j] = data[aux1:j], data[:aux1]
    else:
        aux1 = i-1
        aux2 = aux1 + n - j
        data[aux1:aux2], data[aux2:] = data[j:], data[aux1:j]
print(abs(data[0] - data[-1]))
print(*data)








View More Similar Problems

Sparse Arrays

There is a collection of input strings and a collection of query strings. For each query string, determine how many times it occurs in the list of input strings. Return an array of the results. Example: strings=['ab', 'ab', 'abc'] queries=['ab', 'abc', 'bc'] There are instances of 'ab', 1 of 'abc' and 0 of 'bc'. For each query, add an element to the return array, results=[2,1,0]. Fun

View Solution →

Array Manipulation

Starting with a 1-indexed array of zeros and a list of operations, for each operation add a value to each of the array element between two given indices, inclusive. Once all operations have been performed, return the maximum value in the array. Example: n=10 queries=[[1,5,3], [4,8,7], [6,9,1]] Queries are interpreted as follows: a b k 1 5 3 4 8 7 6 9 1 Add the valu

View Solution →

Print the Elements of a Linked List

This is an to practice traversing a linked list. Given a pointer to the head node of a linked list, print each node's data element, one per line. If the head pointer is null (indicating the list is empty), there is nothing to print. Function Description: Complete the printLinkedList function in the editor below. printLinkedList has the following parameter(s): 1.SinglyLinkedListNode

View Solution →

Insert a Node at the Tail of a Linked List

You are given the pointer to the head node of a linked list and an integer to add to the list. Create a new node with the given integer. Insert this node at the tail of the linked list and return the head node of the linked list formed after inserting this new node. The given head pointer may be null, meaning that the initial list is empty. Input Format: You have to complete the SinglyLink

View Solution →

Insert a Node at the head of a Linked List

Given a pointer to the head of a linked list, insert a new node before the head. The next value in the new node should point to head and the data value should be replaced with a given value. Return a reference to the new head of the list. The head pointer given may be null meaning that the initial list is empty. Function Description: Complete the function insertNodeAtHead in the editor below

View Solution →

Insert a node at a specific position in a linked list

Given the pointer to the head node of a linked list and an integer to insert at a certain position, create a new node with the given integer as its data attribute, insert this node at the desired position and return the head node. A position of 0 indicates head, a position of 1 indicates one node away from the head and so on. The head pointer given may be null meaning that the initial list is e

View Solution →