QHEAP1


Problem Statement :


This question is designed to help you get a better understanding of basic heap operations.
You will be given queries of  types:

" 1 v " - Add an element  to the heap.
" 2 v " - Delete the element  from the heap.
"3" - Print the minimum of all the elements in the heap.
NOTE: It is guaranteed that the element to be deleted will be there in the heap. Also, at any instant, only distinct elements will be in the heap.

Input Format

The first line contains the number of queries, Q.
Each of the next Q lines contains a single query of any one of the 3 above mentioned types.


Output Format

For each query of type 3, print the minimum value on a single line.


Solution :



title-img 


                            Solution in C :

In C ++ :






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


int main() {
    vector<int> vec;
    int q;
    cin >> q;
    int minval=1000000007;
    bool flag=false;
    for (int i=0; i<q; i++)
    {
        int a,v;
        cin >> a;
        if (a==1)
        {
            cin >> v;
            vec.push_back(v);
            minval=min(minval,v);
        }
        if (a==2)
        {
            cin >> v;
            if (v==minval)
                flag=true;
            for (int j=0; j<vec.size(); j++)
            {
                if (vec[j]==v)
                {
                    vec.erase(vec.begin()+j);
                }
            }
        }
        if (a==3)
        {
            if (flag)
            {
                minval=1000000007;
                for (int j=0; j<vec.size();j++)
                {
                    minval=min(minval,vec[j]);    
                }
                flag = false;
            }
            cout << minval << endl;
        }
    }
    /*
    priority_queue<int> pq;
    int q;
    cin >> q;
    for (int i=0; i<q; i++)
    {
        int a,v;
        cin >> a;
        if (a==1)
        {
            cin >> v;
            pq.push(-1*v);
        }
        if (a==2)
        {
            cin >> v;
            pq.pop();
        }
        if (a==3)
        {
            cout << pq.top()*-1 << endl;
            //pq.pop();
        }
    }
    */
    return 0;
}








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 s = new Scanner(System.in);
        int n = s.nextInt();
        PriorityQueue<Integer> pq = new PriorityQueue<Integer>();
        for (int i=0;i<n;i++) {
            int cmd = s.nextInt();
            switch (cmd) {
                case 1:
                    int val = s.nextInt();
                    pq.add(val);
                    break;
                case 2:
                    val = s.nextInt();
                    pq.remove(val);
                    break;
                case 3:
                    val = pq.peek();
                    System.out.println(val);
                    break;
            }
        }
        s.close();
    }
}









In C :





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

int arr[1000000];
int curr=-1;

int mindex(int left,int right){
    return arr[left]<arr[right]?left:right;
}

int find_index(int y){
    for(int i=0;i<=curr;i++)
        if(arr[i]==y)
        return i;
        
        return -1;
}

void heapifyUp(int index){
    int parent=index/2;
    if(arr[parent]<arr[index] || (parent==index))
        return;
    else{
        int temp=arr[parent];
        arr[parent]=arr[index];
        arr[index]=temp;
        heapifyUp(parent);
    }
return;
}

void heapifyDown(int index){
    int left=2*index;
    int right=2*index+1;
    if(left<=curr && right<=curr){
        if(arr[index]<arr[left] && arr[index]<arr[right])
            return;
        else{
            int min_index=mindex(left,right);
            int temp=arr[min_index];
            arr[min_index]=arr[index];
            arr[index]=temp;
            heapifyDown(min_index);
        }
    }
}

void insert(int y){
    arr[++curr]=y;
    heapifyUp(curr);
    /*for(int i=0;i<=curr;i++)
        printf("%d\t",arr[i]);
    printf("\n");*/
}

void delet(int y){
int index=find_index(y);
    //printf("found index:%d\t",index);
    if(index!=-1){
    arr[index]=arr[curr--];
        heapifyDown(index);
    }
    /*for(int i=0;i<=curr;i++)
        printf("%d\t",arr[i]);
    printf("\n");
*/
}


int main() {

    int n,x,y;
    scanf("%d",&n);
    
    while(n--){
        scanf("%d",&x);
        if(x==1){
            scanf("%d",&y);
            insert(y);
        }
        if(x==2){
            scanf("%d",&y);
            delet(y);
        }
        if(x==3){
            if(curr>=0)
                printf("%d\n",arr[0]);
        }
    }
    
    return 0;
}









In Python3 :





import heapq
m = []
N = int(input())
di = dict()
for i in range(N):
    b = input().split()
    if b[0] == '1':
        b = int(b[1])
        di[b] = 0
        heapq.heappush(m,b)
    elif b[0] == '2':
        b = int(b[1])
        di[b] = 1
    else:
        while di[m[0]] == 1:
            heapq.heappop(m)
        print(m[0])








View More Similar Problems

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 →

Delete a Node

Delete the node at a given position in a linked list and return a reference to the head node. The head is at position 0. The list may be empty after you delete the node. In that case, return a null value. Example: list=0->1->2->3 position=2 After removing the node at position 2, list'= 0->1->-3. Function Description: Complete the deleteNode function in the editor below. deleteNo

View Solution →

Print in Reverse

Given a pointer to the head of a singly-linked list, print each data value from the reversed list. If the given list is empty, do not print anything. Example head* refers to the linked list with data values 1->2->3->Null Print the following: 3 2 1 Function Description: Complete the reversePrint function in the editor below. reversePrint has the following parameters: Sing

View Solution →

Reverse a linked list

Given the pointer to the head node of a linked list, change the next pointers of the nodes so that their order is reversed. The head pointer given may be null meaning that the initial list is empty. Example: head references the list 1->2->3->Null. Manipulate the next pointers of each node in place and return head, now referencing the head of the list 3->2->1->Null. Function Descriptio

View Solution →

Compare two linked lists

You’re given the pointer to the head nodes of two linked lists. Compare the data in the nodes of the linked lists to check if they are equal. If all data attributes are equal and the lists are the same length, return 1. Otherwise, return 0. Example: list1=1->2->3->Null list2=1->2->3->4->Null The two lists have equal data attributes for the first 3 nodes. list2 is longer, though, so the lis

View Solution →

Merge two sorted linked lists

This challenge is part of a tutorial track by MyCodeSchool Given pointers to the heads of two sorted linked lists, merge them into a single, sorted linked list. Either head pointer may be null meaning that the corresponding list is empty. Example headA refers to 1 -> 3 -> 7 -> NULL headB refers to 1 -> 2 -> NULL The new list is 1 -> 1 -> 2 -> 3 -> 7 -> NULL. Function Description C

View Solution →