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

Left Rotation

A left rotation operation on an array of size n shifts each of the array's elements 1 unit to the left. Given an integer, d, rotate the array that many steps left and return the result. Example: d=2 arr=[1,2,3,4,5] After 2 rotations, arr'=[3,4,5,1,2]. Function Description: Complete the rotateLeft function in the editor below. rotateLeft has the following parameters: 1. int d

View Solution →

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 →