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])








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