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 :
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
Array-DS
An array is a type of data structure that stores elements of the same type in a contiguous block of memory. In an array, A, of size N, each memory location has some unique index, i (where 0<=i<N), that can be referenced as A[i] or Ai. Reverse an array of integers. Note: If you've already solved our C++ domain's Arrays Introduction challenge, you may want to skip this. Example: A=[1,2,3
View Solution →2D Array-DS
Given a 6*6 2D Array, arr: 1 1 1 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 An hourglass in A is a subset of values with indices falling in this pattern in arr's graphical representation: a b c d e f g There are 16 hourglasses in arr. An hourglass sum is the sum of an hourglass' values. Calculate the hourglass sum for every hourglass in arr, then print t
View Solution →Dynamic Array
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.
View Solution →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 →