Maximum Element
Problem Statement :
You have an empty sequence, and you will be given N queries. Each query is one of these three types: 1 x -Push the element x into the stack. 2 -Delete the element present at the top of the stack. 3 -Print the maximum element in the stack. Input Format The first line of input contains an integer, N . The next N lines each contain an above mentioned query. (It is guaranteed that each query is valid.) Constraints 1 < = N < = 10 ^5 1 < = x < = 10^9 1 <= type <= 3 Output Format For each type 3 query, print the maximum element in the stack on a new line.
Solution :
Solution in C :
In C ++ :
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#include <stack>
using namespace std;
int main() {
stack<int> max;
stack<int> s;
max.push(0);
int n;
cin>>n;
while (n--){
int a;
cin>>a;
if(a==1){
cin>>a;
if(a>=max.top()) max.push(a);
s.push(a);
}
else if(a==2){
if(s.top()==max.top()) max.pop();
s.pop();
}
else if(a==3) cout<<max.top()<<endl;
}
return 0;
}
In Java :
import java.io.*;
import java.util.*;
public class Solution {
private static void getMaxElementFromStack()
{
Stack<Integer> stack = new Stack<Integer>();
Stack<Integer> onlyMaxs = new Stack<Integer>();
Scanner sc = new Scanner(System.in);
int N = Integer.parseInt(sc.nextLine());
int temp = 0;
while(sc.hasNext())
{
String type[] = sc.nextLine().split(" ");
switch(type[0])
{
case "1":
temp = Integer.parseInt(type[1]);
stack.push(temp);
if(onlyMaxs.isEmpty() || onlyMaxs.peek() <= temp)
onlyMaxs.push(temp);
break;
case "2":
temp = stack.pop();
if(temp == onlyMaxs.peek())
onlyMaxs.pop();
break;
case "3":
System.out.println(onlyMaxs.peek());
}
N--;
}
while(N-- > 0)
System.out.println(onlyMaxs.peek());
}
public static void main(String[] args) {
getMaxElementFromStack();
}
}
In C :
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
typedef struct _node node;
struct _node{
long long int data;
node *next;
};
node* getnode(long long int data){
node *newnode=(node*)malloc(sizeof(node));
newnode->data=data;
newnode->next=NULL;
return newnode;
}
node * push(node *top,int long long data){
node *newnode=getnode(data);
if(top == NULL)
return newnode;
newnode->next=top;
return newnode;
}
node * delete(node *top){
node *temp=top;
top=top->next;
free(temp);
return top;
}
void move(node *fromTop,node *toTop) {
int data=fromTop->data;
delete(fromTop);
push(toTop,data);
}
void printMax(node *maxtop){
printf("%lld\n",maxtop->data);
}
int main() {
int choice;
long long int N;
long long int data;
node *top=NULL,*maxtop=NULL;
scanf("%lld",&N);
while(N) {
scanf("%d",&choice);
if(choice == 1) {
scanf("%lld",&data);
if(top == NULL)
maxtop=push(top,data);
else if(data >= maxtop->data)
maxtop=push(maxtop,data);
top=push(top,data);
}
else if(choice == 2) {
if(top->data == maxtop->data)
maxtop=delete(maxtop);
top=delete(top);
}
else if(choice == 3) {
printMax(maxtop);
}
N--;
}
return 0;
}
In Python3 :
stack = []
max_stack = []
for _ in range(int(input())):
try:
cmd = input()
except:
cmd = '3'
if cmd[0] == '1':
n = int(cmd.split()[1])
stack.append(n)
if len(max_stack) == 0 or max_stack[-1] < n:
max_stack.append(n)
else:
max_stack.append(max_stack[-1])
elif cmd == '2':
stack.pop()
max_stack.pop()
else:
print(max_stack[-1])
View More Similar Problems
Tree Coordinates
We consider metric space to be a pair, , where is a set and such that the following conditions hold: where is the distance between points and . Let's define the product of two metric spaces, , to be such that: , where , . So, it follows logically that is also a metric space. We then define squared metric space, , to be the product of a metric space multiplied with itself: . For
View Solution →Array Pairs
Consider an array of n integers, A = [ a1, a2, . . . . an] . Find and print the total number of (i , j) pairs such that ai * aj <= max(ai, ai+1, . . . aj) where i < j. Input Format The first line contains an integer, n , denoting the number of elements in the array. The second line consists of n space-separated integers describing the respective values of a1, a2 , . . . an .
View Solution →Self Balancing Tree
An AVL tree (Georgy Adelson-Velsky and Landis' tree, named after the inventors) is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to restore this property. We define balance factor for each node as : balanceFactor = height(left subtree) - height(righ
View Solution →Array and simple queries
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. Ty
View Solution →Median Updates
The median M of numbers is defined as the middle number after sorting them in order if M is odd. Or it is the average of the middle two numbers if M is even. You start with an empty number list. Then, you can add numbers to the list, or remove existing numbers from it. After each add or remove operation, output the median. Input: The first line is an integer, N , that indicates the number o
View Solution →Maximum Element
You have an empty sequence, and you will be given N queries. Each query is one of these three types: 1 x -Push the element x into the stack. 2 -Delete the element present at the top of the stack. 3 -Print the maximum element in the stack. Input Format The first line of input contains an integer, N . The next N lines each contain an above mentioned query. (It is guaranteed that each
View Solution →