**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])
```

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