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 :



title-img


                            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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