Maximum Element | Editorial by suraj279

You can solve that using a single stack. Just use **stack< pair<int, int> > st;**
Algorithm is pretty straight forward.
**PUSH : **In pair first is to store integer value we need to insert in stack and second is for the max value till the present element.
```
void push(int key){
	pair <int, int> temp;
	if(st.empty()){
		st.push( make_pair( key , key) ) // Stack is empty so the max value till now is the current one.
	}
	else{
		temp = st.top();
		// Max value in stack till top is (st.top()).second or temp.second.
		if(temp.second < a){
			st.push( make_pair( a, a ) ); // a is max value in stack.
		}
		else{
			st.push( make_pair( a, temp.second ) );
		}
	}
	return;
}

```

So every time max value will be at **(st.top()).second .**
All other operations like **pop()** and **top () **are same as for usual  stack < int > 
In case anyone want to see my solution for this problem.
```
#include<bits/stdc++.h>

using namespace std;

#define fast ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
typedef long long ll;
typedef pair<int, int> ii;
typedef vector<ii> vii;
typedef vector<int> vi;
#define INF 1000000000
#define pb push_back
#define mk make_pair

#define flu(i, s, e) for(int i = s; i < e; i++)
#define fld(i, s, e) for(int i = s; i >= e; i--)

int main () {
	fast;

int t, a; scanf("%d", &t);
	stack<ii> st;
	ii temp;
	while(t--){
		scanf("%d", &a);
		if(a == 1){
			scanf("%d", &a);
			if(st.empty()) st.push(mk(a, a));
			else{
				temp = st.top();
				temp.second = max(temp.second, a);
				st.push(mk(a, temp.second));
			}
		}
		else if(a == 2) st.pop();
		else printf("%d\n", (st.top()).second);
	}

return 0;
}
```

Maximum Element | Editorial by suraj279

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