# Array Pairs

### Problem Statement :

```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 .```

### Solution :

```                            ```Solution in C :

In C++ :

#include <bits/stdc++.h>
using namespace std;

#define rep(i,n) for(int i=0;i<n;i++)
#define ll long long int
#define f first
#define s second
#define pi pair<ll,ll>
#define pii pair<pi,ll>
#define f first
#define s second
#define pb push_back
#define mod 1000000007
#define mp make_pair
#define pb push_back
#define rep(i,n) for(int i=0;i<n;i++)

int N;
int A[1000011];
int L[1000011];
int R[1000011];
vector<int>g[1000011];
ll bt[1000011];
int maxn;
void update(int ind, int val) {
while(ind <= maxn) {
bt[ind] += val;
ind += (ind & -ind);
}
}
ll query(int ind) {
ll ans = 0;
while(ind > 0) {
ans += bt[ind];
ind -= (ind & -ind);
}
return ans;
}
vector<int>V;
int find_ind(int x) {
if(V.back() <= x) return V.size();
return upper_bound(V.begin(), V.end(), x) - V.begin();
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);

cin >> N;
set<int>S;
unordered_map<int, int>M;

for(int i = 1; i <= N; i++) {
cin >> A[i];
assert(A[i] >= 1 and A[i] <= 1000000000);
S.insert(A[i]);
}
vector<pi>window;
for(int i = 1; i <= N; i++) {
while(window.size() > 0 and window.back().f < A[i]) window.pop_back();
if(window.size() == 0) L[i] = 1;
else {
L[i] = window.back().s + 1;
}
window.pb(mp(A[i], i));
}
window.clear();
for(int i = N; i >= 1; i--) {
while(window.size() > 0 and window.back().f <= A[i]) window.pop_back();
if(window.size() == 0) R[i] = N;
else {
R[i] = window.back().s - 1;
}
window.pb(mp(A[i], i));
}

for(int i = 1; i <= N; i++) {
if(i - L[i] <= R[i] - i) {
for(int j = L[i]; j < i; j++) {
g[i - 1].pb(-A[i] / A[j]);
g[R[i]].pb(A[i] / A[j]);
//S.insert(A[i]/A[j]);
}

g[i].pb(-1);
g[R[i]].pb(1);
} else {

for(int j = i + 1; j <= R[i]; j++) {
g[L[i] - 1].pb(-A[i] / A[j]);
g[i].pb(A[i] / A[j]);
//S.insert(A[i]/A[j]);
}

g[L[i] - 1].pb(-1);
g[i - 1].pb(1);
}
}
maxn = S.size() + 2;
int cnt = 1;
for(set<int>::iterator it = S.begin(); it != S.end(); it++) {
M[*it] = cnt++;
}
ll ans = 0;
int r;
V = vector<int>(S.begin(), S.end());
for(int i = 1; i <= N; i++) {
update(M[A[i]], 1);
for(int j = 0; j < g[i].size(); j++) {
r = find_ind(abs(g[i][j]));
if(g[i][j] < 0) {
ans -= query(r);
} else {
ans += query(r);
}
}
}
cout << ans;
}```
```

## 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

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

## 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

## Balanced Brackets

A bracket is considered to be any one of the following characters: (, ), {, }, [, or ]. Two brackets are considered to be a matched pair if the an opening bracket (i.e., (, [, or {) occurs to the left of a closing bracket (i.e., ), ], or }) of the exact same type. There are three types of matched pairs of brackets: [], {}, and (). A matching pair of brackets is not balanced if the set of bra

## Equal Stacks

ou have three stacks of cylinders where each cylinder has the same diameter, but they may vary in height. You can change the height of a stack by removing and discarding its topmost cylinder any number of times. Find the maximum possible height of the stacks such that all of the stacks are exactly the same height. This means you must remove zero or more cylinders from the top of zero or more of

## Game of Two Stacks

Alexa has two stacks of non-negative integers, stack A = [a0, a1, . . . , an-1 ] and stack B = [b0, b1, . . . , b m-1] where index 0 denotes the top of the stack. Alexa challenges Nick to play the following game: In each move, Nick can remove one integer from the top of either stack A or stack B. Nick keeps a running sum of the integers he removes from the two stacks. Nick is disqualified f