Jim and the Skyscrapers

Problem Statement :

Jim has invented a new flying object called HZ42. HZ42 is like a broom and can only fly horizontally, independent of the environment. One day, Jim started his flight from Dubai's highest skyscraper, traveled some distance and landed on another skyscraper of same height! So much fun! But unfortunately, new skyscrapers have been built recently.

Let us describe the problem in one dimensional space. We have in total N skyscrapers aligned from left to right. The th skyscraper has a height of hi. A flying route can be described as ( i , j )  with i =/ j, which means, Jim starts his HZ42 at the top of the skyscraper i and lands on the skyscraper . Since HZ42 can only fly horizontally, Jim will remain at the height hi only. Thus the path ( i, j ) can be valid, only if each of the skyscrapers i, i + 1, . . ,  j - 1,  j  is not strictly greater than  and if the height of the skyscraper he starts from and arrives on have the same height. Formally, ( i , j ) is valid iff  and  hi = hj.

Help Jim in counting the number of valid paths represented by ordered pairs .

Input Format

The first line contains N, the number of skyscrapers. The next line contains N space separated integers representing the heights of the skyscrapers.

Output Format

Print an integer that denotes the number of valid routes.


1   <=  N  <=  3 * 10^5 and no skyscraper will have height greater than 10^6 and less than 1.

Solution :


                            Solution in C :

In C ++  :

#include <cstdio>
#include <vector>
#include <stack>
#include <algorithm>
using namespace std;

const int NMAX = 300010, CMAX = 1000010;

int N, V[NMAX], Right[NMAX];
stack<int> S;
vector<int> Val[CMAX];
long long Ans;

int main()
  //  freopen("c.in", "r", stdin);
   // freopen("c.out", "w", stdout);

    scanf("%i", &N);
    for(int i = 1; i <= N; ++ i)
        scanf("%i", &V[i]);

    V[N + 1] = CMAX;

    S.push(N + 1);
    for(int i = N; i; -- i)
        while(!S.empty() && V[S.top()] <= V[i]) S.pop();
        Right[i] = S.top();

    for(int i = CMAX - 1; i >= 1; -- i)
        int Ptr = 0;
        for(int j = 0; j < Val[i].size(); )
            while(Ptr < Val[i].size() && Val[i][Ptr] <= Right[ Val[i][j] ]) Ptr ++;
            int Len = Ptr - j;
            Ans += 1LL * Len * (Len - 1);
            j = Ptr;

    printf("%lld", Ans);

In Java :

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        Scanner in=new Scanner(System.in);
        int n=in.nextInt();
        TreeMap<Integer,Integer> T=new TreeMap<Integer,Integer>();
        long output=0L;
        for (int i=0;i<n;i++){
            int val=in.nextInt();
            //add this new val to T
            if (T.get(val)==null){T.put(val,1);}
            else {
                int num=T.get(val);
            //kill old values smaller than this new val
            while (T.lowerKey(val)!=null){
                int k=T.lowerKey(val);
                //add to output
                int thisNum=T.get(k);
        for (int k:T.keySet()){
            //add to output
            long thisNum=(long)T.get(k);

In C :

int main()
    int n,t,i,j,a[300001],b[300001],c[1000002];//change
    long long d;
    return 0;

In Python3 :

N = int(input())
h = [int(i) for i in input().split()]

count = 0

s = []
for i in range(0,N):
    while len(s) > 0 and s[-1][0] < h[i]:
    if len(s) > 0 and s[-1][0] == h[i]:
        count += s[-1][1]
        s[-1][1] += 1

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