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.

Constraints

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


Solution :



title-img 


                            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]);
        Val[V[i]].push_back(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();
        S.push(i);
    }

    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);
                T.put(val,num+1);
            }
            //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);
                output+=thisNum*(thisNum-1);
                //remove
                T.remove(k);
            }
        }
        //finisher
        for (int k:T.keySet()){
            //add to output
            long thisNum=(long)T.get(k);
            output+=thisNum*(thisNum-1);
            
        }
        
        System.out.println(output);
        
    }
}








In C :





#include<stdio.h>
int main()
{
    int n,t,i,j,a[300001],b[300001],c[1000002];//change
    long long d;
    scanf("%d",&n);
    for(i=0;i<n;i++)
    {
        scanf("%d",&a[i]);
    }
    for(i=0;i<=1000000;i++)
    {
        c[i]=0;
    }
    t=0;
    d=0;
    b[0]=a[0];
    c[a[0]]=1;
    for(i=1;i<n;i++)
    {
        if(a[i]>b[t])
        {
            while(a[i]>b[t]&&t>=0)
            {
                c[b[t]]=0;
                t--;
            }
            t++;
            b[t]=a[i];
            c[a[i]]++;
        }
        else
        {
            t++;
            b[t]=a[i];
            c[a[i]]++;
        }
        if(c[a[i]]!=0)
        {
            d+=c[a[i]]-1;
        }
    }
    printf("%lld",2*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]:
        s.pop()
    if len(s) > 0 and s[-1][0] == h[i]:
        count += s[-1][1]
        s[-1][1] += 1
    else:
        s.append([h[i],1])
    
print(2*count)








View More Similar Problems

Compare two linked lists

You’re given the pointer to the head nodes of two linked lists. Compare the data in the nodes of the linked lists to check if they are equal. If all data attributes are equal and the lists are the same length, return 1. Otherwise, return 0. Example: list1=1->2->3->Null list2=1->2->3->4->Null The two lists have equal data attributes for the first 3 nodes. list2 is longer, though, so the lis

View Solution →

Merge two sorted linked lists

This challenge is part of a tutorial track by MyCodeSchool Given pointers to the heads of two sorted linked lists, merge them into a single, sorted linked list. Either head pointer may be null meaning that the corresponding list is empty. Example headA refers to 1 -> 3 -> 7 -> NULL headB refers to 1 -> 2 -> NULL The new list is 1 -> 1 -> 2 -> 3 -> 7 -> NULL. Function Description C

View Solution →

Get Node Value

This challenge is part of a tutorial track by MyCodeSchool Given a pointer to the head of a linked list and a specific position, determine the data value at that position. Count backwards from the tail node. The tail is at postion 0, its parent is at 1 and so on. Example head refers to 3 -> 2 -> 1 -> 0 -> NULL positionFromTail = 2 Each of the data values matches its distance from the t

View Solution →

Delete duplicate-value nodes from a sorted linked list

This challenge is part of a tutorial track by MyCodeSchool You are given the pointer to the head node of a sorted linked list, where the data in the nodes is in ascending order. Delete nodes and return a sorted list with each distinct value in the original list. The given head pointer may be null indicating that the list is empty. Example head refers to the first node in the list 1 -> 2 -

View Solution →

Cycle Detection

A linked list is said to contain a cycle if any node is visited more than once while traversing the list. Given a pointer to the head of a linked list, determine if it contains a cycle. If it does, return 1. Otherwise, return 0. Example head refers 1 -> 2 -> 3 -> NUL The numbers shown are the node numbers, not their data values. There is no cycle in this list so return 0. head refer

View Solution →

Find Merge Point of Two Lists

This challenge is part of a tutorial track by MyCodeSchool Given pointers to the head nodes of 2 linked lists that merge together at some point, find the node where the two lists merge. The merge point is where both lists point to the same node, i.e. they reference the same memory location. It is guaranteed that the two head nodes will be different, and neither will be NULL. If the lists share

View Solution →