Poisonous Plants

Problem Statement :

There are a number of plants in a garden. Each of the plants has been treated with some amount of pesticide. After each day, if any plant has more pesticide than the plant on its left, being weaker than the left one, it dies.

You are given the initial values of the pesticide in each of the plants. Determine the number of days after which no plant dies, i.e. the time after which there is no plant with more pesticide content than the plant to its left.

Example

p = [ 3, 6, 2, 7, 5  ] // pesticide levels

Use a 1-indexed array. On day  1, plants  2 and 4 die leaving p' = [ 3, 2, 5 ] . On day ,2, plant 3  in p'  dies leaving p^n  = [ 3, 2 ]. There is no plant with a higher concentration of pesticide than the one to its left, so plants stop dying after day 2.


Function Description
Complete the function poisonousPlants in the editor below.

poisonousPlants has the following parameter(s):

int p[n]: the pesticide levels in each plant
Returns
- int: the number of days until plants no longer die from pesticide

Input Format

The first line contains an integer n, the size of the array p.
The next line contains n  space-separated integers p[ i ] .

Solution :

                            Solution in C :

In C++ :





#include<iostream>
#include<cstdio>
#include<algorithm>
#include<set>
#include<map>
#include<queue>
#include<cassert>
#define PB push_back
#define MP make_pair
#define sz(v) (in((v).size()))
#define forn(i,n) for(in i=0;i<(n);++i)
#define forv(i,v) forn(i,sz(v))
#define fors(i,s) for(auto i=(s).begin();i!=(s).end();++i)
#define all(v) (v).begin(),(v).end()
using namespace std;
typedef long long in;
typedef vector<in> VI;
typedef vector<VI> VVI;
in dct=0;
map<in,in> mar;
set<in> td;
void proc(in id){
  auto it=mar.find(id);
  auto it2=it;
  ++it2;
  mar.erase(it);
  if(it2!=mar.end() && it2!=mar.begin()){
    it=it2;
    --it;
    if(it2->second>it->second)
      td.insert(it2->first);
    else{
      if(td.count(it2->first))
	td.erase(it2->first);
    }
  }
}
VI otd;
int main(){
  ios::sync_with_stdio(0);
  cin.tie(0);
  in n;
  cin>>n;
  in ta;
  forn(i,n){
    cin>>ta;
    mar[i]=ta;
    if(i>0 && mar[i]>mar[i-1])
      td.insert(i);
  }
  while(!td.empty()){
    dct++;
    otd.clear();
    fors(i,td)
      otd.PB(*i);
    td.clear();
    reverse(all(otd));
    forv(i,otd){
      proc(otd[i]);
    }
  }
  cout<<dct<<endl;
  return 0;
}








In Java :





import java.util.Scanner;


public class Solution {
    private Scanner sc = new Scanner(System.in);

    public static void main(String[] args) {
        new Solution().solve();
    }

    private void solve() {
        int n = sc.nextInt();

        int[] p = new int[n];
        for (int i = 0; i < n; ++i) {
            p[i] = sc.nextInt();
        }

        int min = p[0];
        int maxDays = 0;
        for (int i = 1; i < n; ++i) {
            min = Math.min(min, p[i]);
            if (p[i] > p[i - 1]) {
                int last = p[i];
                int k = i + 1;
                int days = 1;
                while (k < n && min < p[k]) {
                    if (p[k] <= last) {
                        last = p[k];
                        ++days;
                    }
                    ++k;
                }

                maxDays = Math.max(maxDays, days);
            }
        }

        System.out.println(maxDays);
    }
}








In C :




#include<stdio.h>
int main(){
    long int n,i,j,min=0,locmin;
    scanf("%ld",&n);
    long long int *p=(long long int *)malloc(sizeof(long long int)*n);
    for(i=0;i<n;i++)
    scanf("%lld",&p[i]);
    
    i=n-2;
    j=n-1;
    
    while(i>=0){
               if(j<n && p[j]>p[i]){
                      locmin=0;
                              while(j<n && (p[j]>p[i] || p[j]<0)){
  
                              
                              //if(p[j-1]<0)
                              if(p[j]>0)
                              p[j]=locmin-1;
                              
                              if(locmin>p[j])
                              locmin = p[j];
                              //else
                              //p[j] = -1;
                              j++;               
                              }  
               }
               j=i;
               i--;
    }
    for(i=0;i<n;i++){
                     if(p[i]<min)
                     min=p[i];
    }
    printf("%ld ",-min);
    free(p);
    return 0;
}








In Python3 :






class plant:
    def __init__(self, pest):
        self.pest = pest
        self.prevkiller = self.nextkiller = None
    __slots__ = 'pest,next,prevkiller,nextkiller'.split(',')
    
nplants = int(input())
plants = [int(pest) for pest in input().split()]
start = current = plant(plants[0])
curkiller = firstkiller = plant(None)
for pest in plants[1:]:
    current.next = newplant = plant(pest)
    if current.pest < newplant.pest:
            curkiller.nextkiller = current
            current.prevkiller = curkiller
            curkiller = current
    current = newplant
last = current.next = plant(-1)
last.prevkiller = curkiller
curkiller.nextkiller = last
day = 0
while last.prevkiller is not firstkiller:
    day += 1
    curkiller = last.prevkiller
    while curkiller is not firstkiller:
        victim = curkiller.next
        if not(hasattr(victim, "next")):
            print(victim.pest, day)
        curkiller.next = victim.next
        if victim.prevkiller == curkiller:
            curkiller.nextkiller = victim.nextkiller
            victim.nextkiller.prevkiller = curkiller
            victim.prevkiller = victim.nextkiller = None
        curkiller = curkiller.prevkiller

    curkiller = last.prevkiller
    while curkiller is not firstkiller:
        prevkiller = curkiller.prevkiller
        if curkiller.pest >= curkiller.next.pest:
            curkiller.nextkiller.prevkiller = prevkiller
            prevkiller.nextkiller = curkiller.nextkiller
            curkiller.prevkiller = curkiller.nextkiller = None
        curkiller = prevkiller
print(day)

View More Similar Problems

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

View Solution →

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

View Solution →

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

View Solution →

Largest Rectangle

Skyline Real Estate Developers is planning to demolish a number of old, unoccupied buildings and construct a shopping mall in their place. Your task is to find the largest solid area in which the mall can be constructed. There are a number of buildings in a certain two-dimensional landscape. Each building has a height, given by . If you join adjacent buildings, they will form a solid rectangle

View Solution →

Simple Text Editor

In this challenge, you must implement a simple text editor. Initially, your editor contains an empty string, S. You must perform Q operations of the following 4 types: 1. append(W) - Append W string to the end of S. 2 . delete( k ) - Delete the last k characters of S. 3 .print( k ) - Print the kth character of S. 4 . undo( ) - Undo the last (not previously undone) operation of type 1 or 2,

View Solution →