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<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;
}
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;
}
Solution in Java :
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 sc = new Scanner(System.in);
int n = sc.nextInt();
int[] p = new int[n+1];
int[] surv = new int[n+1];
int[] index = new int[n+1];
p[0] = Integer.MAX_VALUE;
for (int i = 1; i <= n; i++)
{
p[i] = sc.nextInt();
index[i] = i;
}
int min = p[1];
surv[1] = -1;
int longest = -1;
for (int i = 2; i <= n; i++)
{
if (p[i] > min)
{
int j = i -1;
int max = 0;
while (j >= 1 && surv[j] != -1)
{
if (p[j] < p[i])
break;
max = Math.max(surv[j] + 1, max);
j = index[j];
}
surv[i] = max;
index[i] = j;
longest = Math.max(longest, max);
}
else
{
surv[i] = -1;
min = p[i];
}
}
System.out.println(longest+1);
}
}
Solution in Python :
In Python3 :
class Plant:
def __init__(self, pest):
self.p =pest
self.d = 0 #days passed before it reaches this point
input()
plants = [Plant(int(i)) for i in input().split()]
stack = [plants[-1]]
days = 0
for i in range(len(plants)-2,-1,-1):
if len(stack)==0 or plants[i].p >=stack[-1].p:
stack.append(plants[i])
else:
local = 0
while len(stack)>0 and plants[i].p < stack[-1].p:
local = max(local+1, stack[-1].d)
stack.pop()
plants[i].d = local
days = max(local, days)
stack.append(plants[i])
print(days)
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