### 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;
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)
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)```
```

## Poisonous Plants

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 plan

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