# 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)```
```

## Insert a Node at the Tail of a Linked List

You are given the pointer to the head node of a linked list and an integer to add to the list. Create a new node with the given integer. Insert this node at the tail of the linked list and return the head node of the linked list formed after inserting this new node. The given head pointer may be null, meaning that the initial list is empty. Input Format: You have to complete the SinglyLink

Given a pointer to the head of a linked list, insert a new node before the head. The next value in the new node should point to head and the data value should be replaced with a given value. Return a reference to the new head of the list. The head pointer given may be null meaning that the initial list is empty. Function Description: Complete the function insertNodeAtHead in the editor below

## Insert a node at a specific position in a linked list

Given the pointer to the head node of a linked list and an integer to insert at a certain position, create a new node with the given integer as its data attribute, insert this node at the desired position and return the head node. A position of 0 indicates head, a position of 1 indicates one node away from the head and so on. The head pointer given may be null meaning that the initial list is e

## Delete a Node

Delete the node at a given position in a linked list and return a reference to the head node. The head is at position 0. The list may be empty after you delete the node. In that case, return a null value. Example: list=0->1->2->3 position=2 After removing the node at position 2, list'= 0->1->-3. Function Description: Complete the deleteNode function in the editor below. deleteNo

## Print in Reverse

Given a pointer to the head of a singly-linked list, print each data value from the reversed list. If the given list is empty, do not print anything. Example head* refers to the linked list with data values 1->2->3->Null Print the following: 3 2 1 Function Description: Complete the reversePrint function in the editor below. reversePrint has the following parameters: Sing