There are multiple solutions to this problem, each with different thought processes but the same end result. Two are presented below. An operation of type $1$ refers to a single walk with pesticide of type $1$, while an operation of type $2$ refers to a single walk with pesticide of type $2$.

Solution 1: Difference Arrays

Let $\text{diff}(a)$ be the difference array of $a$. In particular, $\text{diff}(a) = [a_1, a_2 - a_1, \ldots, a_n - a_{n-1}]$. Let's examine how each type of operation affects $\text{diff}(a)$.

Assume we do an operation of type 1. The value at indices less than $h$ remain unchanged, so the corresponding values of $\text{diff}(a)$ stay the same as well. Because $a_h$ increases by $1$ and $a_{h-1}$ remains unchanged, $\text{diff}(a)_h$ is the first value that changes, and it increases by $1$. Similarly, because $a_{h+1}$ increases by $2$ and $a_h$ increases by $1$, $\text{diff}(a)_{h+1}$ increases by $1$. Extending this argument, an operation of type $1$ simply increments a suffix of $\text{diff}(a)$, while and operation of type $2$ decrements a suffix.

This transformation can be done one more time. Let's examine how $\text{diff}(\text{diff}(a))$ changes after each operation. By doing a similar analysis as before, we can see that the only index that changes is $\text{diff}(\text{diff}(a))_h$, and it either increases or decreases by $1$ depending on the operation type. Finding the minimum operations to make $\text{diff}(\text{diff}(a))$ all $0$'s is now straightforward, as it is equal to $\sum_{i=1}^n |\text{diff}(\text{diff}(a))_i|$, because all of the indices are independent.

Rohin Garg's code:

#include <bits/stdc++.h>
using namespace std;
 
using ll = long long;
 
vector<ll> diff(vector<ll> a) {
    vector<ll> b(a.size());
    for (int i = 0; i < (int) b.size(); i++) {
        b[i] = a[i];
        if (i > 0)
            b[i] -= a[i-1];
    }
    return b;
}

int main() {
    ios::sync_with_stdio(false); cin.tie(0);
   
    int n; cin >> n;
    vector<ll> a(n);
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
    
    a = diff(diff(a));
    ll ans = 0;
    for (int i = 0; i < n; i++)
        ans += abs(a[i]);
        
    cout << ans << '\n';
}

Solution 2: Optimized Simulation

The only way to alter the number of bacteria on patch $1$ would be for Farmer John to use $h = 1$. This motivates the idea of first calculating the operations that must be done to make $a_1 = 0$, then $a_2 = 0$, and so on.

Note that you'll never do both types of operations with the same $h$ value, as they would just undo each other, and doing neither instead of both would get you the same final configuration in a smaller number of operations.

This means that we know exactly what set of operations we'll do with $h = 1$: If $a_i$ is positive, we'll do $a_i$ operations of type $2$, otherwise we'll do $-a_i$ operations of type $1$. Then, we can ignore $a_1$, as it can no longer be changed by any future operation, and continue this process (making sure to update all array elements when doing an operation). This immediately lends an $\mathcal{O}(n^2)$ solution, as at each index you have to update $\mathcal{O}(n)$ other indices.

What remains is to optimize the simulation of each operation. Let's start by considering a simpler version of the problem, where an operation adds $1$ to the indices after it, instead of $1, 2, \ldots$ In this version, notice that the value of an index is only changed by the operations that occurred at another index before it, and all that matters is the difference between the number of operations of type $1$ and operations of type $2$. Because of this, instead of directly updating the values, you can just keep track of the number of operations that have occurred, and update the value when it is iterated over.

In this version, you can do something similar, except you have to keep track of two things: the difference between the number of operations of type $1$ and the number of operations of type $2$, as well as the total amount the operations add. As an example, let's say that there were $3$ more operations of type $1$ than $2$ at indices $\leq 5$, and those operations increased the value of $a_5$ by $4$. Then, because the contribution of each operation of type $1$ increases by $1$, and the contribution of each element of type $2$ decreases by $1$, the contribution of all of the operations increases by exactly $3$. Therefore, the value of $a_6$ will increase by exactly $4 + 3 = 7$ due to the operations. In this way, we can calculate the contribution of the operations without having to directly simulate them, lending us an $\mathcal{O}(n)$ solution.

Rohin Garg's code:

#include <bits/stdc++.h>
using namespace std;
 
using ll = long long;
 
int main() {
    ios::sync_with_stdio(false); cin.tie(0);
   
    int n; cin >> n;
    vector<ll> a(n);
    for (int i = 0; i < n; i++) {
        cin >> a[i];
    }
   
    ll ans = 0;
    ll contribution = 0;
    ll cnt_ops = 0;
    for (int i = 0; i < n; i++) {
        contribution += cnt_ops;
        a[i] += contribution;
        
        ll cur_ops = -a[i];
        ans += abs(cur_ops);
        cnt_ops += cur_ops;
        contribution += cur_ops;
    }
    
    cout << ans << '\n';
}