<p>You are given a <strong>0-indexed</strong> array of strings <code>garbage</code> where <code>garbage[i]</code> represents the assortment of garbage at the <code>i<sup>th</sup></code> house. <code>garbage[i]</code> consists only of the characters <code>&#39;M&#39;</code>, <code>&#39;P&#39;</code> and <code>&#39;G&#39;</code> representing one unit of metal, paper and glass garbage respectively. Picking up <strong>one</strong> unit of any type of garbage takes <code>1</code> minute.</p>

<p>You are also given a <strong>0-indexed</strong> integer array <code>travel</code> where <code>travel[i]</code> is the number of minutes needed to go from house <code>i</code> to house <code>i + 1</code>.</p>

<p>There are three garbage trucks in the city, each responsible for picking up one type of garbage. Each garbage truck starts at house <code>0</code> and must visit each house <strong>in order</strong>; however, they do <strong>not</strong> need to visit every house.</p>

<p>Only <strong>one</strong> garbage truck may be used at any given moment. While one truck is driving or picking up garbage, the other two trucks <strong>cannot</strong> do anything.</p>

<p>Return<em> the <strong>minimum</strong> number of minutes needed to pick up all the garbage.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> garbage = [&quot;G&quot;,&quot;P&quot;,&quot;GP&quot;,&quot;GG&quot;], travel = [2,4,3]
<strong>Output:</strong> 21
<strong>Explanation:</strong>
The paper garbage truck:
1. Travels from house 0 to house 1
2. Collects the paper garbage at house 1
3. Travels from house 1 to house 2
4. Collects the paper garbage at house 2
Altogether, it takes 8 minutes to pick up all the paper garbage.
The glass garbage truck:
1. Collects the glass garbage at house 0
2. Travels from house 0 to house 1
3. Travels from house 1 to house 2
4. Collects the glass garbage at house 2
5. Travels from house 2 to house 3
6. Collects the glass garbage at house 3
Altogether, it takes 13 minutes to pick up all the glass garbage.
Since there is no metal garbage, we do not need to consider the metal garbage truck.
Therefore, it takes a total of 8 + 13 = 21 minutes to collect all the garbage.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> garbage = [&quot;MMM&quot;,&quot;PGM&quot;,&quot;GP&quot;], travel = [3,10]
<strong>Output:</strong> 37
<strong>Explanation:</strong>
The metal garbage truck takes 7 minutes to pick up all the metal garbage.
The paper garbage truck takes 15 minutes to pick up all the paper garbage.
The glass garbage truck takes 15 minutes to pick up all the glass garbage.
It takes a total of 7 + 15 + 15 = 37 minutes to collect all the garbage.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= garbage.length &lt;= 10<sup>5</sup></code></li>
	<li><code>garbage[i]</code> consists of only the letters <code>&#39;M&#39;</code>, <code>&#39;P&#39;</code>, and <code>&#39;G&#39;</code>.</li>
	<li><code>1 &lt;= garbage[i].length &lt;= 10</code></li>
	<li><code>travel.length == garbage.length - 1</code></li>
	<li><code>1 &lt;= travel[i] &lt;= 100</code></li>
</ul>

class Solution:
    def garbageCollection(self, garbage: List[str], travel: List[int]) -> int:
        # Initialize an array to store the cumulative travel time and an array to store the last occurrence of each type of garbage
        fib, memo = [0], [0] * 3

        # Calculate the cumulative travel time for each house
        for time in travel: fib.append(fib[-1] + time)
            
        
        # Iterate over the garbage array in reverse order #we want to find the last occurence of each garbage type.
        for i in range(len(garbage)-1, -1, -1):
            v = garbage[i]
            # If the current house has metal garbage and it's the first occurrence, store its index
            if not memo[0] and "M" in v: memo[0] = i
            # If the current house has paper garbage and it's the first occurrence, store its index
            if not memo[1] and "P" in v: memo[1] = i
            # If the current house has glass garbage and it's the first occurrence, store its index
            if not memo[2] and "G" in v: memo[2] = i

            # If we have found at least one occurrence of each type of garbage, stop the iteration
            if all(memo): break

        # Return the total amount of garbage (1 minute for each)
        # plus the sum of the travel times to the last occurrence of each type of garbage
        return len("".join(garbage)) + (fib[memo[0]] + fib[memo[1]] + fib[memo[2]])
676