LeetCode #2391 — MEDIUM

Minimum Amount of Time to Collect Garbage

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given a 0-indexed array of strings garbage where garbage[i] represents the assortment of garbage at the i^th house. garbage[i] consists only of the characters 'M', 'P' and 'G' representing one unit of metal, paper and glass garbage respectively. Picking up one unit of any type of garbage takes 1 minute.

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

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

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

Return the minimum number of minutes needed to pick up all the garbage.

Example 1:

Input: garbage = ["G","P","GP","GG"], travel = [2,4,3]
Output: 21
Explanation:
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.

Example 2:

Input: garbage = ["MMM","PGM","GP"], travel = [3,10]
Output: 37
Explanation:
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.

Constraints:

2 <= garbage.length <= 10⁵
garbage[i] consists of only the letters 'M', 'P', and 'G'.
1 <= garbage[i].length <= 10
travel.length == garbage.length - 1
1 <= travel[i] <= 100

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed array of strings garbage where garbage[i] represents the assortment of garbage at the ith house. garbage[i] consists only of the characters 'M', 'P' and 'G' representing one unit of metal, paper and glass garbage respectively. Picking up one unit of any type of garbage takes 1 minute. You are also given a 0-indexed integer array travel where travel[i] is the number of minutes needed to go from house i to house i + 1. There are three garbage trucks in the city, each responsible for picking up one type of garbage. Each garbage truck starts at house 0 and must visit each house in order; however, they do not need to visit every house. Only one garbage truck may be used at any given moment. While one truck is driving or picking up garbage, the other two trucks cannot do anything. Return the minimum number of minutes needed to pick up all the garbage.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

["G","P","GP","GG"]
[2,4,3]

Example 2

["MMM","PGM","GP"]
[3,10]

Step 02

Core Insight

What unlocks the optimal approach

Where can we save time? By not visiting all the houses.
For each type of garbage, find the house with the highest index that has at least 1 unit of this type of garbage.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2391: Minimum Amount of Time to Collect Garbage
class Solution {
    public int garbageCollection(String[] garbage, int[] travel) {
        Map<Character, Integer> last = new HashMap<>(3);
        int ans = 0;
        for (int i = 0; i < garbage.length; ++i) {
            String s = garbage[i];
            ans += s.length();
            for (char c : s.toCharArray()) {
                last.put(c, i);
            }
        }
        int ts = 0;
        for (int i = 1; i <= travel.length; ++i) {
            ts += travel[i - 1];
            for (int j : last.values()) {
                if (i == j) {
                    ans += ts;
                }
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2391: Minimum Amount of Time to Collect Garbage
func garbageCollection(garbage []string, travel []int) (ans int) {
	last := map[byte]int{}
	for i, s := range garbage {
		ans += len(s)
		for j := range s {
			last[s[j]] = i
		}
	}
	ts := 0
	for i := 1; i <= len(travel); i++ {
		ts += travel[i-1]
		for _, j := range last {
			if i == j {
				ans += ts
			}
		}
	}
	return
}

# Accepted solution for LeetCode #2391: Minimum Amount of Time to Collect Garbage
class Solution:
    def garbageCollection(self, garbage: List[str], travel: List[int]) -> int:
        last = {}
        ans = 0
        for i, s in enumerate(garbage):
            ans += len(s)
            for c in s:
                last[c] = i
        ts = 0
        for i, t in enumerate(travel, 1):
            ts += t
            ans += sum(ts for j in last.values() if i == j)
        return ans

// Accepted solution for LeetCode #2391: Minimum Amount of Time to Collect Garbage
use std::collections::HashMap;

impl Solution {
    pub fn garbage_collection(garbage: Vec<String>, travel: Vec<i32>) -> i32 {
        let mut last: HashMap<char, usize> = HashMap::new();
        let mut ans = 0;
        for (i, s) in garbage.iter().enumerate() {
            ans += s.len() as i32;
            for c in s.chars() {
                last.insert(c, i);
            }
        }
        let mut ts = 0;
        for (i, t) in travel.iter().enumerate() {
            ts += t;
            for &j in last.values() {
                if i + 1 == j {
                    ans += ts;
                }
            }
        }
        ans
    }
}

// Accepted solution for LeetCode #2391: Minimum Amount of Time to Collect Garbage
function garbageCollection(garbage: string[], travel: number[]): number {
    const last: Map<string, number> = new Map();
    let ans = 0;
    for (let i = 0; i < garbage.length; ++i) {
        const s = garbage[i];
        ans += s.length;
        for (const c of s) {
            last.set(c, i);
        }
    }
    let ts = 0;
    for (let i = 1; i <= travel.length; ++i) {
        ts += travel[i - 1];
        for (const [_, j] of last) {
            if (i === j) {
                ans += ts;
            }
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(k)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.