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2391. Minimum Amount of Time to Collect Garbage

Description

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.

 

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

Solutions

Solution 1: Hash Table + Prefix Sum

According to the problem description, each garbage truck starts from house $0$, collects one type of garbage, and moves forward in order until it reaches the house index where this type of garbage last appears.

Therefore, we can use a hash table $\textit{last}$ to record the house index where each type of garbage last appears. We assume that the $i$-th type of garbage last appears in the $j$-th house, then the driving time required for the $i$-th truck is $\textit{travel}[0] + \textit{travel}[1] + \cdots + \textit{travel}[j-1]$. Note, if $j = 0$, no driving time is needed. We accumulate the driving time of all vehicles, add the total collection time of each type of garbage, and we can get the answer.

The time complexity is $O(n)$, and the space complexity is $O(k)$, where $n$ and $k$ are the number and types of garbage, respectively. In this problem, $k = 3$.

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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

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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;
    }
}

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class Solution {
public:
    int garbageCollection(vector<string>& garbage, vector<int>& travel) {
        unordered_map<char, int> last;
        int ans = 0;
        for (int i = 0; i < garbage.size(); ++i) {
            auto& s = garbage[i];
            ans += s.size();
            for (char& c : s) {
                last[c] = i;
            }
        }
        int ts = 0;
        for (int i = 1; i <= travel.size(); ++i) {
            ts += travel[i - 1];
            for (auto& [_, j] : last) {
                if (i == j) {
                    ans += ts;
                }
            }
        }
        return ans;
    }
};

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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
}

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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;
}

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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
    }
}

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public class Solution {
    public int GarbageCollection(string[] garbage, int[] travel) {
        Dictionary<char, int> last = new Dictionary<char, int>();
        int ans = 0;
        for (int i = 0; i < garbage.Length; ++i) {
            ans += garbage[i].Length;
            foreach (char c in garbage[i]) {
                last[c] = i;
            }
        }
        int ts = 0;
        for (int i = 1; i <= travel.Length; ++i) {
            ts += travel[i - 1];
            foreach (int j in last.Values) {
                if (i == j) {
                    ans += ts;
                }
            }
        }
        return ans;
    }
}

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