[LeetCode] 1235. Maximum Profit in Job Scheduling

1235. Maximum Profit in Job Scheduling

• Hardness: \(\color{red}\textsf{Hard}\)
• Ralated Topics: Array、Binary Search、Dynamic Programming、Sorting

一、題目

We have n jobs, where every job is scheduled to be done from startTime[i] to endTime[i], obtaining a profit of profit[i].
You’re given the startTime, endTime and profit arrays, return the maximum profit you can take such that there are no two jobs in the subset with overlapping time range.
If you choose a job that ends at time X you will be able to start another job that starts at time X.

Example 1:

• Input: startTime = [1,2,3,3], endTime = [3,4,5,6], profit = [50,10,40,70]
• Output: 120
• Explanation: The subset chosen is the first and fourth job.
  Time range [1-3]+[3-6] , we get profit of 120 = 50 + 70.

Example 2:

• Input: startTime = [1,2,3,4,6], endTime = [3,5,10,6,9], profit = [20,20,100,70,60]
• Output: 150
• Explanation: The subset chosen is the first, fourth and fifth job.
  Profit obtained 150 = 20 + 70 + 60.

Example 3:

• Input: The subset chosen is the first, fourth and fifth job. Profit obtained 150 = 20 + 70 + 60.
• Output: 6

Constraints:

• 1 <= startTime.length == endTime.length == profit.length <= 5 * 10^4
• 1 <= startTime[i] < endTime[i] <= 10^9
• 1 <= profit[i] <= 10^4

二、分析

• 在思考這一題，首先要先有 coin change 的思維，也就是動態規劃：
  • 我們將 dp[n] 定義為在時間 n 時的最大利益。
  • 所以當時間點 i 的最大利益會等於 max(dp[i-1], dp[i - time_cost] + profit
  • 以範例 1 為例即：
    • dp[0] = 0
    • dp[1] = 0
    • dp[2] = 0
    • dp[3] = 50 = max(dp[1]+50, dp[2])
    • dp[4] = 50 = dp[3]
    • dp[5] = 90 = max(dp[3]+40, dp[4])
    • …
  • 其中我們可以發現，有可能發生改變的時間點都是在每一個工作的 endTime，也就是說我們只要針對每個 endTime 去記錄即可，其中我們可將 dp[i - time_cost] 改為搜尋小於 startTime 的最大值，即：
    • dp[0] = 0
    • dp[3] = 50 = max(dp[0], dp[3])
    • dp[4] = 10 = max(dp[0]+10, dp[4]) 我們只記錄當下最大利益，故不記錄
    • dp[5] = 90 = max(dp[3]+40, dp[5])
    • dp[6] = 120 = max(dp[3]+70, dp[6]
• 我們可以使用 map 這個資料結構，將所有 trigger point 依 endTime 排序後，逐步更新。
  • 其中注意 upper_bound(x) 這個函式，會找大於 x 的位子，而且我們要找的是比小於等於當前 startTime 的資料，故我們找的是 upper_bound(x)-1。
  • 由於時間 t = 0 時不會有收益，我們可以加入 {0,0}，這樣可以省去解決 iterator out of range(it 指向 -1) 的情形。

三、解題

1. DP + Binary Search

• Time complexity: \(O(n\log n)\)
• Space complexity: \(O(n)\)

int jobScheduling(vector<int>& startTime, vector<int>& endTime, vector<int>& profit) {
    map<int,int> dp;
    vector<vector<int>> job;
    int n = startTime.size();
    for (int i = 0; i < n; i++) {
        job.push_back({endTime[i], startTime[i], profit[i]});
    }
    sort(job.begin(), job.end());   // sort by endTime
    dp.insert({0,0});               // 省去處理 out of range
    int res = 0;
    for (int i = 0; i < n; i++) {
        auto it = dp.upper_bound(job[i][1]);    // > startTime
        it--;                                   // <= startTime
        int last = it->second;
        int val = last + job[i][2];             // 由當前最大收益往上累積
        int pos = job[i][0];
        if (val < res) continue;  // 若當前最大收益比歷史最大收益還小，則跳過不記錄

        dp[pos] = max(dp[pos], val);            // 更新當前最大收益
        res = max(dp[pos], res);                // 更新歷史最大收益
    }
    return res;
}

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