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Weekly Contest 359 Q3
Array
Hash Table
Binary Search
Dynamic Programming
Sorting

2830. Maximize the Profit as the Salesman

中文文档

Description

You are given an integer n representing the number of houses on a number line, numbered from 0 to n - 1.

Additionally, you are given a 2D integer array offers where offers[i] = [starti, endi, goldi], indicating that ith buyer wants to buy all the houses from starti to endi for goldi amount of gold.

As a salesman, your goal is to maximize your earnings by strategically selecting and selling houses to buyers.

Return the maximum amount of gold you can earn.

Note that different buyers can't buy the same house, and some houses may remain unsold.

 

Example 1:

Input: n = 5, offers = [[0,0,1],[0,2,2],[1,3,2]]
Output: 3
Explanation: There are 5 houses numbered from 0 to 4 and there are 3 purchase offers.
We sell houses in the range [0,0] to 1st buyer for 1 gold and houses in the range [1,3] to 3rd buyer for 2 golds.
It can be proven that 3 is the maximum amount of gold we can achieve.

Example 2:

Input: n = 5, offers = [[0,0,1],[0,2,10],[1,3,2]]
Output: 10
Explanation: There are 5 houses numbered from 0 to 4 and there are 3 purchase offers.
We sell houses in the range [0,2] to 2nd buyer for 10 golds.
It can be proven that 10 is the maximum amount of gold we can achieve.

 

Constraints:

  • 1 <= n <= 105
  • 1 <= offers.length <= 105
  • offers[i].length == 3
  • 0 <= starti <= endi <= n - 1
  • 1 <= goldi <= 103

Solutions

Solution 1: Sorting + Binary Search + Dynamic Programming

We sort all the purchase offers by $end$ in ascending order, and then use dynamic programming to solve the problem.

Define $f[i]$ to represent the maximum amount of gold we can get from the first $i$ purchase offers. The answer is $f[n]$.

For $f[i]$, we can choose not to sell the $i$th purchase offer, in which case $f[i] = f[i - 1]$; or we can choose to sell the $i$th purchase offer, in which case $f[i] = f[j] + gold_i$, where $j$ is the largest index that satisfies $end_j \leq start_i$.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Here, $n$ is the number of purchase offers.

Python3

class Solution:
    def maximizeTheProfit(self, n: int, offers: List[List[int]]) -> int:
        offers.sort(key=lambda x: x[1])
        f = [0] * (len(offers) + 1)
        g = [x[1] for x in offers]
        for i, (s, _, v) in enumerate(offers, 1):
            j = bisect_left(g, s)
            f[i] = max(f[i - 1], f[j] + v)
        return f[-1]

Java

class Solution {
    public int maximizeTheProfit(int n, List<List<Integer>> offers) {
        offers.sort((a, b) -> a.get(1) - b.get(1));
        n = offers.size();
        int[] f = new int[n + 1];
        int[] g = new int[n];
        for (int i = 0; i < n; ++i) {
            g[i] = offers.get(i).get(1);
        }
        for (int i = 1; i <= n; ++i) {
            var o = offers.get(i - 1);
            int j = search(g, o.get(0));
            f[i] = Math.max(f[i - 1], f[j] + o.get(2));
        }
        return f[n];
    }

    private int search(int[] nums, int x) {
        int l = 0, r = nums.length;
        while (l < r) {
            int mid = (l + r) >> 1;
            if (nums[mid] >= x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }
}

C++

class Solution {
public:
    int maximizeTheProfit(int n, vector<vector<int>>& offers) {
        sort(offers.begin(), offers.end(), [](const vector<int>& a, const vector<int>& b) {
            return a[1] < b[1];
        });
        n = offers.size();
        vector<int> f(n + 1);
        vector<int> g;
        for (auto& o : offers) {
            g.push_back(o[1]);
        }
        for (int i = 1; i <= n; ++i) {
            auto o = offers[i - 1];
            int j = lower_bound(g.begin(), g.end(), o[0]) - g.begin();
            f[i] = max(f[i - 1], f[j] + o[2]);
        }
        return f[n];
    }
};

Go

func maximizeTheProfit(n int, offers [][]int) int {
	sort.Slice(offers, func(i, j int) bool { return offers[i][1] < offers[j][1] })
	n = len(offers)
	f := make([]int, n+1)
	g := []int{}
	for _, o := range offers {
		g = append(g, o[1])
	}
	for i := 1; i <= n; i++ {
		j := sort.SearchInts(g, offers[i-1][0])
		f[i] = max(f[i-1], f[j]+offers[i-1][2])
	}
	return f[n]
}

TypeScript

function maximizeTheProfit(n: number, offers: number[][]): number {
    offers.sort((a, b) => a[1] - b[1]);
    n = offers.length;
    const f: number[] = Array(n + 1).fill(0);
    const g = offers.map(x => x[1]);
    const search = (x: number) => {
        let l = 0;
        let r = n;
        while (l < r) {
            const mid = (l + r) >> 1;
            if (g[mid] >= x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    };
    for (let i = 1; i <= n; ++i) {
        const j = search(offers[i - 1][0]);
        f[i] = Math.max(f[i - 1], f[j] + offers[i - 1][2]);
    }
    return f[n];
}