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Biweekly Contest 61 Q3
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Binary Search
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2008. Maximum Earnings From Taxi

中文文档

Description

There are n points on a road you are driving your taxi on. The n points on the road are labeled from 1 to n in the direction you are going, and you want to drive from point 1 to point n to make money by picking up passengers. You cannot change the direction of the taxi.

The passengers are represented by a 0-indexed 2D integer array rides, where rides[i] = [starti, endi, tipi] denotes the ith passenger requesting a ride from point starti to point endi who is willing to give a tipi dollar tip.

For each passenger i you pick up, you earn endi - starti + tipi dollars. You may only drive at most one passenger at a time.

Given n and rides, return the maximum number of dollars you can earn by picking up the passengers optimally.

Note: You may drop off a passenger and pick up a different passenger at the same point.

 

Example 1:

Input: n = 5, rides = [[2,5,4],[1,5,1]]
Output: 7
Explanation: We can pick up passenger 0 to earn 5 - 2 + 4 = 7 dollars.

Example 2:

Input: n = 20, rides = [[1,6,1],[3,10,2],[10,12,3],[11,12,2],[12,15,2],[13,18,1]]
Output: 20
Explanation: We will pick up the following passengers:
- Drive passenger 1 from point 3 to point 10 for a profit of 10 - 3 + 2 = 9 dollars.
- Drive passenger 2 from point 10 to point 12 for a profit of 12 - 10 + 3 = 5 dollars.
- Drive passenger 5 from point 13 to point 18 for a profit of 18 - 13 + 1 = 6 dollars.
We earn 9 + 5 + 6 = 20 dollars in total.

 

Constraints:

  • 1 <= n <= 105
  • 1 <= rides.length <= 3 * 104
  • rides[i].length == 3
  • 1 <= starti < endi <= n
  • 1 <= tipi <= 105

Solutions

Solution 1: Memoization Search + Binary Search

First, we sort $rides$ in ascending order by $start$. Then we design a function $dfs(i)$, which represents the maximum tip that can be obtained from accepting orders starting from the $i$-th passenger. The answer is $dfs(0)$.

The calculation process of the function $dfs(i)$ is as follows:

For the $i$-th passenger, we can choose to accept or not to accept the order. If we don't accept the order, the maximum tip that can be obtained is $dfs(i + 1)$. If we accept the order, we can use binary search to find the first passenger encountered after the drop-off point of the $i$-th passenger, denoted as $j$. The maximum tip that can be obtained is $dfs(j) + end_i - start_i + tip_i$. Take the larger of the two. That is:

$$ dfs(i) = \max(dfs(i + 1), dfs(j) + end_i - start_i + tip_i) $$

Where $j$ is the smallest index that satisfies $start_j \ge end_i$, which can be obtained by binary search.

In this process, we can use memoization search to save the answer of each state to avoid repeated calculations.

The time complexity is $O(m \times \log m)$, and the space complexity is $O(m)$. Here, $m$ is the length of $rides$.

Python3

class Solution:
    def maxTaxiEarnings(self, n: int, rides: List[List[int]]) -> int:
        @cache
        def dfs(i: int) -> int:
            if i >= len(rides):
                return 0
            st, ed, tip = rides[i]
            j = bisect_left(rides, ed, lo=i + 1, key=lambda x: x[0])
            return max(dfs(i + 1), dfs(j) + ed - st + tip)

        rides.sort()
        return dfs(0)

Java

class Solution {
    private int m;
    private int[][] rides;
    private Long[] f;

    public long maxTaxiEarnings(int n, int[][] rides) {
        Arrays.sort(rides, (a, b) -> a[0] - b[0]);
        m = rides.length;
        f = new Long[m];
        this.rides = rides;
        return dfs(0);
    }

    private long dfs(int i) {
        if (i >= m) {
            return 0;
        }
        if (f[i] != null) {
            return f[i];
        }
        int[] r = rides[i];
        int st = r[0], ed = r[1], tip = r[2];
        int j = search(ed, i + 1);
        return f[i] = Math.max(dfs(i + 1), dfs(j) + ed - st + tip);
    }

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

C++

class Solution {
public:
    long long maxTaxiEarnings(int n, vector<vector<int>>& rides) {
        sort(rides.begin(), rides.end());
        int m = rides.size();
        long long f[m];
        memset(f, -1, sizeof(f));
        function<long long(int)> dfs = [&](int i) -> long long {
            if (i >= m) {
                return 0;
            }
            if (f[i] != -1) {
                return f[i];
            }
            auto& r = rides[i];
            int st = r[0], ed = r[1], tip = r[2];
            int j = lower_bound(rides.begin() + i + 1, rides.end(), ed, [](auto& a, int val) { return a[0] < val; }) - rides.begin();
            return f[i] = max(dfs(i + 1), dfs(j) + ed - st + tip);
        };
        return dfs(0);
    }
};

Go

func maxTaxiEarnings(n int, rides [][]int) int64 {
	sort.Slice(rides, func(i, j int) bool { return rides[i][0] < rides[j][0] })
	m := len(rides)
	f := make([]int64, m)
	var dfs func(int) int64
	dfs = func(i int) int64 {
		if i >= m {
			return 0
		}
		if f[i] == 0 {
			st, ed, tip := rides[i][0], rides[i][1], rides[i][2]
			j := sort.Search(m, func(j int) bool { return rides[j][0] >= ed })
			f[i] = max(dfs(i+1), int64(ed-st+tip)+dfs(j))
		}
		return f[i]
	}
	return dfs(0)
}

TypeScript

function maxTaxiEarnings(n: number, rides: number[][]): number {
    rides.sort((a, b) => a[0] - b[0]);
    const m = rides.length;
    const f: number[] = Array(m).fill(-1);
    const search = (x: number, l: number): number => {
        let r = m;
        while (l < r) {
            const mid = (l + r) >> 1;
            if (rides[mid][0] >= x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    };
    const dfs = (i: number): number => {
        if (i >= m) {
            return 0;
        }
        if (f[i] === -1) {
            const [st, ed, tip] = rides[i];
            const j = search(ed, i + 1);
            f[i] = Math.max(dfs(i + 1), dfs(j) + ed - st + tip);
        }
        return f[i];
    };
    return dfs(0);
}

Solution 2: Dynamic Programming + Binary Search

We can change the memoization search in Solution 1 to dynamic programming.

First, sort $rides$, this time we sort by $end$ in ascending order. Then define $f[i]$, which represents the maximum tip that can be obtained from the first $i$ passengers. Initially, $f[0] = 0$, and the answer is $f[m]$.

For the $i$-th passenger, we can choose to accept or not to accept the order. If we don't accept the order, the maximum tip that can be obtained is $f[i-1]$. If we accept the order, we can use binary search to find the last passenger whose drop-off point is not greater than $start_i$ before the $i$-th passenger gets on the car, denoted as $j$. The maximum tip that can be obtained is $f[j] + end_i - start_i + tip_i$. Take the larger of the two. That is:

$$ f[i] = \max(f[i - 1], f[j] + end_i - start_i + tip_i) $$

Where $j$ is the largest index that satisfies $end_j \le start_i$, which can be obtained by binary search.

The time complexity is $O(m \times \log m)$, and the space complexity is $O(m)$. Here, $m$ is the length of $rides$.

Similar problems:

Python3

class Solution:
    def maxTaxiEarnings(self, n: int, rides: List[List[int]]) -> int:
        rides.sort(key=lambda x: x[1])
        f = [0] * (len(rides) + 1)
        for i, (st, ed, tip) in enumerate(rides, 1):
            j = bisect_left(rides, st + 1, hi=i, key=lambda x: x[1])
            f[i] = max(f[i - 1], f[j] + ed - st + tip)
        return f[-1]

Java

class Solution {
    public long maxTaxiEarnings(int n, int[][] rides) {
        Arrays.sort(rides, (a, b) -> a[1] - b[1]);
        int m = rides.length;
        long[] f = new long[m + 1];
        for (int i = 1; i <= m; ++i) {
            int[] r = rides[i - 1];
            int st = r[0], ed = r[1], tip = r[2];
            int j = search(rides, st + 1, i);
            f[i] = Math.max(f[i - 1], f[j] + ed - st + tip);
        }
        return f[m];
    }

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

C++

class Solution {
public:
    long long maxTaxiEarnings(int n, vector<vector<int>>& rides) {
        sort(rides.begin(), rides.end(), [](const vector<int>& a, const vector<int>& b) { return a[1] < b[1]; });
        int m = rides.size();
        vector<long long> f(m + 1);
        for (int i = 1; i <= m; ++i) {
            auto& r = rides[i - 1];
            int st = r[0], ed = r[1], tip = r[2];
            auto it = lower_bound(rides.begin(), rides.begin() + i, st + 1, [](auto& a, int val) { return a[1] < val; });
            int j = distance(rides.begin(), it);
            f[i] = max(f[i - 1], f[j] + ed - st + tip);
        }
        return f.back();
    }
};

Go

func maxTaxiEarnings(n int, rides [][]int) int64 {
	sort.Slice(rides, func(i, j int) bool { return rides[i][1] < rides[j][1] })
	m := len(rides)
	f := make([]int64, m+1)
	for i := 1; i <= m; i++ {
		r := rides[i-1]
		st, ed, tip := r[0], r[1], r[2]
		j := sort.Search(m, func(j int) bool { return rides[j][1] >= st+1 })
		f[i] = max(f[i-1], f[j]+int64(ed-st+tip))
	}
	return f[m]
}

TypeScript

function maxTaxiEarnings(n: number, rides: number[][]): number {
    rides.sort((a, b) => a[1] - b[1]);
    const m = rides.length;
    const f: number[] = Array(m + 1).fill(0);
    const search = (x: number, r: number): number => {
        let l = 0;
        while (l < r) {
            const mid = (l + r) >> 1;
            if (rides[mid][1] >= x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    };
    for (let i = 1; i <= m; ++i) {
        const [st, ed, tip] = rides[i - 1];
        const j = search(st + 1, i);
        f[i] = Math.max(f[i - 1], f[j] + ed - st + tip);
    }
    return f[m];
}