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 <= 1051 <= rides.length <= 3 * 104rides[i].length == 31 <= starti < endi <= n1 <= 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 $rides$.
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| 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)
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| 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;
}
}
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| 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);
}
};
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| 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)
}
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| 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);
}
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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:
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| 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]
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| 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;
}
}
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| 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();
}
};
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| 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]
}
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| 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];
}
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