Skip to content

feat: update solutions to lc problem: No.2073 #3209

New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

Sign up for GitHub

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

Merged
merged 1 commit into from
Jul 6, 2024
Merged

feat: update solutions to lc problem: No.2073 #3209

Show file tree
Hide file tree
Changes from all commits
Commits
Show all changes
1 commit
Select commit
File filter

Filter by extension

Filter by extension
Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
65 changes: 22 additions & 43 deletions solution/2000-2099/2073.Time Needed to Buy Tickets/README.md
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -67,7 +67,13 @@ tags:

<!-- solution:start -->

### 方法一
### 方法一：一次遍历

根据题目描述，当第 $k$ 个人完成购票时，在第 $k$ 个人前面的所有人，购买的票数都不会超过第 $k$ 个人购买的票数，而在第 $k$ 个人后面的所有人，购买的票数都不会超过第 $k$ 个人购买的票数减 $1$。

因此，我们可以遍历整个队伍，对于第 $i$ 个人，如果 $i \leq k$，购票时间为 $\min(\text{tickets}[i], \text{tickets}[k])$，否则购票时间为 $\min(\text{tickets}[i], \text{tickets}[k] - 1)$。我们将所有人的购票时间相加即可。

时间复杂度 $O(n)$，其中 $n$ 为队伍的长度。空间复杂度 $O(1)$。

<!-- tabs:start -->

Expand All @@ -77,11 +83,8 @@ tags:
class Solution:
def timeRequiredToBuy(self, tickets: List[int], k: int) -> int:
ans = 0
for i, t in enumerate(tickets):
if i <= k:
ans += min(tickets[k], t)
else:
ans += min(tickets[k] - 1, t)
for i, x in enumerate(tickets):
ans += min(x, tickets[k] if i <= k else tickets[k] - 1)
return ans
```

Expand All @@ -91,12 +94,8 @@ class Solution:
class Solution {
public int timeRequiredToBuy(int[] tickets, int k) {
int ans = 0;
for (int i = 0; i < tickets.length; i++) {
if (i <= k) {
ans += Math.min(tickets[k], tickets[i]);
} else {
ans += Math.min(tickets[k] - 1, tickets[i]);
}
for (int i = 0; i < tickets.length; ++i) {
ans += Math.min(tickets[i], i <= k ? tickets[k] : tickets[k] - 1);
}
return ans;
}
Expand All @@ -111,11 +110,7 @@ public:
int timeRequiredToBuy(vector<int>& tickets, int k) {
int ans = 0;
for (int i = 0; i < tickets.size(); ++i) {
if (i <= k) {
ans += min(tickets[k], tickets[i]);
} else {
ans += min(tickets[k] - 1, tickets[i]);
}
ans += min(tickets[i], i <= k ? tickets[k] : tickets[k] - 1);
}
return ans;
}
Expand All @@ -125,42 +120,26 @@ public:
#### Go

```go
func timeRequiredToBuy(tickets []int, k int) int {
ans := 0
for i, t := range tickets {
if i <= k {
ans += min(tickets[k], t)
} else {
ans += min(tickets[k]-1, t)
func timeRequiredToBuy(tickets []int, k int) (ans int) {
for i, x := range tickets {
t := tickets[k]
if i > k {
t--
}
ans += min(x, t)
}
return ans
return
}
```

#### TypeScript

```ts
function timeRequiredToBuy(tickets: number[], k: number): number {
const n = tickets.length;
let target = tickets[k] - 1;
let ans = 0;
// round1
for (let i = 0; i < n; i++) {
let num = tickets[i];
if (num <= target) {
ans += num;
tickets[i] = 0;
} else {
ans += target;
tickets[i] -= target;
}
}

// round2
for (let i = 0; i <= k; i++) {
let num = tickets[i];
ans += num > 0 ? 1 : 0;
const n = tickets.length;
for (let i = 0; i < n; ++i) {
ans += Math.min(tickets[i], i <= k ? tickets[k] : tickets[k] - 1);
}
return ans;
}
Expand Down
65 changes: 22 additions & 43 deletions solution/2000-2099/2073.Time Needed to Buy Tickets/README_EN.md
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -67,7 +67,13 @@ The person at&nbsp;position 0 has successfully bought 5 tickets and it took 4 +

<!-- solution:start -->

### Solution 1
### Solution 1: Single Pass

According to the problem description, when the $k^{th}$ person finishes buying tickets, all the people in front of the $k^{th}$ person will not buy more tickets than the $k^{th}$ person, and all the people behind the $k^{th}$ person will not buy more tickets than the $k^{th}$ person minus $1$.

Therefore, we can traverse the entire queue. For the $i^{th}$ person, if $i \leq k$, the time to buy tickets is $\min(\text{tickets}[i], \text{tickets}[k])$; otherwise, the time to buy tickets is $\min(\text{tickets}[i], \text{tickets}[k] - 1)$. We sum the buying time for all people to get the result.

The time complexity is $O(n)$, where $n$ is the length of the queue. The space complexity is $O(1)$.

<!-- tabs:start -->

Expand All @@ -77,11 +83,8 @@ The person at&nbsp;position 0 has successfully bought 5 tickets and it took 4 +
class Solution:
def timeRequiredToBuy(self, tickets: List[int], k: int) -> int:
ans = 0
for i, t in enumerate(tickets):
if i <= k:
ans += min(tickets[k], t)
else:
ans += min(tickets[k] - 1, t)
for i, x in enumerate(tickets):
ans += min(x, tickets[k] if i <= k else tickets[k] - 1)
return ans
```

Expand All @@ -91,12 +94,8 @@ class Solution:
class Solution {
public int timeRequiredToBuy(int[] tickets, int k) {
int ans = 0;
for (int i = 0; i < tickets.length; i++) {
if (i <= k) {
ans += Math.min(tickets[k], tickets[i]);
} else {
ans += Math.min(tickets[k] - 1, tickets[i]);
}
for (int i = 0; i < tickets.length; ++i) {
ans += Math.min(tickets[i], i <= k ? tickets[k] : tickets[k] - 1);
}
return ans;
}
Expand All @@ -111,11 +110,7 @@ public:
int timeRequiredToBuy(vector<int>& tickets, int k) {
int ans = 0;
for (int i = 0; i < tickets.size(); ++i) {
if (i <= k) {
ans += min(tickets[k], tickets[i]);
} else {
ans += min(tickets[k] - 1, tickets[i]);
}
ans += min(tickets[i], i <= k ? tickets[k] : tickets[k] - 1);
}
return ans;
}
Expand All @@ -125,42 +120,26 @@ public:
#### Go

```go
func timeRequiredToBuy(tickets []int, k int) int {
ans := 0
for i, t := range tickets {
if i <= k {
ans += min(tickets[k], t)
} else {
ans += min(tickets[k]-1, t)
func timeRequiredToBuy(tickets []int, k int) (ans int) {
for i, x := range tickets {
t := tickets[k]
if i > k {
t--
}
ans += min(x, t)
}
return ans
return
}
```

#### TypeScript

```ts
function timeRequiredToBuy(tickets: number[], k: number): number {
const n = tickets.length;
let target = tickets[k] - 1;
let ans = 0;
// round1
for (let i = 0; i < n; i++) {
let num = tickets[i];
if (num <= target) {
ans += num;
tickets[i] = 0;
} else {
ans += target;
tickets[i] -= target;
}
}

// round2
for (let i = 0; i <= k; i++) {
let num = tickets[i];
ans += num > 0 ? 1 : 0;
const n = tickets.length;
for (let i = 0; i < n; ++i) {
ans += Math.min(tickets[i], i <= k ? tickets[k] : tickets[k] - 1);
}
return ans;
}
Expand Down
6 changes: 1 addition & 5 deletions solution/2000-2099/2073.Time Needed to Buy Tickets/Solution.cpp
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -3,11 +3,7 @@ class Solution {
int timeRequiredToBuy(vector<int>& tickets, int k) {
int ans = 0;
for (int i = 0; i < tickets.size(); ++i) {
if (i <= k) {
ans += min(tickets[k], tickets[i]);
} else {
ans += min(tickets[k] - 1, tickets[i]);
}
ans += min(tickets[i], i <= k ? tickets[k] : tickets[k] - 1);
}
return ans;
}
Expand Down
15 changes: 7 additions & 8 deletions solution/2000-2099/2073.Time Needed to Buy Tickets/Solution.go
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,11 +1,10 @@
func timeRequiredToBuy(tickets []int, k int) int {
ans := 0
for i, t := range tickets {
if i <= k {
ans += min(tickets[k], t)
} else {
ans += min(tickets[k]-1, t)
func timeRequiredToBuy(tickets []int, k int) (ans int) {
for i, x := range tickets {
t := tickets[k]
if i > k {
t--
}
ans += min(x, t)
}
return ans
return
}
8 changes: 2 additions & 6 deletions solution/2000-2099/2073.Time Needed to Buy Tickets/Solution.java
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,12 +1,8 @@
class Solution {
public int timeRequiredToBuy(int[] tickets, int k) {
int ans = 0;
for (int i = 0; i < tickets.length; i++) {
if (i <= k) {
ans += Math.min(tickets[k], tickets[i]);
} else {
ans += Math.min(tickets[k] - 1, tickets[i]);
}
for (int i = 0; i < tickets.length; ++i) {
ans += Math.min(tickets[i], i <= k ? tickets[k] : tickets[k] - 1);
}
return ans;
}
Expand Down
7 changes: 2 additions & 5 deletions solution/2000-2099/2073.Time Needed to Buy Tickets/Solution.py
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,9 +1,6 @@
class Solution:
def timeRequiredToBuy(self, tickets: List[int], k: int) -> int:
ans = 0
for i, t in enumerate(tickets):
if i <= k:
ans += min(tickets[k], t)
else:
ans += min(tickets[k] - 1, t)
for i, x in enumerate(tickets):
ans += min(x, tickets[k] if i <= k else tickets[k] - 1)
return ans
21 changes: 3 additions & 18 deletions solution/2000-2099/2073.Time Needed to Buy Tickets/Solution.ts
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -1,23 +1,8 @@
function timeRequiredToBuy(tickets: number[], k: number): number {
const n = tickets.length;
let target = tickets[k] - 1;
let ans = 0;
// round1
for (let i = 0; i < n; i++) {
let num = tickets[i];
if (num <= target) {
ans += num;
tickets[i] = 0;
} else {
ans += target;
tickets[i] -= target;
}
}

// round2
for (let i = 0; i <= k; i++) {
let num = tickets[i];
ans += num > 0 ? 1 : 0;
const n = tickets.length;
for (let i = 0; i < n; ++i) {
ans += Math.min(tickets[i], i <= k ? tickets[k] : tickets[k] - 1);
}
return ans;
}
Loading