Time Needed to Buy Tickets - Leetcode 2073 - Python

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Problem Link: leetcode.com/problems/time-ne...
0:00 - Read the problem
0:19 - Drawing Explanation
9:56 - Coding Explanation
leetcode 2073
#neetcode #leetcode #python

[@yang5843]

The problem had a solution that was more optimal than the simulation so I knew you were going to make a video.
Here's the java solution
class Solution {
public int timeRequiredToBuy(int[] tickets, int k) {
int rc = 0;
for (int i=0;i

[@user-vu4ng4rb8k]

please make a playlist for leetcode database problems

[@indianundergraddiaries]

Here's my Python code, I feel this maybe more intuitive and straight forward
class Solution:
def timeRequiredToBuy(self, tickets: List[int], k: int) -> int:
count = 0
while True:
for i in range(len(tickets)):
if tickets[k] == 0:
return count
elif tickets[i] == 0:
continue
tickets[i] -= 1
count += 1

[@AbhijeetMuneshwar]

Thanks for this easy explaination 🙏🏼

[@MP-ny3ep]

Great explanation as always. Thank you

[@rauffares3558]

Great Explanation

[@yuriaragao1910]

Pretty neat solution! I solved it with a different logic, maybe yours is more straightforward, but I will share here anyways:
The idea is to start res as the length of the queue multiplied by the number of tickets the kth person wants to buy (this is the upper bound of the result), then we traverse the tickets array decreasing res checking two conditions:
1. If tickets[i] is smaller than tickets[k] then decrease res by their difference (we counted tickets[i] too many times)
2. if i > k and tickets[i] >= tickets[k] decrease the result by 1, since the ith person is behind the kth person in the queue
Here is the solution in Python:
class Solution:
def timeRequiredToBuy(self, tickets: List[int], k: int) -> int:
res = len(tickets) * tickets[k]
for i in range(len(tickets)):
if tickets[i] < tickets[k]:
res += tickets[i] - tickets[k]
elif i > k:
res -= 1
return res

[@dilmurodabdusamadov5572]

I tried to come up with mathy approach but it didn't fit well :)
class Solution:
def timeRequiredToBuy(self, tickets: List[int], k: int) -> int:
more, less = 0, 0
for i in range(len(tickets)):
if tickets[i] < tickets[k]:
less += tickets[i]
else:
more += 1
return tickets[k] * more + less
Thank you for the explanation tho!

[@adrianharo6586]

I just did this on Java. Using a while and nested for loop

[@comatosesperrow]

I feel like there's a O(1) math solution somewhere here. Any ideas?

[@DebopriyoBasu]

Great explanation!
Here is my JavaScript solution using the same logic:
const timeRequiredToBuy = (tickets, k) => {
let time = 0;
const len = tickets.length;
for (let i = 0; i < len; i++) {
if (i

[@CrabGuyy]

amazing, in javascript you could also do it in a more functional way with the "reduce" function of arrays, if you didn't know i would suggest you look into it since its pretty useful for this case

[@deathbombs]

I solved 450 qs, couldn't optimize past simulation. Have tips or im just dumb?

[@yang5843]

Reflect on the problem, every person in front of k has to be able to buy student[k] tickets, and every person after k must be able to buy student[k] - 1 tickets.

[@dragon_id_]

did every1 use "nums" instead of tickets at first !!

[@servantofthelord8147]

same lol

[@s8x.]

how did u learn everything

[@spsc07]

someone already said life lessons and hashmap

[@s8x.]

@@spsc07 neet

[@user-zj8vp3no5l]

Here is my solution in java : class Solution {
public int timeRequiredToBuy(int[] tickets, int k) {
int l=tickets.length;
int val=tickets[k];
int time=0;
for(int i=0;i

[@ayushw]

JS One-liner:
var timeRequiredToBuy = function(tickets, k) {
return tickets.reduce((acc, cur, i) => acc + Math.min(cur, tickets[k] - (i > k)), 0);
};