Time Needed to Buy Tickets - Leetcode 2073 - 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

[@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

[@Logeshwar-s7m]

please make a playlist for leetcode database problems

[@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!

[@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

[@rauffares3558]

Great Explanation

[@AbhijeetMuneshwar]

Thanks for this easy explaination 🙏🏼

[@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

[@MP-ny3ep]

Great explanation as always. Thank you

[@deepakkanshetti1413]

Really neat solution 👍

[@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?

[@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.

[@s8x.]

how did u learn everything

[@spsc07]

someone already said life lessons and hashmap

[@s8x.]

@@spsc07 neet

[@MaazMalik-c9o]

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

[@dragon_id_]

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

[@servantofthelord8147]

same lol

[@ayushw]

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