Time Needed to Buy Tickets
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Problem Statement Solution 1: Brute Force Approach Solution 2: Direct Time Calculation with Conditional Adjustments Solution 3: Circular Queue Simulation for Ticket Purchasing Solution 4: Circular Queue Simulation with Index Reset Solution 5: Queue - Index & Value Appending Solution 6: Simulation with Solution 7: Optimized Time Calculation with Conditional Adjustments Solution 8: Queue-Based Simulation with Ticket Decrement" More Problems Solutions
Problem Statement
In a bustling queue for tickets, there are n people standing in line, each holding a different number of tickets they wish to purchase. The queue operates in a sequential manner where the 0th person is at the front and the (n - 1)th person is at the end.
Each person can only buy one ticket per second and, if they still have tickets left to buy, they return to the end of the line instantly. Once a person has bought all their tickets, they leave the queue.
Given an array tickets where tickets[i] represents the number of tickets the ith person wants to buy, and an integer k indicating the position of a specific person in the queue (0-indexed), your task is to determine how much time it will take for the person at position k to finish buying all their tickets.
Example 1
Input: tickets = [2,3,2] , k = 2
Output: 6
Example 2
Input: tickets = [5,1,1,1] , k = 0
Output: 8
Solution 1: Brute Force Approach
function timeRequiredToBuyBruteForce(tickets, k) {
let time = 0;
while (tickets[k] > 0) {
for (let i = 0; i < tickets.length; i++) {
if (tickets[k] === 0) break;
if (tickets[i] > 0) {
tickets[i]--;
time++;
}
}
}
return time;
}
const tickets1 = [2,3,2];
const k1 = 2;
timeRequiredToBuyBruteForce(tickets1,k1); //output: 6
const tickets2 = [5,1,1,1];
const k2 = 0;
timeRequiredToBuyBruteForce(tickets2,k2); //output: 8
Solution 2: Direct Time Calculation with Conditional Adjustments
function timeRequiredToBuyConditionalArgs(tickets, k) {
let res = 0;
let target = tickets[k];
for (let i = 0; i < tickets.length; i++) {
if (i <= k) {
res += Math.min(tickets[i], target);
} else {
if (target <= tickets[i]) {
res += Math.min(tickets[i], target) - 1;
} else {
res += Math.min(tickets[i], target);
}
}
}
return res;
}
const tickets1 = [2,3,2];
const k1 = 2;
timeRequiredToBuyConditionalArgs(tickets1,k1); //output: 6
const tickets2 = [5,1,1,1];
const k2 = 0;
timeRequiredToBuyConditionalArgs(tickets2,k2); //output: 8
Solution 3: Circular Queue Simulation for Ticket Purchasing
function timeRequiredToBuyCircularQueue(tickets, k) {
let sec = 0;
let i = 0;
while (tickets[k] > 0) {
if (i === k) {
tickets[i]--;
sec++;
i++;
} else if (i < tickets.length && tickets[i] === 0 && i !== k) {
i++;
continue;
} else if (tickets[k] > 0 && i < tickets.length && tickets[i] > 0 && i !== k) {
tickets[i]--;
sec++;
i++;
}
if (i === tickets.length) {
i = 0;
}
}
return sec;
}
const tickets1 = [2,3,2];
const k1 = 2;
timeRequiredToBuyCircularQueue(tickets1,k1); //output: 6
const tickets2 = [5,1,1,1];
const k2 = 0;
timeRequiredToBuyCircularQueue(tickets2,k2); //output: 8
Solution 4: Circular Queue Simulation with Index Reset
function timeRequiredToBuyCircularResetIndex(tickets, k) {
let time = 0;
let i = 0;
while (tickets[k] > 0) {
if (tickets[i] > 0) {
tickets[i]--;
time++;
}
i++;
if (i === tickets.length) {
i = 0;
}
}
return time;
}
const tickets1 = [2,3,2];
const k1 = 2;
timeRequiredToBuyCircularResetIndex(tickets1,k1); //output: 6
const tickets2 = [5,1,1,1];
const k2 = 0;
timeRequiredToBuyCircularResetIndex(tickets2,k2); //output: 8
Solution 5: Queue - Index & Value Appending
function timeRequiredToBuyIndexAppending(tickets, k) {
const queue = [];
for (let i = 0; i < tickets.length; i++) {
queue.push([i, tickets[i]]);
}
let timeTaken = 0;
while (queue.length > 0) {
const [index, numTickets] = queue.shift();
if (numTickets > 0) {
queue.push([index, numTickets - 1]);
timeTaken++;
}
if (index === k && numTickets - 1 === 0) {
break;
}
}
return timeTaken;
}
const tickets1 = [2,3,2];
const k1 = 2;
timeRequiredToBuyIndexAppending(tickets1,k1); //output: 6
const tickets2 = [5,1,1,1];
const k2 = 0;
timeRequiredToBuyIndexAppending(tickets2,k2); //output: 8
Solution 6: Simulation with
function timeRequiredToBuySequentialEarlyTermination (tickets, k) {
let count = 0;
while (tickets[k] > 0) {
for (let i = 0; i < tickets.length; i++) {
if (tickets[k] === 0) break;
if (tickets[i] === 0) continue;
tickets[i]--;
count++;
}
}
return count;
}
const tickets1 = [2,3,2];
const k1 = 2;
timeRequiredToBuySequentialEarlyTermination(tickets1,k1); //output: 6
const tickets2 = [5,1,1,1];
const k2 = 0;
timeRequiredToBuySequentialEarlyTermination(tickets2,k2); //output: 8
Solution 7: Optimized Time Calculation with Conditional Adjustments
function timeRequiredToBuyConditionalAdjustments (tickets, k) {
let time = 0;
const n = tickets.length;
const tk = tickets[k];
for (let i = 0; i < n; i++) {
if (tickets[i] >= tk) {
if (i > k) {
time += tk - 1;
} else {
time += tk;
}
} else {
time += tickets[i];
}
if (tickets[k] === 0) break;
}
return time;
}
const tickets1 = [2,3,2];
const k1 = 2;
timeRequiredToBuyConditionalAdjustments(tickets1,k1); //output: 6
const tickets2 = [5,1,1,1];
const k2 = 0;
timeRequiredToBuyConditionalAdjustments(tickets2,k2); //output: 8
Solution 8: Queue-Based Simulation with Ticket Decrement"
function timeRequiredToBuyTicketDecrement (tickets, k) {
const queue = [];
// Initialize the queue with each person's ticket count and their index
for (let i = 0; i < tickets.length; i++) {
queue.push([tickets[i], i]);
}
let timeTaken = 0;
while (queue.length > 0) {
timeTaken++;
const [ticketCount, index] = queue.shift();
if (ticketCount - 1 <= 0 && index === k) {
return timeTaken;
}
if (ticketCount - 1 > 0) {
queue.push([ticketCount - 1, index]);
}
}
return -1; // This line will never be reached with the given problem constraints
}
const tickets1 = [2,3,2];
const k1 = 2;
timeRequiredToBuyTicketDecrement(tickets1,k1); //output: 6
const tickets2 = [5,1,1,1];
const k2 = 0;
timeRequiredToBuyTicketDecrement(tickets2,k2); //output: 8