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Posts tagged as “simulation”

花花酱 LeetCode 1496. Path Crossing

Given a string path, where path[i] = 'N''S''E' or 'W', each representing moving one unit north, south, east, or west, respectively. You start at the origin (0, 0) on a 2D plane and walk on the path specified by path.

Return True if the path crosses itself at any point, that is, if at any time you are on a location you’ve previously visited. Return False otherwise.

Example 1:

Input: path = "NES"
Output: false 
Explanation: Notice that the path doesn't cross any point more than once.

Example 2:

Input: path = "NESWW"
Output: true
Explanation: Notice that the path visits the origin twice.

Constraints:

  • 1 <= path.length <= 10^4
  • path will only consist of characters in {'N', 'S', 'E', 'W}

Solution: Simulation + Hashtable

Time complexity: O(n)
Space complexity: O(n)

C++

花花酱 LeetCode 1486. XOR Operation in an Array

Given an integer n and an integer start.

Define an array nums where nums[i] = start + 2*i (0-indexed) and n == nums.length.

Return the bitwise XOR of all elements of nums.

Example 1:

Input: n = 5, start = 0
Output: 8
Explanation: Array nums is equal to [0, 2, 4, 6, 8] where (0 ^ 2 ^ 4 ^ 6 ^ 8) = 8.
Where "^" corresponds to bitwise XOR operator.

Example 2:

Input: n = 4, start = 3
Output: 8
Explanation: Array nums is equal to [3, 5, 7, 9] where (3 ^ 5 ^ 7 ^ 9) = 8.

Example 3:

Input: n = 1, start = 7
Output: 7

Example 4:

Input: n = 10, start = 5
Output: 2

Constraints:

  • 1 <= n <= 1000
  • 0 <= start <= 1000
  • n == nums.length

Solution: Simulation

Time complexity: O(n)
Space complexity: O(1)

C++

花花酱 LeetCode 1475. Final Prices With a Special Discount in a Shop

Given the array prices where prices[i] is the price of the ith item in a shop. There is a special discount for items in the shop, if you buy the ith item, then you will receive a discount equivalent to prices[j] where j is the minimum index such that j > i and prices[j] <= prices[i], otherwise, you will not receive any discount at all.

Return an array where the ith element is the final price you will pay for the ith item of the shop considering the special discount.

Example 1:

Input: prices = [8,4,6,2,3]
Output: [4,2,4,2,3]
Explanation: 
For item 0 with price[0]=8 you will receive a discount equivalent to prices[1]=4, therefore, the final price you will pay is 8 - 4 = 4. 
For item 1 with price[1]=4 you will receive a discount equivalent to prices[3]=2, therefore, the final price you will pay is 4 - 2 = 2. 
For item 2 with price[2]=6 you will receive a discount equivalent to prices[3]=2, therefore, the final price you will pay is 6 - 2 = 4. 
For items 3 and 4 you will not receive any discount at all.

Example 2:

Input: prices = [1,2,3,4,5]
Output: [1,2,3,4,5]
Explanation: In this case, for all items, you will not receive any discount at all.

Example 3:

Input: prices = [10,1,1,6]
Output: [9,0,1,6]

Constraints:

  • 1 <= prices.length <= 500
  • 1 <= prices[i] <= 10^3

Solution 1: Simulation

Time complexity: O(n^2)
Space complexity: O(1)

C++

Solution 2: Monotonic Stack

Use a stack to store monotonically increasing items, when the current item is cheaper than the top of the stack, we get the discount and pop that item. Repeat until the current item is no longer cheaper or the stack becomes empty.

Time complexity: O(n)
Space complexity: O(n)

C++

index version

C++

花花酱 LeetCode 1472. Design Browser History

You have a browser of one tab where you start on the homepage and you can visit another url, get back in the history number of steps or move forward in the history number of steps.

Implement the BrowserHistory class:

  • BrowserHistory(string homepage) Initializes the object with the homepage of the browser.
  • void visit(string url) visits url from the current page. It clears up all the forward history.
  • string back(int steps) Move steps back in history. If you can only return x steps in the history and steps > x, you will return only x steps. Return the current url after moving back in history at most steps.
  • string forward(int steps) Move steps forward in history. If you can only forward x steps in the history and steps > x, you will forward only x steps. Return the current url after forwarding in history at most steps.

Example:

Input:
["BrowserHistory","visit","visit","visit","back","back","forward","visit","forward","back","back"]
[["leetcode.com"],["google.com"],["facebook.com"],["youtube.com"],[1],[1],[1],["linkedin.com"],[2],[2],[7]]
Output:
[null,null,null,null,"facebook.com","google.com","facebook.com",null,"linkedin.com","google.com","leetcode.com"]

Explanation: BrowserHistory browserHistory = new BrowserHistory("leetcode.com"); browserHistory.visit("google.com"); // You are in "leetcode.com". Visit "google.com" browserHistory.visit("facebook.com"); // You are in "google.com". Visit "facebook.com" browserHistory.visit("youtube.com"); // You are in "facebook.com". Visit "youtube.com" browserHistory.back(1); // You are in "youtube.com", move back to "facebook.com" return "facebook.com" browserHistory.back(1); // You are in "facebook.com", move back to "google.com" return "google.com" browserHistory.forward(1); // You are in "google.com", move forward to "facebook.com" return "facebook.com" browserHistory.visit("linkedin.com"); // You are in "facebook.com". Visit "linkedin.com" browserHistory.forward(2); // You are in "linkedin.com", you cannot move forward any steps. browserHistory.back(2); // You are in "linkedin.com", move back two steps to "facebook.com" then to "google.com". return "google.com" browserHistory.back(7); // You are in "google.com", you can move back only one step to "leetcode.com". return "leetcode.com"

Constraints:

  • 1 <= homepage.length <= 20
  • 1 <= url.length <= 20
  • 1 <= steps <= 100
  • homepage and url consist of  ‘.’ or lower case English letters.
  • At most 5000 calls will be made to visitback, and forward.

Solution: Vector

Time complexity:
visit: Amortized O(1)
back: O(1)
forward: O(1)

C++

花花酱 LeetCode 1470. Shuffle the Array

Given the array nums consisting of 2n elements in the form [x1,x2,...,xn,y1,y2,...,yn].

Return the array in the form [x1,y1,x2,y2,...,xn,yn].

Example 1:

Input: nums = [2,5,1,3,4,7], n = 3
Output: [2,3,5,4,1,7] 
Explanation: Since x1=2, x2=5, x3=1, y1=3, y2=4, y3=7 then the answer is [2,3,5,4,1,7].

Example 2:

Input: nums = [1,2,3,4,4,3,2,1], n = 4
Output: [1,4,2,3,3,2,4,1]

Example 3:

Input: nums = [1,1,2,2], n = 2
Output: [1,2,1,2]

Constraints:

  • 1 <= n <= 500
  • nums.length == 2n
  • 1 <= nums[i] <= 10^3

Solution: Simulation

Time complexity: O(n)
Space complexity: O(n)

C++