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花花酱 LeetCode 1642. Furthest Building You Can Reach

You are given an integer array heights representing the heights of buildings, some bricks, and some ladders.

You start your journey from building 0 and move to the next building by possibly using bricks or ladders.

While moving from building i to building i+1 (0-indexed),

  • If the current building’s height is greater than or equal to the next building’s height, you do not need a ladder or bricks.
  • If the current building’s height is less than the next building’s height, you can either use one ladder or (h[i+1] - h[i]) bricks.

Return the furthest building index (0-indexed) you can reach if you use the given ladders and bricks optimally.

Example 1:

Input: heights = [4,2,7,6,9,14,12], bricks = 5, ladders = 1
Output: 4
Explanation: Starting at building 0, you can follow these steps:
- Go to building 1 without using ladders nor bricks since 4 >= 2.
- Go to building 2 using 5 bricks. You must use either bricks or ladders because 2 < 7.
- Go to building 3 without using ladders nor bricks since 7 >= 6.
- Go to building 4 using your only ladder. You must use either bricks or ladders because 6 < 9.
It is impossible to go beyond building 4 because you do not have any more bricks or ladders.

Example 2:

Input: heights = [4,12,2,7,3,18,20,3,19], bricks = 10, ladders = 2
Output: 7

Example 3:

Input: heights = [14,3,19,3], bricks = 17, ladders = 0
Output: 3


  • 1 <= heights.length <= 105
  • 1 <= heights[i] <= 106
  • 0 <= bricks <= 109
  • 0 <= ladders <= heights.length

Solution 0: DFS

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

AC but should be TLE

Solution 1: Binary Search + Greedy

Guess we can reach to m, sort the height differences from 0~m. Use ladders for larger values and use bricks for smallest values left.

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


Solution 2: Min heap

Use a min heap to store all the height differences ( > 0) so far, if heap size is greater than ladders, which means we have to use bricks, extract the smallest value and subtract the bricks.

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


花花酱 LeetCode 1094. Car Pooling

You are driving a vehicle that has capacity empty seats initially available for passengers.  The vehicle only drives east (ie. it cannot turn around and drive west.)

Given a list of tripstrip[i] = [num_passengers, start_location, end_location] contains information about the i-th trip: the number of passengers that must be picked up, and the locations to pick them up and drop them off.  The locations are given as the number of kilometers due east from your vehicle’s initial location.

Return true if and only if it is possible to pick up and drop off all passengers for all the given trips. 

Example 1:

Input: trips = [[2,1,5],[3,3,7]], capacity = 4
Output: false

Example 2:

Input: trips = [[2,1,5],[3,3,7]], capacity = 5
Output: true

Example 3:

Input: trips = [[2,1,5],[3,5,7]], capacity = 3
Output: true

Example 4:

Input: trips = [[3,2,7],[3,7,9],[8,3,9]], capacity = 11
Output: true

Solution1: Min heap

Sort events by location

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


Solution 2: Preprocessing

Time complexity: O(n)
Space complexity: O(1000)


花花酱 LeetCode 778. Swim in Rising Water

On an N x N grid, each square grid[i][j] represents the elevation at that point (i,j).

Now rain starts to fall. At time t, the depth of the water everywhere is t. You can swim from a square to another 4-directionally adjacent square if and only if the elevation of both squares individually are at most t. You can swim infinite distance in zero time. Of course, you must stay within the boundaries of the grid during your swim.

You start at the top left square (0, 0). What is the least time until you can reach the bottom right square (N-1, N-1)?

Example 1:

Input: [[0,2],[1,3]]
Output: 3
At time 0, you are in grid location (0, 0).
You cannot go anywhere else because 4-directionally adjacent neighbors have a higher elevation than t = 0.

You cannot reach point (1, 1) until time 3.
When the depth of water is 3, we can swim anywhere inside the grid.

Example 2:

Input: [[0,1,2,3,4],[24,23,22,21,5],[12,13,14,15,16],[11,17,18,19,20],[10,9,8,7,6]]
Output: 16
 0  1  2  3  4
24 23 22 21  5
12 13 14 15 16
11 17 18 19 20
10  9  8  7  6

The final route is marked in bold.
We need to wait until time 16 so that (0, 0) and (4, 4) are connected.


  1. 2 <= N <= 50.
  2. grid[i][j] is a permutation of [0, …, N*N – 1].

Solution 1: Dijkstra’s Algorithm

Time complexity: O(n^2*logn)
Space complexity: O(n^2)


Solution 2: Binary Search + BFS

Time complexity: O(2logn * n^2)
Space complexity: O(n^2)


花花酱 LeetCode 703. Kth Largest Element in a Stream


Design a class to find the kth largest element in a stream. Note that it is the kth largest element in the sorted order, not the kth distinct element.

Your KthLargest class will have a constructor which accepts an integer k and an integer array nums, which contains initial elements from the stream. For each call to the method KthLargest.add, return the element representing the kth largest element in the stream.


int k = 3;
int[] arr = [4,5,8,2];
KthLargest kthLargest = new KthLargest(3, arr);
kthLargest.add(3);   // returns 4
kthLargest.add(5);   // returns 5
kthLargest.add(10);  // returns 5
kthLargest.add(9);   // returns 8
kthLargest.add(4);   // returns 8

You may assume that nums‘ length ≥ k-1 and k ≥ 1.

Solution: BST / Min Heap

Time complexity: O(nlogk)

Space complexity: O(k)

C++ / BST

C++ / Min Heap



花花酱 LeetCode 239. Sliding Window Maximum



Given an array nums, there is a sliding window of size k which is moving from the very left of the array to the very right. You can only see the k numbers in the window. Each time the sliding window moves right by one position.

For example,
Given nums = [1,3,-1,-3,5,3,6,7], and k = 3.

Therefore, return the max sliding window as [3,3,5,5,6,7].

You may assume k is always valid, ie: 1 ≤ k ≤ input array’s size for non-empty array.

Follow up:
Could you solve it in linear time?




Solution 1: Brute Force

Time complexity: O((n – k + 1) * k)

Space complexity: O(1)




Solution 2: BST

Time complexity: O((n – k + 1) * logk)

Space complexity: O(k)


Solution 3: Monotonic Queue

Time complexity: O(n)

Space complexity: O(k)


C++ V2

C++ V3



Python3 V2