Given an array of points where points[i] = [x<sub>i</sub>, y<sub>i</sub>] represents a point on the X-Y plane and an integer k, return the k closest points to the origin (0, 0).
The distance between two points on the X-Y plane is the Euclidean distance (i.e., √(x<sub>1</sub> - x<sub>2</sub>)<sup>2</sup> + (y<sub>1</sub> - y<sub>2</sub>)<sup>2</sup>).
You may return the answer in any order. The answer is guaranteed to be unique (except for the order that it is in).
Example 1:
Input: points = [[1,3],[-2,2]], k = 1 Output: [[-2,2]] Explanation: The distance between (1, 3) and the origin is sqrt(10). The distance between (-2, 2) and the origin is sqrt(8). Since sqrt(8) < sqrt(10), (-2, 2) is closer to the origin. We only want the closest k = 1 points from the origin, so the answer is just [[-2,2]].
Example 2:
Input: points = [[3,3],[5,-1],[-2,4]], k = 2 Output: [[3,3],[-2,4]] Explanation: The answer [[-2,4],[3,3]] would also be accepted.
defspiralOrder(self, matrix: List[List[int]]) -> List[int]: ifnot matrix ornot matrix[0]: return [] res = [] top, bottom, left, right = 0, len(matrix) - 1, 0, len(matrix[0]) - 1 while top <= bottom and left <= right: for i inrange(left, right + 1): res.append(matrix[top][i]) top += 1 for i inrange(top, bottom + 1): res.append(matrix[i][right]) # remember [i[[right] not [right][i] right -= 1 if top <= bottom: for i inrange(right, left - 1, -1): res.append(matrix[bottom][i]) bottom -= 1 if left <= right: for i inrange(bottom, top - 1, -1): res.append(matrix[i][left]) left += 1 return res
Say you have an array for which the ith element is the price of a given stock on day i.
If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.
Example 1:
Input: [7, 1, 5, 3, 6, 4]
Output: 5
max. difference = 6-1 = 5 (not 7-1 = 6, as selling price needs to be larger than buying price)
Example 2:
Input: [7, 6, 4, 3, 1]
Output: 0
In this case, no transaction is done, i.e. max profit = 0.
In an infinite chess board with coordinates from -infinity to +infinity, you have a knight at square [0, 0].
A knight has 8 possible moves it can make, as illustrated below. Each move is two squares in a cardinal direction, then one square in an orthogonal direction.
Return the minimum number of steps needed to move the knight to the square [x, y]. It is guaranteed the answer exists.
Example 1:
Input: x = 2, y = 1
Output: 1
Explanation: [0, 0] → [2, 1]
Because x and y are constrained to be in range[-300, 300], we can use BFS to find the minimum steps needed to reach target(x, y). Furthermore, we can only consider the case that x >=0 && y >=0 since the chess board is symmetric. The bfs implementation is pretty straightforward. There are two important points you need to be careful with.
Pruning. We can limit the search dimension within 310 * 310. Any moves that lead to a position that is outside this box will not yield an optimal result.
2. Initially, you used a Set of type int[] to track visited positions. This caused TLE because you didn’t overwrite the hashCode and equals methods for int[]. As a result, Set uses the default hashCode and equals method when checking if an element is already in the set. For equals(), The default implementation provided by the JDK is based on memory location — two objects are equal if and only if they are stored in the same memory address. For a comprehensive reading, refer to https://dzone.com/articles/working-with-hashcode-and-equals-in-java
for direction in directions: neighbor = (node[0] + direction[0], node[1] + direction[1]) if neighbor in visited: continue queue.append(neighbor) visited.add(neighbor) distance[neighbor] = distance[node] + 1