Magical Python: I thought I knew something about Python, and then realized I know nothing about Python at all.
There are N students in a class. Some of them are friends, while some are not. Their friendship is transitive in nature. For example, if A is a direct friend of B, and B is a direct friend of C, then A is an indirect friend of C. And we defined a friend circle is a group of students who are direct or indirect friends.
Given a N*N matrix M representing the friend relationship between students in the class. If M[i][j] = 1, then the ith and jth students are direct friends with each other, otherwise not. And you have to output the total number of friend circles among all the students.
Example 1:
Input:
[[1,1,0],
[1,1,0],
[0,0,1]]
Output: 2
Explanation:The 0th and 1st students are direct friends, so they are in a friend circle.
The 2nd student himself is in a friend circle. So return 2.
Example 2:
Input:
[[1,1,0],
[1,1,1],
[0,1,1]]
Output: 1
Explanation:The 0th and 1st students are direct friends, the 1st and 2nd students are direct friends,
so the 0th and 2nd students are indirect friends. All of them are in the same friend circle, so return 1.
Magical Python
Solution 1, using a SciPy function:
import scipy.sparse
class Solution(object):
def findCircleNum(self, M):
return scipy.sparse.csgraph.connected_components(M)[0]
Solution 2, compute the transitive closure of the (boolean) matrix and count the number of different rows:
import numpy as np
class Solution(object):
def findCircleNum(self, M):
return len(set(map(tuple, (np.matrix(M, dtype='bool')**len(M)).A)))
There is a list of sorted integers from 1 to n. Starting from left to right, remove the first number and every other number afterward until you reach the end of the list.
Repeat the previous step again, but this time from right to left, remove the right most number and every other number from the remaining numbers.
We keep repeating the steps again, alternating left to right and right to left, until a single number remains.
Find the last number that remains starting with a list of length n.
Example:
Input:
n = 9,
1 2 3 4 5 6 7 8 9
2 4 6 8
2 6
6
Output:
6
class Solution(object):
def lastRemaining(self, n):
"""
:type n: int
:rtype: int
"""
nums = range(1, n+1)
while len(nums) > 1:
nums = nums[1::2][::-1]
return nums[0]
Nice code , but... Time Limit Exceeded ,Last executed input: 100000000