sadrok

# NODE CLASS PROVIDED
class Node:
    def __init__(self,data):
        self.right=self.left=None
        self.data = data
               
class Solution:
    # INSERT METHOD PROVIDED
    def insert(self,root,data):
        if root==None:

sadrok

"""
I found some problems with my initial solution including a huge
flaw in the final logic.
"""

BRACKETS = dict(['()', '{}', '[]', "<>"])
CLOSING_BRACKETS = set(BRACKETS.values())

def brackets(test_str):
    bracket_stack = []

sadrok

"""
anagram_1 was my initial naive approach to the simple anagram-check problem.
anagram_2 is a more compact version which should provide the same results.
However out of interests sake I ran timeit tests against both functions and got surprising results:

Testing anagram_1 against a small set of words resulted in about 21usec going over each loop
>python -m timeit -s "from anagram_test import test, anagram_1" -- "test(anagram_1)"
20000 loops, best of 5: 14.9 usec per loop

But anagram_2, the more compact version, was surprisingly slower!

sadrok

const uint8_t numColumns = 5;
const uint8_t frames[16 * numColumns] PROGMEM = {
  0x82, 0x7C, 0x7C, 0x7C, 0x82,   // digit 0
  0xFE, 0xBC, 0x00, 0xFC, 0xFE,   // digit 1
  0xB8, 0x74, 0x6C, 0x6C, 0x9C,   // digit 2
  0xBA, 0x7C, 0x6C, 0x6C, 0x92,   // digit 3
  0xE6, 0xD6, 0xB6, 0x00, 0xF6,   // digit 4
  0x1A, 0x6C, 0x6C, 0x6C, 0x72,   // digit 5  
  0xC2, 0xAC, 0x6C, 0x6C, 0xF2,   // digit 6
  0x7E, 0x70, 0x6E, 0x5E, 0x3E,   // digit 7