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make-the-string-great.py
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# 1544. Make The String Great
# 🟢 Easy
#
# https://leetcode.com/problems/make-the-string-great/
#
# Tags: String - Stack
import timeit
# Iterate over the elements of the input string pushing them into a
# stack, if the element that we visit is the uppercase or lowercase
# counterpart of the last element on the stack, pop that element.
#
# Time complexity: O(n) - Where n is the number of characters in the
# input.
# Space complexity: O(n) - The stack could have n elements where n is
# the number of characters of the input.
#
# Runtime: 53 ms, faster than 76.90%
# Memory Usage: 13.8 MB, less than 61.93%
class StackPushPop:
def makeGood(self, s: str) -> str:
stack = []
for c in s:
if stack and (
(c.islower() and c.upper() == stack[-1])
or (c.isupper() and c.lower() == stack[-1])
):
stack.pop()
else:
stack.append(c)
return "".join(stack)
# Similar approach but simplify the if statement using character's ASCII
# values.
#
# Time complexity: O(n) - Where n is the number of characters in the
# input.
# Space complexity: O(n) - The stack could have n elements where n is
# the number of characters of the input.
#
# Runtime: 24 ms, faster than 99.91%
# Memory Usage: 13.8 MB, less than 61.93%
class StackAndOrd:
def makeGood(self, s: str) -> str:
stack = []
for c in s:
if stack and abs(ord(c) - ord(stack[-1])) == 32:
stack.pop()
else:
stack.append(c)
return "".join(stack)
def test():
executors = [
StackPushPop,
StackAndOrd,
]
tests = [
["s", "s"],
["Pp", ""],
["abBAcC", ""],
["leEeetcode", "leetcode"],
]
for executor in executors:
start = timeit.default_timer()
for _ in range(1):
for col, t in enumerate(tests):
sol = executor()
result = sol.makeGood(t[0])
exp = t[1]
assert result == exp, (
f"\033[93m» {result} <> {exp}\033[91m for"
+ f" test {col} using \033[1m{executor.__name__}"
)
stop = timeit.default_timer()
used = str(round(stop - start, 5))
cols = "{0:20}{1:10}{2:10}"
res = cols.format(executor.__name__, used, "seconds")
print(f"\033[92m» {res}\033[0m")
test()