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longpress.py
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longpress.py
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# Input: name = "alex", typed = "aaleex"
# Output: true
# Explanation: 'a' and 'e' in 'alex' were long pressed.
# Input: name = "saeed", typed = "ssaaedd"
# Output: false
# Explanation: 'e' must have been pressed twice, but it wasn't in the typed output.
#
#
def isLongPressedName(name: str, typed: str) -> bool:
c = 0
tar_map = {}
for i in name:
if i != tar_map.get(len(tar_map), [0])[0]:
c += 1
tar_map[c] = [i, 1]
else:
tar_map[c][1] += 1
c1 = 0
ty_map = {}
for i in typed:
if i != ty_map.get(len(ty_map), [0])[0]:
c1 += 1
ty_map[c1] = [i, 1]
else:
ty_map[c1][1] += 1
if c != c1:
return False
lent = max(c, c1)
for i in range(lent):
if ty_map[i + 1][0] != tar_map[i + 1][0]:
return False
if ty_map[i + 1][1] < tar_map[i + 1][1]:
return False
return True
# Optimized two pointer Solution
# def isLongPressedName(name: str, typed: str) -> bool:
# if name == typed:
# return True
# i = 0
# for j in range(len(typed)):
# if i < len(name) and name[i] == typed[j]:
# i += 1
# elif i and name[i - 1] == typed[j]:
# continue
# else:
# return False
# return i == len(name)
#
print(isLongPressedName("saeed", "ssaaedd"))