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O(m*nlogn) solution

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This commit is contained in:
Jacob Delgado 2023-09-07 20:07:43 -07:00
parent 4e83f1d5ea
commit 5551dcc4fe
1 changed files with 12 additions and 41 deletions
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53
NeetCodeRoadmap/Arrays_and_Hashing/49_Group_Anagrams.py
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@ -1,50 +1,21 @@
def validAnagram(s: str, t: str) -> bool:
seen = {}
check = {}
count = 0
for letter in s:
if letter in seen:
seen[letter] += 1
else:
seen[letter] = 1
for letter in t:
if letter in check:
check[letter] += 1
else:
check[letter] = 1
if seen != check:
return False
else:
return True
def groupAnagrams(strs: list[str])->list[str]:
"""
:type strs: List[str]
:rtype: List[List[str]]
"""
final = []
seen = []
# Fill first list with matching anagrams
for i in range(len(strs)):
grouped = []
print(f'i is: {strs[i]}')
if i == len(strs)-1 and strs[i] not in grouped:
grouped.append(strs[i])
for j in range(i+1, len(strs)):
print(f'j is: {strs[j]}')
if validAnagram(strs[i], strs[j]) == True:
if strs[i] not in grouped and strs[i] not in seen:
grouped.append(strs[i])
seen.append(strs[i])
if strs[j] not in grouped and strs[j] not in seen:
grouped.append(strs[j])
seen.append(strs[j])
if grouped:
final.append(grouped)
strs_table = {}
return final
for string in strs:
sorted_string = ''.join(sorted(string))
if sorted_string not in strs_table:
strs_table[sorted_string] = []
strs_table[sorted_string].append(string)
return list(strs_table.values())
print(groupAnagrams(["eat","tea","tan","ate","nat","bat"]))
# print(groupAnagrams(["eat","tea","tan","ate","nat","bat"]))
print(groupAnagrams(["",""]))