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Paper- Saejin #53

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120 changes: 111 additions & 9 deletions hash_practice/exercises.py
Original file line number Diff line number Diff line change
Expand Up @@ -2,18 +2,55 @@
def grouped_anagrams(strings):
""" This method will return an array of arrays.
Each subarray will have strings which are anagrams of each other
Time Complexity: ?
Space Complexity: ?
Time Complexity: On
Space Complexity: On
"""
Comment on lines 2 to 7

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👍

pass

result = []
hash = {}

for word in strings:
new_word = "".join(sorted(word))
if new_word in hash:
hash[new_word].append(word)
else:
hash[new_word] = [word]

for value in hash.values():
result.append(value)

return result




def top_k_frequent_elements(nums, k):
""" This method will return the k most common elements
In the case of a tie it will select the first occuring element.
Time Complexity: ?
Space Complexity: ?
Time Complexity: Onk
Space Complexity: On
Comment on lines 27 to +31

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👍

"""
pass
if not nums or k <= 0:
return []

hash = {}
result = []

for n in nums:
if n not in hash:
hash[n] = 1
else:
hash[n] += 1

while len(result) < k:
biggest = max(hash, key=hash.get)
result.append(biggest)

del hash[biggest]

return result




def valid_sudoku(table):

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👍 Correct!

Expand All @@ -22,8 +59,73 @@ def valid_sudoku(table):
Each element can either be a ".", or a digit 1-9
The same digit cannot appear twice or more in the same
row, column or 3x3 subgrid
Time Complexity: ?
Space Complexity: ?
Time Complexity: O1 bc it's always 9x9?
Space Complexity: O1 bc the hash is always #1-9?
"""
pass
hash = {}
for n in range(1, 10):
hash[str(n)] = 0

#check rows first
for row in table:
for sq in row:
if sq != ".":
if hash[sq] == 0:
hash[sq] = 1
elif hash[sq] != 0:
return False
#wipe clean
for n in range(1, 10):
hash[str(n)] = 0

#wipe clean
for n in range(1, 10):
hash[str(n)] = 0

#check columns
row = 0
col = 0
for col in range(9):
for row in range(9):
sq = table[row][col]
if sq != ".":
if hash[sq] == 0:
hash[sq] = 1
elif hash[sq] != 0:
return False
#wipe clean
for n in range(1, 10):
hash[str(n)] = 0

#wipe clean
for n in range(1, 10):
hash[str(n)] = 0

#check 3x3 tables
r = 0
c = 0

while (r < 9 and c < 9):
for row in range(r, r+3):
for col in range(c, c+3):
sq = table[row][col]

if sq != ".":
if hash[sq] == 0:
hash[sq] = 1
elif hash[sq] != 0:
return False

#go to next block
# reset r, c
r += 3
c += 3
# wipe hash clean
for n in range(1, 10):
hash[str(n)] = 0

return True