Hard
Edit Distance — Python
Full explanation · Time O(m * n) · Space O(m + n)
# Time: O(n * m)
# Space: O(n + m)
class Solution(object):
# @return an integer
def minDistance(self, word1, word2):
if len(word1) < len(word2):
return self.minDistance(word2, word1)
distance = [i for i in xrange(len(word2) + 1)]
for i in xrange(1, len(word1) + 1):
pre_distance_i_j = distance[0]
distance[0] = i
for j in xrange(1, len(word2) + 1):
insert = distance[j - 1] + 1
delete = distance[j] + 1
replace = pre_distance_i_j
if word1[i - 1] != word2[j - 1]:
replace += 1
pre_distance_i_j = distance[j]
distance[j] = min(insert, delete, replace)
return distance[-1]
# Time: O(n * m)
# Space: O(n * m)
class Solution2(object):
# @return an integer
def minDistance(self, word1, word2):
distance = [[i] for i in xrange(len(word1) + 1)]
distance[0] = [j for j in xrange(len(word2) + 1)]
for i in xrange(1, len(word1) + 1):
for j in xrange(1, len(word2) + 1):
insert = distance[i][j - 1] + 1
delete = distance[i - 1][j] + 1
replace = distance[i - 1][j - 1]
if word1[i - 1] != word2[j - 1]:
replace += 1
distance[i].append(min(insert, delete, replace))
return distance[-1][-1]