Medium
Count Pairs of Equal Substrings With Minimum Difference — Python
Full explanation · Time O(n) · Space O(1)
# Time: O(n)
# Space: O(1)
class Solution(object):
def countQuadruples(self, firstString, secondString):
"""
:type firstString: str
:type secondString: str
:rtype: int
"""
lookup1 = [-1]*26
for i in reversed(xrange(len(firstString))):
lookup1[ord(firstString[i])-ord('a')] = i
lookup2 = [-1]*26
for i in xrange(len(secondString)):
lookup2[ord(secondString[i])-ord('a')] = i
result, diff = 0, float("inf")
for i in xrange(26):
if lookup1[i] == -1 or lookup2[i] == -1:
continue
if lookup1[i]-lookup2[i] < diff:
diff = lookup1[i]-lookup2[i]
result = 0
result += int(lookup1[i]-lookup2[i] == diff)
return result