1638. Count Substrings That Differ by One Character
Description
Given two strings s
and t
, find the number of ways you can choose a non-empty substring of s
and replace a single character by a different character such that the resulting substring is a substring of t
. In other words, find the number of substrings in s
that differ from some substring in t
by exactly one character.
For example, the underlined substrings in "computer"
and "computation"
only differ by the 'e'
/'a'
, so this is a valid way.
Return the number of substrings that satisfy the condition above.
A substring is a contiguous sequence of characters within a string.
Example 1:
Input: s = "aba", t = "baba" Output: 6 Explanation: The following are the pairs of substrings from s and t that differ by exactly 1 character: ("aba", "baba") ("aba", "baba") ("aba", "baba") ("aba", "baba") ("aba", "baba") ("aba", "baba") The underlined portions are the substrings that are chosen from s and t.
Example 2:
Input: s = "ab", t = "bb" Output: 3 Explanation: The following are the pairs of substrings from s and t that differ by 1 character: ("ab", "bb") ("ab", "bb") ("ab", "bb") The underlined portions are the substrings that are chosen from s and t.
Constraints:
1 <= s.length, t.length <= 100
s
andt
consist of lowercase English letters only.
Solution
count-substrings-that-differ-by-one-character.py
class Solution:
def countSubstrings(self, s: str, t: str) -> int:
if s == t: return 0
res = 0
for i in range(len(s)):
for z in range(i+1):
c1 = s[i-z:i+1]
l = len(c1)
for j in range(len(t)-l+1):
c2 = t[j:j+l]
if c1 != c2 and len(c1) == len(c2) and sum(c1[i] != c2[i] for i in range(l)) == 1:
res += 1
return res