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 <= 100sandtconsist of lowercase English letters only.
class Solution:
def countSubstrings(self, s: str, t: str) -> int:
ans = 0
for i, a in enumerate(s):
for j, b in enumerate(t):
if a != b:
l = r = 1
while i >= l and j >= l and s[i - l] == t[j - l]:
l += 1
while i + r < len(s) and j + r < len(t) and s[i + r] == t[j + r]:
r += 1
ans += l * r
return ansclass Solution {
public int countSubstrings(String s, String t) {
int m = s.length(), n = t.length();
int ans = 0;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (s.charAt(i) != t.charAt(j)) {
int l = 1, r = 1;
while (i - l >= 0 && j - l >= 0 && s.charAt(i - l) == t.charAt(j - l)) {
++l;
}
while (i + r < m && j + r < n && s.charAt(i + r) == t.charAt(j + r)) {
++r;
}
ans += l * r;
}
}
}
return ans;
}
}class Solution {
public:
int countSubstrings(string s, string t) {
int m = s.size(), n = t.size();
int ans = 0;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (s[i] == t[j]) {
continue;
}
int l = 1, r = 1;
while (i - l >= 0 && j - l >= 0 && s[i - l] == t[j - l]) {
++l;
}
while (i + r < m && j + r < n && s[i + r] == t[j + r]) {
++r;
}
ans += l * r;
}
}
return ans;
}
};func countSubstrings(s string, t string) (ans int) {
for i, a := range s {
for j, b := range t {
if a != b {
l, r := 1, 1
for i >= l && j >= l && s[i-l] == t[j-l] {
l++
}
for i+r < len(s) && j+r < len(t) && s[i+r] == t[j+r] {
r++
}
ans += l * r
}
}
}
return
}