# 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 and t consist of lowercase English letters only.

## Solutions

• class Solution {
public int countSubstrings(String s, String t) {
int ans = 0;
int m = s.length(), n = t.length();
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (s.charAt(i) != t.charAt(j)) {
int l = 0, r = 0;
while (i - l > 0 && j - l > 0 && s.charAt(i - l - 1) == t.charAt(j - l - 1)) {
++l;
}
while (i + r + 1 < m && j + r + 1 < n
&& s.charAt(i + r + 1) == t.charAt(j + r + 1)) {
++r;
}
ans += (l + 1) * (r + 1);
}
}
}
return ans;
}
}

• class Solution {
public:
int countSubstrings(string s, string t) {
int ans = 0;
int m = s.size(), n = t.size();
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (s[i] != t[j]) {
int l = 0, r = 0;
while (i - l > 0 && j - l > 0 && s[i - l - 1] == t[j - l - 1]) {
++l;
}
while (i + r + 1 < m && j + r + 1 < n && s[i + r + 1] == t[j + r + 1]) {
++r;
}
ans += (l + 1) * (r + 1);
}
}
}
return ans;
}
};

• class Solution:
def countSubstrings(self, s: str, t: str) -> int:
ans = 0
m, n = len(s), len(t)
for i, a in enumerate(s):
for j, b in enumerate(t):
if a != b:
l = r = 0
while i > l and j > l and s[i - l - 1] == t[j - l - 1]:
l += 1
while (
i + r + 1 < m and j + r + 1 < n and s[i + r + 1] == t[j + r + 1]
):
r += 1
ans += (l + 1) * (r + 1)
return ans


• func countSubstrings(s string, t string) (ans int) {
m, n := len(s), len(t)
for i, a := range s {
for j, b := range t {
if a != b {
l, r := 0, 0
for i > l && j > l && s[i-l-1] == t[j-l-1] {
l++
}
for i+r+1 < m && j+r+1 < n && s[i+r+1] == t[j+r+1] {
r++
}
ans += (l + 1) * (r + 1)
}
}
}
return
}