## 1638. Count Substrings That Differ by One Character

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.
```
Example 3:
```Input: s = "a", t = "a"
Output: 0
```

Example 4:

```Input: s = "abe", t = "bbc"
Output: 10
```

Constraints:

• `1 <= s.length, t.length <= 100`
• `s` and `t` consist of lowercase English letters only.

## Rust Solution

``````struct Solution;

impl Solution {
fn count_substrings(s: String, t: String) -> i32 {
let n = s.len();
let m = t.len();
let s: Vec<char> = s.chars().collect();
let t: Vec<char> = t.chars().collect();
let mut res = 0;
for i in 0..n {
for j in 0..m {
let k = (n - i).min(m - j);
let mut diff = 0;
for d in 0..k {
if s[i + d] != t[j + d] {
diff += 1;
}
if diff == 1 {
res += 1;
}
}
}
}
res
}
}

#[test]
fn test() {
let s = "aba".to_string();
let t = "baba".to_string();
let res = 6;
assert_eq!(Solution::count_substrings(s, t), res);
let s = "ab".to_string();
let t = "bb".to_string();
let res = 3;
assert_eq!(Solution::count_substrings(s, t), res);
let s = "a".to_string();
let t = "a".to_string();
let res = 0;
assert_eq!(Solution::count_substrings(s, t), res);
let s = "abe".to_string();
let t = "bbc".to_string();
let res = 10;
assert_eq!(Solution::count_substrings(s, t), res);
}
``````

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