## 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 `"`

and __compute__r"`"`

only differ by the __computa__tion"`'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:6Explanation: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:3Explanation: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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