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 <= 100sandtconsist 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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