carloscn
commented
1 year ago Analysis
fn delete_dup_elements(vec:&mut Vec<&str>)
{
let mut i = 0;
while i < vec.len() {
let current_item = &vec[i].clone();
if vec.iter().filter(|&&item| item == *current_item).count() > 1 {
vec.retain(|&item| item != *current_item);
} else {
i += 1;
}
}
}
pub fn count_words(words1: Vec<&str>, words2: Vec<&str>) -> i32
{
let mut ret:i32 = 0;
if words1.len() < 1 || words2.len() < 1 {
return ret;
}
let mut wp1 = words1.clone();
let mut wp2 = words2.clone();
delete_dup_elements(&mut wp1);
delete_dup_elements(&mut wp2);
for e in &wp1 {
for k in &wp2 {
if *e == *k {
ret += 1;
}
}
}
return ret;
}
Description
Given two string arrays words1 and words2, return the number of strings that appear exactly once in each of the two arrays.
Example 1:
Input: words1 = ["leetcode","is","amazing","as","is"], words2 = ["amazing","leetcode","is"] Output: 2 Explanation:
Example 2:
Input: words1 = ["b","bb","bbb"], words2 = ["a","aa","aaa"] Output: 0 Explanation: There are no strings that appear in each of the two arrays.
Example 3:
Input: words1 = ["a","ab"], words2 = ["a","a","a","ab"] Output: 1 Explanation: The only string that appears exactly once in each of the two arrays is "ab".
Constraints:
1 <= words1.length, words2.length <= 1000 1 <= words1[i].length, words2[j].length <= 30 words1[i] and words2[j] consists only of lowercase English letters.