LeetCode Entry
3042. Count Prefix and Suffix Pairs I
Count prefix-suffix matched pairs
3042. Count Prefix and Suffix Pairs I easy
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Problem TLDR
Count prefix-suffix matched pairs #easy
Intuition
The brute-force is accepted.
More interesting solutions are:
- Trie: traverse each word forwards and backwards, if suffix trie has the same word as prefix trie, add frequency
- HashMap: just store words in a frequency HashMap and traverse it on each new word
- Robin-Karp rolling hash & KMP/z-function: same idea as with Trie, but check rolling hash to match visited hashes, then do quick-match with KMP/z-function
Approach
- on a smaller input the O(n^2) solutions are faster
- we can use a single Trie with the key of
(prefix-letetr, suffix-letter)
Complexity
-
Time complexity: \(O(n^2w^2)\), or O(nw) for more optimal
-
Space complexity: \(O(1)\) or O(n)
Code
fun countPrefixSuffixPairs(words: Array<String>) =
(0..<words.size).flatMap { i -> (i + 1..<words.size).map { i to it }}
.count { (i, j) -> words[j].startsWith(words[i]) && words[j].endsWith(words[i])}
pub fn count_prefix_suffix_pairs(words: Vec<String>) -> i32 {
#[derive(Default)] struct T(usize, i32, HashMap<u8, T>);
let (mut tf, mut tb, mut res) = (T::default(), T::default(), 0);
for (p, w) in words.iter().map(|w| w.as_bytes()).enumerate() {
let (mut f, mut b) = (&mut tf, &mut tb);
for i in 0..w.len() {
let (cf, cb) = (w[i], w[w.len() - i - 1]);
f = f.2.entry(cf).or_default();
b = b.2.entry(cb).or_default();
if f.0 > 0 && f.0 == b.0 { res += f.1 }
}
f.0 = p + 1; b.0 = p + 1; f.1 += 1; b.1 += 1
}
res
}
int countPrefixSuffixPairs(vector<string>& words) {
unordered_map<string, int> m; int res = 0; m[words[0]] = 1;
for (int i = 1; i < words.size(); ++m[words[i++]])
for (auto& [prev, freq] : m)
if (words[i].starts_with(prev) && words[i].ends_with(prev))
res += freq;
return res;
}