3042. Count Prefix and Suffix Pairs I
Description
You are given a 0-indexed string array words.
Let's define a boolean function isPrefixAndSuffix that takes two strings, str1 and str2:
isPrefixAndSuffix(str1, str2)returnstrueifstr1is both a prefix and a suffix ofstr2, andfalseotherwise.
For example, isPrefixAndSuffix("aba", "ababa") is true because "aba" is a prefix of "ababa" and also a suffix, but isPrefixAndSuffix("abc", "abcd") is false.
Return an integer denoting the number of index pairs (i, j) such that i < j, and isPrefixAndSuffix(words[i], words[j]) is true.
Example 1:
Input: words = ["a","aba","ababa","aa"]
Output: 4
Explanation: In this example, the counted index pairs are:
i = 0 and j = 1 because isPrefixAndSuffix("a", "aba") is true.
i = 0 and j = 2 because isPrefixAndSuffix("a", "ababa") is true.
i = 0 and j = 3 because isPrefixAndSuffix("a", "aa") is true.
i = 1 and j = 2 because isPrefixAndSuffix("aba", "ababa") is true.
Therefore, the answer is 4.
Example 2:
Input: words = ["pa","papa","ma","mama"]
Output: 2
Explanation: In this example, the counted index pairs are:
i = 0 and j = 1 because isPrefixAndSuffix("pa", "papa") is true.
i = 2 and j = 3 because isPrefixAndSuffix("ma", "mama") is true.
Therefore, the answer is 2.
Example 3:
Input: words = ["abab","ab"]
Output: 0
Explanation: In this example, the only valid index pair is i = 0 and j = 1, and isPrefixAndSuffix("abab", "ab") is false.
Therefore, the answer is 0.
Constraints:
1 <= words.length <= 501 <= words[i].length <= 10words[i]consists only of lowercase English letters.
Solutions
Solution 1: Enumeration
We can enumerate all index pairs $(i, j)$, where $i < j$, and then determine whether words[i] is a prefix or suffix of words[j]. If it is, we increment the count.
The time complexity is $O(n^2 \times m)$, where $n$ and $m$ are the length of words and the maximum length of the strings, respectively.
Python3
class Solution:
def countPrefixSuffixPairs(self, words: List[str]) -> int:
ans = 0
for i, s in enumerate(words):
for t in words[i + 1 :]:
ans += t.endswith(s) and t.startswith(s)
return ans
Java
class Solution {
public int countPrefixSuffixPairs(String[] words) {
int ans = 0;
int n = words.length;
for (int i = 0; i < n; ++i) {
String s = words[i];
for (int j = i + 1; j < n; ++j) {
String t = words[j];
if (t.startsWith(s) && t.endsWith(s)) {
++ans;
}
}
}
return ans;
}
}
C++
class Solution {
public:
int countPrefixSuffixPairs(vector<string>& words) {
int ans = 0;
int n = words.size();
for (int i = 0; i < n; ++i) {
string s = words[i];
for (int j = i + 1; j < n; ++j) {
string t = words[j];
if (t.find(s) == 0 && t.rfind(s) == t.length() - s.length()) {
++ans;
}
}
}
return ans;
}
};
Go
func countPrefixSuffixPairs(words []string) (ans int) {
for i, s := range words {
for _, t := range words[i+1:] {
if strings.HasPrefix(t, s) && strings.HasSuffix(t, s) {
ans++
}
}
}
return
}
TypeScript
function countPrefixSuffixPairs(words: string[]): number {
let ans = 0;
for (let i = 0; i < words.length; ++i) {
const s = words[i];
for (const t of words.slice(i + 1)) {
if (t.startsWith(s) && t.endsWith(s)) {
++ans;
}
}
}
return ans;
}
Solution 2: Trie
We can treat each string $s$ in the string array as a list of character pairs, where each character pair $(s[i], s[m - i - 1])$ represents the $i$th character pair of the prefix and suffix of string $s$.
We can use a trie to store all the character pairs, and then for each string $s$, we search for all the character pairs $(s[i], s[m - i - 1])$ in the trie, and add their counts to the answer.
The time complexity is $O(n \times m)$, and the space complexity is $O(n \times m)$. Here, $n$ and $m$ are the lengths of words and the maximum length of the strings, respectively.
Python3
class Node:
__slots__ = ["children", "cnt"]
def __init__(self):
self.children = {}
self.cnt = 0
class Solution:
def countPrefixSuffixPairs(self, words: List[str]) -> int:
ans = 0
trie = Node()
for s in words:
node = trie
for p in zip(s, reversed(s)):
if p not in node.children:
node.children[p] = Node()
node = node.children[p]
ans += node.cnt
node.cnt += 1
return ans
Java
class Node {
Map<Integer, Node> children = new HashMap<>();
int cnt;
}
class Solution {
public int countPrefixSuffixPairs(String[] words) {
int ans = 0;
Node trie = new Node();
for (String s : words) {
Node node = trie;
int m = s.length();
for (int i = 0; i < m; ++i) {
int p = s.charAt(i) * 32 + s.charAt(m - i - 1);
node.children.putIfAbsent(p, new Node());
node = node.children.get(p);
ans += node.cnt;
}
++node.cnt;
}
return ans;
}
}
C++
class Node {
public:
unordered_map<int, Node*> children;
int cnt;
Node()
: cnt(0) {}
};
class Solution {
public:
int countPrefixSuffixPairs(vector<string>& words) {
int ans = 0;
Node* trie = new Node();
for (const string& s : words) {
Node* node = trie;
int m = s.length();
for (int i = 0; i < m; ++i) {
int p = s[i] * 32 + s[m - i - 1];
if (node->children.find(p) == node->children.end()) {
node->children[p] = new Node();
}
node = node->children[p];
ans += node->cnt;
}
++node->cnt;
}
return ans;
}
};
Go
type Node struct {
children map[int]*Node
cnt int
}
func countPrefixSuffixPairs(words []string) (ans int) {
trie := &Node{children: make(map[int]*Node)}
for _, s := range words {
node := trie
m := len(s)
for i := 0; i < m; i++ {
p := int(s[i])*32 + int(s[m-i-1])
if _, ok := node.children[p]; !ok {
node.children[p] = &Node{children: make(map[int]*Node)}
}
node = node.children[p]
ans += node.cnt
}
node.cnt++
}
return
}