LeetCode 2416: Sum of Prefix Scores of Strings
LeetCode 2416 Solution Explanation
Explanation:
We can solve this problem efficiently by iterating through each word in the array and counting the number of strings that have the current word as a prefix. We can achieve this by creating a Trie data structure where each node represents a character in the word. By traversing the Trie for each word, we can find the count of strings that have the current word as a prefix. Finally, we calculate the sum of scores for all the prefixes of each word.
Algorithm:
- Create a Trie data structure to store the words.
- Insert all words into the Trie.
- Traverse the Trie for each word and count the number of strings with the current word as a prefix.
- Calculate the sum of scores for all prefixes of each word.
- Return the array of scores.
Time Complexity:
- Building the Trie: O(n * m) where n is the number of words and m is the average length of a word.
- Traversing the Trie for each word: O(n * m)
Overall, the time complexity is O(n * m).
Space Complexity:
- Trie data structure: O(n * m)
Overall, the space complexity is O(n * m).
LeetCode 2416 Solutions in Java, C++, Python
class TrieNode {
TrieNode[] children;
int count;
public TrieNode() {
children = new TrieNode[26];
count = 0;
}
}
class Solution {
public int[] prefixScores(String[] words) {
TrieNode root = new TrieNode();
int[] scores = new int[words.length];
for (String word : words) {
TrieNode node = root;
for (char c : word.toCharArray()) {
if (node.children[c - 'a'] == null) {
node.children[c - 'a'] = new TrieNode();
}
node = node.children[c - 'a'];
node.count++;
}
}
for (int i = 0; i < words.length; i++) {
TrieNode node = root;
String word = words[i];
for (char c : word.toCharArray()) {
node = node.children[c - 'a'];
scores[i] += node.count;
}
}
return scores;
}
}Interactive Code Editor for LeetCode 2416
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