EASY AND BEGINNER FRIENDLY PYTHON SOLUTION🔥🔥🔥
Intuition\nThe problem requires calculating the prefix score of each word, which depends on how many other words share the same prefix at each step. A Trie (pre
# Intuition\nThe problem requires calculating the prefix score of each word, which depends on how many other words share the same prefix at each step. A Trie (prefix tree) is ideal for such tasks because it allows efficient prefix matching. By building a Trie with all the words and keeping track of how many words pass through each node (each character in the Trie), we can easily determine how many words share a particular prefix. The goal is to calculate the prefix score by summing the number of words that share the same prefix for each character in the word.\n\n# Approach\n1. Trie Construction: First, we build a Trie where each node represents a character. As we insert a word character by character, we traverse the Trie, and for each node we visit, we increment a (prefix_count) to track how many words share the prefix that leads to that node. This allows us to later know how many words share the same prefix up to any point in the word.\n\n2. Insert Words into Trie: For each word in the list, we insert it into the Trie. Each time we encounter a character that hasn\'t been added yet, we create a new node. As we traverse the word, we increment the (prefix_count) of each node along the way, reflecting the number of words that have passed through that node.\n\n3. Calculate Prefix Scores: After building the Trie, for each word, we traverse it again from the root of the Trie. For each character, we sum the (prefix_count) at the corresponding node, which gives us the number of words that share the prefix up to that point. This sum is the prefix score for the word.\n\n4. Return Results: We repeat the prefix score calculation for every word in the list and store the results in a list, which is returned as the final output.\n\n# Complexity\n- Time complexity: \uD835\uDC42(\uD835\uDC41\xD7\uD835\uDC3F)\n\n- Space complexity: \uD835\uDC42(\uD835\uDC41\xD7\uD835\uDC3F)\n\n# Code\n```python []\nclass TrieNode:\n    def __init__(self):\n        self.children = {}\n        self.prefix_count = 0\n\nclass Solution(object):\n\n    def __init__(self):\n        self.root = TrieNode()\n\n    def insert(self, word):\n        node = self.root\n        for char in word:\n            if char not in node.children:\n                node.children[char] = TrieNode()\n            node = node.children[char]\n            node.prefix_count += 1\n\n    def sumPrefixScores(self, words):\n        \n        for word in words:\n            self.insert(word)\n\n        scores = []\n\n        for word in words:\n            score = 0\n            node = self.root\n\n            for char in word:\n                if char in node.children:\n                    node = node.children[char]\n                    score += node.prefix_count\n            scores.append(score)\n        \n        return scores\n\n```
['Python']