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Commit 2a6e17f

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Add medium problems
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‎Leetcode/medium/map-sum-pairs.py‎

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"""
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# MAP SUM PAIRS
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Implement the MapSum class:
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MapSum() Initializes the MapSum object.
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void insert(String key, int val) Inserts the key-val pair into the map. If the key already existed, the original key-value pair will be overridden to the new one.
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int sum(string prefix) Returns the sum of all the pairs' value whose key starts with the prefix.
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Example 1:
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Input
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["MapSum", "insert", "sum", "insert", "sum"]
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[[], ["apple", 3], ["ap"], ["app", 2], ["ap"]]
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Output
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[null, null, 3, null, 5]
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Explanation
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MapSum mapSum = new MapSum();
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mapSum.insert("apple", 3);
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mapSum.sum("ap"); // return 3 (apple = 3)
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mapSum.insert("app", 2);
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mapSum.sum("ap"); // return 5 (apple + app = 3 + 2 = 5)
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Constraints:
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1 <= key.length, prefix.length <= 50
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key and prefix consist of only lowercase English letters.
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1 <= val <= 1000
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At most 50 calls will be made to insert and sum.
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"""
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class MapSum:
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def __init__(self):
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"""
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Initialize your data structure here.
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"""
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self.root = {}
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self.endSymbol = "*"
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def insert(self, key: str, val: int) -> None:
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current = self.root
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for letter in key:
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if letter not in current:
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current[letter] = {}
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current = current[letter]
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current[self.endSymbol] = val
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def sum(self, prefix: str) -> int:
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current = self.root
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for letter in prefix:
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if letter not in current:
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return 0
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current = current[letter]
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return self.getSum(current)
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def getSum(self, trieNode):
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currentSum = 0
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for letter in trieNode:
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if letter == self.endSymbol:
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currentSum += trieNode[letter]
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else:
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currentSum += self.getSum(trieNode[letter])
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return currentSum
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# Your MapSum object will be instantiated and called as such:
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# obj = MapSum()
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# obj.insert(key,val)
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# param_2 = obj.sum(prefix)
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"""
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# TOP K FREQUENT WORDS
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Given a non-empty list of words, return the k most frequent elements.
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Your answer should be sorted by frequency from highest to lowest. If two words have the same frequency, then the word with the lower alphabetical order comes first.
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Example 1:
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Input: ["i", "love", "leetcode", "i", "love", "coding"], k = 2
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Output: ["i", "love"]
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Explanation: "i" and "love" are the two most frequent words.
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Note that "i" comes before "love" due to a lower alphabetical order.
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Example 2:
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Input: ["the", "day", "is", "sunny", "the", "the", "the", "sunny", "is", "is"], k = 4
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Output: ["the", "is", "sunny", "day"]
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Explanation: "the", "is", "sunny" and "day" are the four most frequent words,
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with the number of occurrence being 4, 3, 2 and 1 respectively.
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Note:
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You may assume k is always valid, 1 ≤ k ≤ number of unique elements.
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Input words contain only lowercase letters.
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Follow up:
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Try to solve it in O(n log k) time and O(n) extra space.
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"""
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class Solution:
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def topKFrequent(self, words: List[str], k: int) -> List[str]:
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frequency = {}
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trie = Trie()
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for word in words:
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freq = trie.insert(word)
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if freq == 1:
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frequency[word] = 1
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else:
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frequency[word] += 1
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wordList = []
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for word in frequency:
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wordList.append((word, frequency[word]))
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wordList.sort(key = lambda x: (-x[1], x[0]))
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return (word[0] for word in wordList[:k])
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class Trie:
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def __init__(self):
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self.root = {}
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self.endSymbol = "*"
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def insert(self, word):
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current = self.root
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for letter in word:
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if letter not in current:
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current[letter] = {}
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current = current[letter]
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if self.endSymbol not in current:
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current[self.endSymbol] = 0
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current[self.endSymbol] += 1
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return current[self.endSymbol]
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