Python - Algorithm #2

Python Algorithm

LeetCode

347. Top K Frequent Elements

• Given an integer array [nums] and an integer k, return the k most frequent elements. You may return the answer in any order.
• Constraints:
  • 1 ≤ nums.length ≤ 10⁵
  • -10⁴ ≤ nums[i] ≤ 10⁴
  • k is in the range [1, the number of unique elements in the array].
  • It is guaranteed that the answer is unique.

class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        answer = []
        # 1. nums 배열에 등장한 횟수 집계
        nums_dict = {}
        for num in nums:
            if num not in nums_dict:
                nums_dict[num] = 0
            nums_dict[num] += 1
        # 2. 많이 등장한 횟수별로 정렬
        sorted_keys = sorted(nums_dict, key=lambda x: nums_dict[x], reverse=True)
        # 3. sorted_keys에서 k개만큼 answer에 append
        for i in range(k):
            answer.append(sorted_keys[i])
        return answer

from collections import Counter

class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
    	# 1. nums 배열에 등장한 횟수 집계
        nums_counter = Counter(nums)
        # 2. 많이 등장한 횟수로 정렬해서 앞에서부터 k개 리턴해주기
        return [x[0] for x in nums_counter.most_common()[:k]]

from collections import Counter

class Solution:
    def topKFrequent(self, nums: List[int], k: int) -> List[int]:
        answer = []
        num_dict = Counter(nums)
        count = sorted(num_dict.items(), key=lambda x: x[1], reverse=True)
        for i in range(k):
            answer.append(count[i][0])
        return answer

• Counter( ) : 주어진 객체의 요소를 키, 요소의 개수를 값으로 가지는 딕셔너리 생성
  • from collections import Counter
• Counter( ).most_common(x) : 최빈값 x개 반환

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2011. Final Value of Variable After Performing Operations

• There is a programming language with only four operations and one variable X:
  • ++X and X++ increments the value of the variable X by 1.
  • --X and X-- decrements the value of the variable X by 1.
• Initially, the value of X is 0.
• Given an array of strings [operations] containing a list of operations, return the final value of X after performing all the operations.
• Constraints:
  • 1 ≤ operations.length ≤ 100
  • operations[i] will be either "++X", "X++", "--X", or "X--".
    

class Solution:
    def finalValueAfterOperations(self, operations: List[str]) -> int:
        answer = 0
        for operation in operations:
            # X++ 혹은 ++X일 때는 +1
            # 아니면 -1
            if operation == "++X" or operation == "X++":
                answer += 1
            else:
                answer -= 1
        return answer

class Solution:
    def finalValueAfterOperations(self, operations: List[str]) -> int:
        answer = 0
        for i in operations:
            if "+" in i:
                answer += 1
            else:
                answer -= 1
        return answer

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https://leetcode.com/problems/top-k-frequent-elements/

https://leetcode.com/problems/final-value-of-variable-after-performing-operations/