Python - Algorithm #56

Python Algorithm

LeetCode

2639. Find the Width of Columns of a Grid

• You are given a 0-indexed m x n integer matrix [grid]. The width of a column is the maximum length of its integers.
  • For example, if grid = [[-10], [3], [12]], the width of the only column is 3 since -10 is of length 3.
• Return an integer array [ans] of size n where ans[i] is the width of the iᵗʰ column.
• The length of an integer x with len digits is equal to len if x is non-negative, and len + 1 otherwise.
• Constraints
  • m == grid.length
  • n == grid[i].length
  • 1 ≤ m, n ≤ 100 
  • -10⁹ ≤ grid[r][c] ≤ 10⁹

class Solution:
    def findColumnWidth(self, grid: List[List[int]]) -> List[int]:
        answer = [0] * len(grid[0])
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                num_len = len(str(grid[i][j]))
                answer[j] = max(answer[j], num_len)
        return answer

class Solution:
    def findColumnWidth(self, grid: List[List[int]]) -> List[int]:
        answer = []
        for i in range(len(grid[0])):
            length = 0
            for j in range(len(grid)):
                if len(str(grid[j][i])) > length:
                    length = len(str(grid[j][i]))
            answer.append(length)
        return answer

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2404. Most Frequent Even Element

• Given an integer array nums, return the most frequent even element.
• If there is a tie, return the smallest one. If there is no such element, return -1.
• Constraints
  • 1 ≤ nums.length ≤ 2000
  • 0 ≤ nums[i] ≤ 10⁵

class Solution:
    def mostFrequentEven(self, nums: List[int]) -> int:
        even_num = [x for x in nums if x % 2 == 0]
        if not even_num:
            return -1
        c = Counter(even_num)
        max_freq = max(list(c.values()))
        # max_freq만큼 등장한 숫자들 찾아서 가장 작은 값을 리턴
        candidates = [num for num, cnt in c.items() if cnt == max_freq]
        return sorted(candidates)[0]

class Solution:
    def mostFrequentEven(self, nums: List[int]) -> int:
        answer = []
        c = Counter(nums)
        c = dict(filter(lambda x: x[0] % 2 == 0, c.items()))
        c = sorted(c.items(), key=lambda x: (-x[1], x[0]))
        if len(c) >= 1:
            return c[0][0]
        else:
            return -1

class Solution:
    def mostFrequentEven(self, nums: List[int]) -> int:
        nums = [num for num in nums if num % 2 == 0]
        if not nums:
            return -1
        c = Counter(nums)
        c = sorted(c.items(), key=lambda x: (-x[1], x[0]))
        return c[0][0]

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https://leetcode.com/problems/find-the-width-of-columns-of-a-grid/

https://leetcode.com/problems/most-frequent-even-element/