자본주의를 살아가는 코린이

Given an integer array nums, move all the even integers at the beginning of the array followed by all the odd integers. Return any array that satisfies this condition. Example 1: Input: nums = [3,1,2,4] Output: [2,4,3,1] Explanation: The outputs [4,2,3,1], [2,4,1,3], and [4,2,1,3] would also be accepted. Example 2: Input: nums = [0] Output: [0] Constraints: 1

Given an integer array nums, move all 0's to the end of it while maintaining the relative order of the non-zero elements. Note that you must do this in-place without making a copy of the array. Example 1: Input: nums = [0,1,0,3,12] Output: [1,3,12,0,0] Example 2: Input: nums = [0] Output: [0] Constraints: 1

Given an array arr, replace every element in that array with the greatest element among the elements to its right, and replace the last element with -1. After doing so, return the array. Example 1: Input: arr = [17,18,5,4,6,1] Output: [18,6,6,6,1,-1] Explanation: - index 0 --> the greatest element to the right of index 0 is index 1 (18). - index 1 --> the greatest element to the right of index 1..

Given an array of integers arr, return true if and only if it is a valid mountain array. Recall that arr is a mountain array if and only if: arr.length >= 3 There exists some i with 0 arr[i + 1] > ... > arr[arr.length - 1] Example 1: Input: arr = [2,1] Output: false Example 2: Input: arr = [3,5,5] Output: false ..

Given an array arr of integers, check if there exist two indices i and j such that : i != j 0

Given an integer array nums sorted in non-decreasing order, remove the duplicates in-place such that each unique element appears only once. The relative order of the elements should be kept the same. Then return the number of unique elements in nums. Consider the number of unique elements of nums to be k, to get accepted, you need to do the following things: Change the array nums such that the f..

input_int, key = map(int, input().split()) result = [] for i in range(1, input_int+1): if input_int % i == 0: result.append(i) if len(result) > key - 1: print(result[key - 1]) else: print(0) 코드 결과는 메모리가 더 작지만, 시간이 4ms 더 길게 나왔습니다. 아래 분의 코드를 사용하면 시간이 44ms로 더 작습니다. https://computer-science-student.tistory.com/574 [파이썬, Python] 백준 2501번 : 약수 구하기 백준 2501번 : 약수 구하기 (문제 바로가기) 내 코드 N, K = map(int, input..

class Solution: def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None: """ Do not return anything, modify nums1 in-place instead. """ n_num1 = [] n_num2 = [] for i in nums1[:m]: n_num1.append(i) for j in nums2: n_num1.append(j) s_list = sorted(n_num1 + n_num2) for x in range(n+m): nums1[x] = s_list[x] 정해진 m,n까지 자르고 그 함수를 합친 후 정렬하는 문제입니다. 더 간단히 하신 분들도 있으니 다른 분들것도 참조하세요.

class Solution: def duplicateZeros(self, arr: List[int]) -> None: """ Do not return anything, modify arr in-place instead. """ result_list = [] for i in range(len(arr)): print(arr[i]) if arr[i] == 0: result_list.append(arr[i]) result_list.append(arr[i]) result = result_list[:len(arr)] for x in range(len(result)): arr[x] = result[x]

배열의 수를 각각 제곱하고 정렬하는 문제입니다. class Solution: def sortedSquares(self, nums: List[int]) -> List[int]: sq_list = [] for i in nums: sq_list.append(i ** 2) result = sorted(sq_list) return result