[Mock] Amazon 10

1710. Maximum Units on a Truck

You are assigned to put some amount of boxes onto one truck. You are given a 2D array boxTypes, where boxTypes[i] = [numberOfBoxesi, numberOfUnitsPerBoxi]:

numberOfBoxesi is the number of boxes of type i.
numberOfUnitsPerBoxi is the number of units in each box of the type i.
You are also given an integer truckSize, which is the maximum number of boxes that can be put on the truck. You can choose any boxes to put on the truck as long as the number of boxes does not exceed truckSize.

Return the maximum total number of units that can be put on the truck.

My Answer 1: Accepted (Runtime: 148 ms - 89.81% / Memory Usage: 14.6 MB - 89.31%)

class Solution:
    def maximumUnits(self, boxTypes: List[List[int]], truckSize: int) -> int:
        boxTypes.sort(key = lambda unit:unit[1], reverse=True)
        
        ans = 0
        for i in range(len(boxTypes)):
            if boxTypes[i][0] > truckSize:
                ans += truckSize * boxTypes[i][1]
                break
            else:
                ans += boxTypes[i][0] * boxTypes[i][1]
                truckSize -= boxTypes[i][0]
        
        return ans

units 값들을 기준으로 오름차순 정렬한 후

박스 개수 * unit 개수 를 ans 에 저장

박스 개수가 truckSize 를 넘으면 truckSize * unit 개수 를 ans 에 저장

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1721. Swapping Nodes in a Linked List

You are given the head of a linked list, and an integer k.

Return the head of the linked list after swapping the values of the kth node from the beginning and the kth node from the end (the list is 1-indexed).

My Answer 1: Accepted (Runtime: 1132 ms - 36.36% / Memory Usage: 48.8 MB - 57.09%)

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def swapNodes(self, head: ListNode, k: int) -> ListNode:
        h = head
        n = 0
        
        while head:
            n += 1
            head = head.next
        
        head = h
        
        i = 0
        l = head
        r = head
        while head:
            i += 1
            
            if i == k:
                l = head
            
            if i == n - k + 1:
                r = head
                
            head = head.next
            
        l.val, r.val = r.val, l.val
        
        return h

우선 총 몇개의 노드가 있는지 n 구하기

앞에서 k 번째 값과 뒤에서 k 번째 값을 찾아서

l, r 에 저장한 후, 서로의 val 값 교환