[Mock] Google 5

681. Next Closest Time

Given a time represented in the format "HH:MM", form the next closest time by reusing the current digits. There is no limit on how many times a digit can be reused.

You may assume the given input string is always valid. For example, "01:34", "12:09" are all valid. "1:34", "12:9" are all invalid.

My Answer 1: Accepted (Runtime: 32 ms - 73.22% / Memory Usage: 14.3 MB - 36.85%)

class Solution:
    def nextClosestTime(self, time: str) -> str:
        # time 에서 숫자만 떼서 저장 => t = [19, 34]
        tmp = time.split(':')
        t = [int(tmp[0]), int(tmp[1])]
        
        # time 의 모든 숫자를 저장 => nums = [1, 9, 3, 4]
        nums = []
        for i in time:
            if i != ":":
                nums.append(int(i))
        
        # 모든 조합 찾기
        every = []
        for i in range(len(nums)):
            for j in range(len(nums)):
                # 중복 제거 + 60 보다 작은 수만 사용
                if nums[i]*10 + nums[j] not in every and nums[i]*10 + nums[j] < 60:
                    every.append(nums[i]*10 + nums[j])
        # 순서대로 정렬
        every.sort()
        
        # 일단 minute 먼저 확인
        for m in every:
            if m > t[1]:    # 주어진 minute 보다 크면 바로 return (every 가 정렬된 상태라 가장 최솟값임)
                if m // 10 == 0:
                    return tmp[0] + ":" + "0" + str(m)
                return tmp[0] + ":" + str(m)
        
        # hour 잡아주기
        for h in every:
            # 주어진 hour 보다 큰 값을 찾기 +_minute 은 최솟값인 every[0] 이용
            if h > t[0] and h < 24:
                strRes = ""
                if h // 10 == 0:
                    strRes += "0"
                strRes += str(h) + ":"
                if every[0] // 10 == 0:
                    strRes += "0"
                strRes += str(every[0])
                return strRes
        
        # 주어진 hour 보다 큰 값이 없을 경우 -> 00 시부터 새로 봐야함
        # => 가장 최솟값인 every[0] 사용해서 hour, minute 지정
        strRes = ""
        if every[0] // 10 == 0:
            strRes += "0"
        strRes += str(every[0]) + ":"
        if every[0] // 10 == 0:
            strRes += "0"
        strRes += str(every[0])
        
        return strRes

좀 조잡..

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683. K Empty Slots

You have n bulbs in a row numbered from 1 to n. Initially, all the bulbs are turned off. We turn on exactly one bulb every day until all bulbs are on after n days.

You are given an array bulbs of length n where bulbs[i] = x means that on the (i+1)th day, we will turn on the bulb at position x where i is 0-indexed and x is 1-indexed.

Given an integer k, return the minimum day number such that there exists two turned on bulbs that have exactly k bulbs between them that are all turned off. If there isn't such day, return -1.

이건 걍 문제만 읽었는데 문제도 이해가 안됨ㅎ

Solution 1: Accepted (Runtime: 984 ms - 88.82% / Memory Usage: 16.6 MB - 88.49%)

class Solution:
    def kEmptySlots(self, bulbs: List[int], k: int) -> int:
        days = [0] * len(bulbs)
        # enumerate => (index, value) | enumerate( , 1) => index + 1 값이 됨
        for day, position in enumerate(bulbs, 1):
            days[position - 1] = day

        ans = float('inf')
        left, right = 0, k+1
        while right < len(days):
            for i in range(left + 1, right):
                if days[i] < days[left] or days[i] < days[right]:
                    left, right = i, i+k+1
                    break
            else:
                ans = min(ans, max(days[left], days[right]))
                left, right = right, right+k+1

        return ans if ans < float('inf') else -1

전에 풀었던 Sliding Window 를 이용

루션이도 이해가 안되네요^^