[Mock] Random 7

전에 풀었던 아마존 문제 였다^^

---

836. Rectangle Overlap

An axis-aligned rectangle is represented as a list [x1, y1, x2, y2], where (x1, y1) is the coordinate of its bottom-left corner, and (x2, y2) is the coordinate of its top-right corner. Its top and bottom edges are parallel to the X-axis, and its left and right edges are parallel to the Y-axis.

Two rectangles overlap if the area of their intersection is positive. To be clear, two rectangles that only touch at the corner or edges do not overlap.

Given two axis-aligned rectangles rec1 and rec2, return true if they overlap, otherwise return false.

My Answer 1: Accepted (Runtime: 32 ms - 51.18% / Memory Usage: 14.4 MB - 12.71%)

class Solution:
    def isRectangleOverlap(self, rec1: List[int], rec2: List[int]) -> bool:
        if rec2[0] == rec2[2] or rec2[1] == rec2[3] or rec1[0] == rec1[2] or rec1[1] == rec1[3]:
            return False
        
        # rec2 의 bottom-left 가 overlap
        if (rec1[0] < rec2[0] and rec1[2] > rec2[0]) and (rec1[1] < rec2[1] and rec1[3] > rec2[1]):
            return True
        # rec2 의 top-right 가 overlap
        if (rec1[0] < rec2[2] and rec1[2] > rec2[2]) and (rec1[1] < rec2[3] and rec1[3] > rec2[3]):
            return True
        # rec2 안에 rec1 이 있을 때
        if rec2[0] <= rec1[0] and rec2[1] <= rec1[1] and rec2[2] >= rec1[2] and rec2[3] >= rec1[3]:
            return True
        # rec2 의 bottom-right 가 overlap
        if (rec1[0] < rec2[2] and rec1[2] > rec2[2]) and (rec1[1] < rec2[1] and rec1[3] > rec2[1]):
            return True
        # rec2 의 top-left 가 overlap
        if (rec1[0] < rec2[0] and rec1[2] > rec2[0]) and (rec1[1] < rec2[3] and rec1[3] > rec2[3]):
            return True
        
        if (rec1[0] < rec2[2] and rec1[2] > rec2[0]) and (rec2[3] > rec1[1] and rec2[1] < rec1[3]):
            return True
        
        return False

저번에도 이렇게 모든 경우를 분류해서 했던 거 같은데...

이런 문제는 다신 만나고 싶지가 않네요...ㅎ

Solution 1: Accepted (Runtime: 32 ms - 51.18% / Memory Usage: 14.3 MB - 12.71%)

class Solution:
    def isRectangleOverlap(self, rec1: List[int], rec2: List[int]) -> bool:
        if rec2[0] == rec2[2] or rec2[1] == rec2[3] or rec1[0] == rec1[2] or rec1[1] == rec1[3]:
            return False
        
        if rec1[0] >= rec2[2] or rec1[2] <= rec2[0] or rec2[3] <= rec1[1] or rec2[1] >= rec1[3]:
            return False

        return True

반대로 False 의 경우를 조건으로 잡으면 더 간단해진다..

---

763. Partition Labels

A string s of lowercase English letters is given. We want to partition this string into as many parts as possible so that each letter appears in at most one part, and return a list of integers representing the size of these parts.

My Answer 1: Accepted (Runtime: 52 ms - 13.48% / Memory Usage: 14.3 MB - 25.44%)

class Solution:
    def partitionLabels(self, s: str) -> List[int]:
        result = []
        
        count = collections.Counter(s)
        
        cnt = 0
        now = []
        for i in range(len(s)):
            if s[i] not in now:
                now.append(s[i])
            count[s[i]] -= 1
            cnt += 1
            
            zero = 0
            for j in range(len(now)):
                if count[now[j]] != 0:
                    zero = 0
                    break
                zero = 1
            
            if zero:
                result.append(cnt)
                now = []
                cnt = 0
        
        return result

각 문자들의 개수를 구함 => count

하나씩 보면서 at most 가 될 때까지 now 에 추가해줌

now 리스트의 값들의 개수가 모두 0 이 되면 result 에 update 후 초기화

Solution 1: Accepted (Runtime: 72 ms - 7.21% / Memory Usage: 14.5 MB - 25.44%)

class Solution:
    def partitionLabels(self, s: str) -> List[int]:
        last = {c: i for i, c in enumerate(s)}
        j = anchor = 0
        ans = []
        for i, c in enumerate(s):
            j = max(j, last[c])
            if i == j:
                ans.append(i - anchor + 1)
                anchor = i + 1
            
        return ans

Greedy

각 문자들의 마지막 위치값을 last 에 저장

j 에는 가장 마지막 인덱스를 update 해주고 그 지점이 나올 때까지 돌려줌 (i == j)

at most 가 되면 ans 에 update

anchor 는 길이 계산에 사용