[Mock] Amazon 3

1176. Diet Plan Performance

A dieter consumes calories[i] calories on the i-th day.

Given an integer k, for every consecutive sequence of k days (calories[i], calories[i+1], ..., calories[i+k-1] for all 0 <= i <= n-k), they look at T, the total calories consumed during that sequence of k days (calories[i] + calories[i+1] + ... + calories[i+k-1]):

• If T < lower, they performed poorly on their diet and lose 1 point;
• If T > upper, they performed well on their diet and gain 1 point;
• Otherwise, they performed normally and there is no change in points.
  Initially, the dieter has zero points. Return the total number of points the dieter has after dieting for calories.length days.

Note that the total points can be negative.

My Answer 1: Time Limit Exceeded (26 / 27 test cases passed.)

class Solution:
    def dietPlanPerformance(self, calories: List[int], k: int, lower: int, upper: int) -> int:
	points = 0
        
        for i in range(len(calories)-k+1):
            if sum(calories[i:i+k]) < lower:
                points -= 1
            if sum(calories[i:i+k]) > upper:
                points += 1
        
        return points

문제대로 했지만.. 내가 봐도... time limit ~

Solution 1: Accepted (Runtime: 180 ms - 100% / Memory Usage: 24.1 MB - 80.83%)

class Solution:
    def dietPlanPerformance(self, calories: List[int], k: int, lower: int, upper: int) -> int:
        sum_t = sum(calories[:k])
        j, ans = 0, 0
        if sum_t < lower: ans = -1
        if sum_t > upper: ans = 1
        for i in range(k, len(calories)):
            sum_t -= calories[j]
            sum_t += calories[i]
            j += 1
            if sum_t < lower: ans -= 1
            if sum_t > upper: ans += 1
        
        return ans

k 길이만큼의 합을 유지

맨 처음에 넣은 값을 빼주고 지금 보는 값을 더해주기

전에 풀었던 Sliding window 랑 유사한가봐요~

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572. Subtree of Another Tree

Given two non-empty binary trees s and t, check whether tree t has exactly the same structure and node values with a subtree of s. A subtree of s is a tree consists of a node in s and all of this node's descendants. The tree s could also be considered as a subtree of itself.

My Answer 1: Accepted (Runtime: 420 ms - 5.00% / Memory Usage: 14.7 MB - 95.88%)

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def isSubtree(self, s: TreeNode, t: TreeNode) -> bool:
        queue = [s]
        subtree = s
        
        # subtree 가 될만한 위치 찾기
        while queue:
            root = queue.pop(0)
            if root:
                queue.append(root.left)
                queue.append(root.right)
                if root.left and root.left.val == t.val:
                    subtree = root.left
                    if self.matchSubtree(subtree, t):
                        return True
                if root.right and root.right.val == t.val:
                    subtree = root.right
                    if self.matchSubtree(subtree, t):
                        return True
        
        if self.matchSubtree(subtree, t):
            return True
        return False
                    
    def matchSubtree(self, subtree, t):
        # subtree 가 맞는지 확인
        subq = [subtree]
        tq = [t]
        while subq:
            subr = subq.pop(0)
            tr = tq.pop(0)
            if subr and tr:
                if subr.val != tr.val:
                    return False
                subq.append(subr.left)
                subq.append(subr.right)
                tq.append(tr.left)
                tq.append(tr.right)
                
            # 둘 중 하나라도 없으면 다른거니까 False
            if subr and not tr or not subr and tr:
                return False
        
        return True