| 題目連結 | LCP 08. 劇情觸發時間 |
|---|---|
| 思路 | 字首和+二分法 |
| 題解連結 | python 字首和+二分法 |
| 關鍵點 | 預處理：前處理得到各個時刻三種資源的累計值（必為升序陣列）；查詢：二分法查詢三種資源需要滿足的時刻，取三者最大值即可得到答案 |
| 時間複雜度 | \(O(n)\) |
| 空間複雜度 | \(O(n)\) |
程式碼實現：
class Solution:
def getTriggerTime(self, increase: List[List[int]], requirements: List[List[int]]) -> List[int]:
n = len(increase)
presum = [
[0] * (n+1) for _ in range(3)
]
for i, (c, r, h) in enumerate(increase):
presum[0][i+1] = presum[0][i] + c
presum[1][i+1] = presum[1][i] + r
presum[2][i+1] = presum[2][i] + h
# found the left-most index `nums[index] >= val`
def lower_bound(nums, val):
left, right = -1, n+1
while left + 1 < right:
mid = (left+right) // 2
if nums[mid] < val:
left = mid
else:
right = mid
return right
m = len(requirements)
answer = [-1] * m
for i, (c, r, h) in enumerate(requirements):
x = lower_bound(presum[0], c)
y = lower_bound(presum[1], r)
z = lower_bound(presum[2], h)
min_index = max(x, y, z)
if min_index <= n:
answer[i] = min_index
return answer
Python-標準庫
class Solution:
def getTriggerTime(self, increase: List[List[int]], requirements: List[List[int]]) -> List[int]:
n = len(increase)
presum = [
[0] * (n+1) for _ in range(3)
]
for i, (c, r, h) in enumerate(increase):
presum[0][i+1] = presum[0][i] + c
presum[1][i+1] = presum[1][i] + r
presum[2][i+1] = presum[2][i] + h
m = len(requirements)
answer = [-1] * m
for i, (c, r, h) in enumerate(requirements):
x = bisect_left(presum[0], c)
y = bisect_left(presum[1], r)
z = bisect_left(presum[2], h)
min_index = max(x, y, z)
if min_index <= n:
answer[i] = min_index
return answer