Contest
1. Q1
Question(Easy)
You are given a 0-indexed array of string words and two integers left and right.
A string is called a vowel string if it starts with a vowel character and ends with a vowel character where vowel characters are ‘a’, ‘e’, ‘i’, ‘o’, and ‘u’.
Return the number of vowel strings words[i] where i belongs to the inclusive range [left, right].
Example 1:
Input: words = [“are”,”amy”,”u”], left = 0, right = 2
Output: 2
Explanation:
- “are” is a vowel string because it starts with ‘a’ and ends with ‘e’.
- “amy” is not a vowel string because it does not end with a vowel.
- “u” is a vowel string because it starts with ‘u’ and ends with ‘u’. The number of vowel strings in the mentioned range is 2.
Example 2:
Input: words = [“hey”,”aeo”,”mu”,”ooo”,”artro”], left = 1, right = 4
Output: 3
Explanation:
- “aeo” is a vowel string because it starts with ‘a’ and ends with ‘o’.
- “mu” is not a vowel string because it does not start with a vowel.
- “ooo” is a vowel string because it starts with ‘o’ and ends with ‘o’.
- “artro” is a vowel string because it starts with ‘a’ and ends with ‘o’. The number of vowel strings in the mentioned range is 3.
Constraints:
- 1 <= words.length <= 1000
- 1 <= words[i].length <= 10
- words[i] consists of only lowercase English letters.
- 0 <= left <= right < words.length
My solution during the contest
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class Solution:
def vowelStrings(self, words: List[str], left: int, right: int) -> int:
vowels = {'a', 'e', 'i', 'o', 'u'}
cnt = 0
for i in range(left, right + 1):
word = words[i]
if word[0] in vowels and word[-1] in vowels:
cnt += 1
return cnt
2. Q2 (Middle) 6316. Rearrange Array to Maximize Prefix Score
You are given a 0-indexed integer array nums. You can rearrange the elements of nums to any order (including the given order).
Let prefix be the array containing the prefix sums of nums after rearranging it. In other words, prefix[i] is the sum of the elements from 0 to i in nums after rearranging it. The score of nums is the number of positive integers in the array prefix.
Return the maximum score you can achieve.
Example 1:
Input: nums = [2,-1,0,1,-3,3,-3] Output: 6 Explanation: We can rearrange the array into nums = [2,3,1,-1,-3,0,-3]. prefix = [2,5,6,5,2,2,-1], so the score is 6. It can be shown that 6 is the maximum score we can obtain.
Example 2:
Input: nums = [-2,-3,0] Output: 0 Explanation: Any rearrangement of the array will result in a score of 0.
Constraints:
- 1 <= nums.length <= 105
- -106 <= nums[i] <= 106
My solution during the contest
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class Solution:
def maxScore(self, nums: List[int]) -> int:
nums.sort(reverse=True)
pre_sum = cnt = 0
for num in nums:
pre_sum += num
if pre_sum > 0:
cnt += 1
else:
break
return cnt
Better solution
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class Solution:
def maxScore(self, nums: List[int]) -> int:
nums.sort(reverse=True)
return sum(s > 0 for s in accumulate(nums))
3. Q3(Middle) Count the Number of Beautiful Subarrays
You are given a 0-indexed integer array nums. In one operation, you can:
Choose two different indices i and j such that 0 <= i, j < nums.length. Choose a non-negative integer k such that the kth bit (0-indexed) in the binary representation of nums[i] and nums[j] is 1. Subtract 2k from nums[i] and nums[j]. A subarray is beautiful if it is possible to make all of its elements equal to 0 after applying the above operation any number of times.
Return the number of beautiful subarrays in the array nums.
A subarray is a contiguous non-empty sequence of elements within an array.
Example 1:
Input: nums = [4,3,1,2,4] Output: 2 Explanation: There are 2 beautiful subarrays in nums: [4,3,1,2,4] and [4,3,1,2,4].
- We can make all elements in the subarray [3,1,2] equal to 0 in the following way:
- Choose [3, 1, 2] and k = 1. Subtract 21 from both numbers. The subarray becomes [1, 1, 0].
- Choose [1, 1, 0] and k = 0. Subtract 20 from both numbers. The subarray becomes [0, 0, 0].
- We can make all elements in the subarray [4,3,1,2,4] equal to 0 in the following way:
- Choose [4, 3, 1, 2, 4] and k = 2. Subtract 22 from both numbers. The subarray becomes [0, 3, 1, 2, 0].
- Choose [0, 3, 1, 2, 0] and k = 0. Subtract 20 from both numbers. The subarray becomes [0, 2, 0, 2, 0].
- Choose [0, 2, 0, 2, 0] and k = 1. Subtract 21 from both numbers. The subarray becomes [0, 0, 0, 0, 0].
Example 2:
Input: nums = [1,10,4] Output: 0 Explanation: There are no beautiful subarrays in nums.
My solution during the contest:
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Better solution
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class Solution:
def beautifulSubarrays(self, nums: List[int]) -> int:
s = list(accumulate(nums, xor, initial=0))
ans, cnt = 0, Counter()
for x in s:
ans += cnt[x]
cnt[x] += 1
return ans
Improvement
XOR + pre sum
4. Q4 (Hard) 6318. Minimum Time to Complete All Tasks
There is a computer that can run an unlimited number of tasks at the same time. You are given a 2D integer array tasks where tasks[i] = [starti, endi, durationi] indicates that the ith task should run for a total of durationi seconds (not necessarily continuous) within the inclusive time range [starti, endi].
You may turn on the computer only when it needs to run a task. You can also turn it off if it is idle.
Return the minimum time during which the computer should be turned on to complete all tasks.
Example 1:
Input: tasks = [[2,3,1],[4,5,1],[1,5,2]] Output: 2 Explanation:
- The first task can be run in the inclusive time range [2, 2].
- The second task can be run in the inclusive time range [5, 5].
- The third task can be run in the two inclusive time ranges [2, 2] and [5, 5]. The computer will be on for a total of 2 seconds. Example 2:
Input: tasks = [[1,3,2],[2,5,3],[5,6,2]] Output: 4 Explanation:
- The first task can be run in the inclusive time range [2, 3].
- The second task can be run in the inclusive time ranges [2, 3] and [5, 5].
- The third task can be run in the two inclusive time range [5, 6]. The computer will be on for a total of 4 seconds.
Constraints:
- 1 <= tasks.length <= 2000
- tasks[i].length == 3
- 1 <= starti, endi <= 2000
- 1 <= durationi <= endi - starti + 1
My solution during the contest
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class Solution:
def findMinimumTime(self, tasks: List[List[int]]) -> int:
tasks.sort(key=lambda t: t[1])
st = [(-2, -2, 0)] # 闭区间左右端点，栈底到栈顶的区间长度的和
for start, end, d in tasks:
_, r, s = st[bisect_left(st, (start,)) - 1]
d -= st[-1][2] - s # 去掉运行中的时间点
if start <= r: # start 在区间 st[i] 内
d -= r - start + 1 # 去掉运行中的时间点
if d <= 0: continue
while end - st[-1][1] <= d: # 剩余的 d 填充区间后缀
l, r, _ = st.pop()
d += r - l + 1 # 合并区间
st.append((end - d + 1, end, st[-1][2] + d))
return st[-1][2]