2389. Longest Subsequence With Limited Sum
Easy
You are given an integer array nums
of length n
, and an integer array queries
of length m
.
Return an array answer
of length m
where answer[i]
is the maximum size of a subsequence that you can take from nums
such that the sum of its elements is less than or equal to queries[i]
.
A subsequence is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements.
Example 1:
Input: nums = [4,5,2,1], queries = [3,10,21] Output: [2,3,4] Explanation: We answer the queries as follows: - The subsequence [2,1] has a sum less than or equal to 3. It can be proven that 2 is the maximum size of such a subsequence, so answer[0] = 2. - The subsequence [4,5,1] has a sum less than or equal to 10. It can be proven that 3 is the maximum size of such a subsequence, so answer[1] = 3. - The subsequence [4,5,2,1] has a sum less than or equal to 21. It can be proven that 4 is the maximum size of such a subsequence, so answer[2] = 4.
Example 2:
Input: nums = [2,3,4,5], queries = [1] Output: [0] Explanation: The empty subsequence is the only subsequence that has a sum less than or equal to 1, so answer[0] = 0.
Constraints:
n == nums.length
m == queries.length
1 <= n, m <= 1000
1 <= nums[i], queries[i] <= 106
class Solution:
def answerQueries(self, nums: List[int], queries: List[int]) -> List[int]:
nums.sort()
accumulated_data = list(accumulate(nums))
return [bisect_right(accumulated_data, q) for q in queries]
# ans = []
# for q in queries:
# way = -1
# s = 0
# for i, num in enumerate(nums):
# s+=num
# if num > q:
# way = i
# break
# elif s > q:
# way = i
# break
# elif s == q:
# way = i+1
# break
# if way == -1:
# way = len(nums)
# ans.append(way)
# return ans