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

Random Note


when you have something circular, most of the time you can have one variableand another can be calculated using size, current value and doing %