2389. Longest Subsequence With Limited Sum

문제)

2389. Longest Subsequence With Limited Sum

 

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

솔루션1)

- 재귀로 풀었더니 TLE 발생

class Solution:
    def answerQueries(self, nums: List[int], queries: List[int]) -> List[int]:        
        nums_len = len(nums)
        
        def recur(idx, subsequence_sum, subsequence_len, target):
            if subsequence_sum > target:
                return subsequence_len - 1
            
            if idx >= nums_len:   
                return subsequence_len
            
            return max(recur(idx + 1, subsequence_sum + nums[idx], subsequence_len + 1, target), 
                       recur(idx + 1, subsequence_sum, subsequence_len, target))
            
        ret = []
        for target in queries:
            ret.append(recur(0, 0, 0, target))
            
        return ret

 

솔루션2)

- nums를 정렬하고 accumulate() 사용해서 누적합 리스트를 생성

- 이진 탐색을 이용해서 문제 해결

 

문제를 풀고나면 이렇게도 풀리는구나.. 너무 허탈...

문제가 막혀있을 때 심플한 방법을 계속 고민하자!

class Solution:
    def answerQueries(self, nums: List[int], queries: List[int]) -> List[int]:        
        # 오름차순 정렬
        nums = sorted(nums)        
        
        # 누적합
        nums = list(accumulate(nums))
        
        # 이진탐색. 실패 시 오른쪽 인덱스 반환
        return [bisect.bisect_right(nums, target) for target in queries]