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] <= 10^6

解題

func answerQueries(nums []int, queries []int) []int {
    sort.Ints(nums)
    sum := make([]int, len(nums))
    sum[0] = nums[0]

    for i:=1; i<len(nums); i++ {
        sum[i] = sum[i - 1] + nums[i]
    }

    ans := make([]int, 0)
    for i1, query := range queries {
        for i, s := range sum {
            if s > query {
                ans = append(ans, i)
                break
            }
        }

        if len(ans) < i1 + 1 {
            ans = append(ans, len(sum))
        }
    }

    return ans
}