# 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 = 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 = 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 = 4.
Example 2:
Input: nums = [2,3,4,5], queries = 
Output: 
Explanation: The empty subsequence is the only subsequence that has a sum less than or equal to 1, so answer = 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 = nums
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
}