README.md

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

Solutions (Rust)

1. Solution

impl Solution {
    pub fn answer_queries(nums: Vec<i32>, queries: Vec<i32>) -> Vec<i32> {
        let mut nums = nums;
        let mut answer = vec![0; queries.len()];

        nums.sort_unstable();
        for i in 1..nums.len() {
            nums[i] += nums[i - 1];
        }

        for i in 0..queries.len() {
            answer[i] = match nums.binary_search(&queries[i]) {
                Ok(j) => j as i32 + 1,
                Err(j) => j as i32,
            };
        }

        answer
    }
}