2389. Longest Subsequence With Limited Sum

Problem

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

Solution (Java)

class Solution {
    public int[] answerQueries(int[] nums, int[] queries) {
        // we can sort the nums because the order of the subsequence does not matter
        Arrays.sort(nums);
        for (int i = 1; i < nums.length; i++) {
            nums[i] = nums[i] + nums[i - 1];
        }
        for (int i = 0; i < queries.length; i++) {
            int j = Arrays.binarySearch(nums, queries[i]);
            if (j < 0) {
                j = -j - 2;
            }
            queries[i] = j + 1;
        }
        return queries;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).

Solution (Javascript)

/**
 * @param {number[]} nums
 * @param {number[]} queries
 * @return {number[]}
 */
var answerQueries = function(nums, queries) {
  nums.sort((a, b) => a - b);
  queries = queries.map((query, idx) => [query, idx]).sort((a, b) => a[0] - b[0]);
  let n = nums.length, m = queries.length;
  let res = Array(m).fill(0);
  let sum = 0;
  for (let j = 0, i = 0; j < m; j++) {
    let [maxSum, index] = queries[j];
    while (i < n && sum + nums[i] <= maxSum) {
      sum += nums[i++];
    }
    res[index] = i;
  }
  return res;
};

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).