Description Link to heading

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] <= 10⁶

Solution Link to heading

We can sort the array first, it doesn’t affect the sum of the subsequence.

Then we can calculate the prefix sum of the sorted array, then we can get the minimum sum of the subsequence whose length is l.

Then for each queries[i], we can use binary search to get the maximum length of the subsequence.

Code Link to heading 

class Solution {
  public:
    int Bfind(vector<int> &prefix, int target) {
        int left = 0, right = prefix.size();
        while (left < right) {
            int mid = left + (right - left) / 2;
            if (prefix[mid] < target) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }
        return left;
    }
    vector<int> answerQueries(vector<int> &nums, vector<int> &queries) {
        std::sort(nums.begin(), nums.end());
        int n = nums.size();
        vector<int> prefix(n + 1);
        for (int i = 1; i <= n; ++i) {
            prefix[i] = prefix[i - 1] + nums[i - 1];
        }
        vector<int> res(queries.size());
        for (int i = 0; i < queries.size(); ++i) {
            res[i] = Bfind(prefix, queries[i] + 1) - 1;
        }
        return res;
    }
};