2389. Longest Subsequence With Limited Sum.cpp

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









// TC -> O(nlogn) + O(mlogn)
// SC -> O(1)


class Solution {
public:
    vector<int> answerQueries(vector<int>& nums, vector<int>& queries) {
        int n = nums.size();
        vector <int> res;
        sort(nums.begin(), nums.end());
        for (int i = 1; i < n; i++) nums[i] += nums[i - 1];
        for (auto &sum: queries) {
            int index = lower_bound(nums.begin(), nums.end(), sum) - nums.begin();
            if (index < n && nums[index] == sum)
                res.push_back(index + 1);
            else res.push_back(index);
        }
        return res;
    }
};