Longest Subsequence With Limited Sum

Problem Statement

In this problem, we are given two integer arrays, nums and queries. The objective is to determine, for each query, the maximum size of a subsequence from nums such that the sum of the subsequence's elements is less than or equal to the query value. A subsequence, by definition, can be formed by removing some or none of the elements from an array without altering the order of the remaining elements. The output should be an array answer, where each answer[i] corresponds to the subsequence size for queries[i].

Examples

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

Approach and Intuition

To understand how to tackle this problem, let's break down the steps needed based on the insights gained from the examples and the constraints:

1. Sort the Array: Begin by sorting nums in ascending order. This sorting assists in checking the smaller elements first, which are more likely to form valid subsequences under the sum constraints posed by queries.

2. Initialize the Answer Array: An answer array of length m (same as queries) should be initialized to record the result for each query.

3. Check Each Query with a Sorted nums:

   • For each query[i], initiate a sum variable to keep track of the running total of the subsequence and a count variable to count the number of elements included in this subsequence.
   • Traverse through the sorted nums and keep adding elements to the sum until adding another element would exceed the value of query[i].
   • Store the count in answer[i] which is the result for this particular query.

4. Edge Considerations:

   • If nums contains only elements larger than the maximum query, the result will have answer[i] = 0 for some i because even the smallest element in nums is larger than the query.
   • All queries are addressed independently. Since arrays query and nums can have maximum lengths up to 1000, the described strategy should perform efficiently under the given constraints.

By sorting nums and independently checking each query, efficient determination of maximum subsequence size for each query is ensured, aligning with the requirements defined in the problem statement. This solution is effective due to its clarity and direct approach towards the problem's needs.

Solutions

class Solution {
public:
    vector<int> processQueries(vector<int>& values, vector<int>& queries) {
        // Sorting values and computing prefix sums in place
        sort(values.begin(), values.end());
        for (int j = 1; j < values.size(); ++j)
            values[j] += values[j - 1];
    
        // Process each query to find how many elements have a sum less than or equal to query
        vector<int> results;
        for (auto q : queries) {
            int pos = upper_bound(values.begin(), values.end(), q) - values.begin();
            results.push_back(pos);
        }
            
        return results;
    }
};