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You are given an integer array nums of length n, and an integer array queries of length m. For each queries[i], find the maximum length of a subsequence that can be obtained from nums such that the sum of the subsequence’s elements does not exceed queries[i].

Return an array answer of length m where answer[i] is the answer to the i-th query.

Clarifying Questions

  1. Can nums contain negative numbers?
    • No, nums only contains non-negative integers.
  2. Can the elements of nums be zero?
    • Yes, elements of nums can be zeros.
  3. Is the order of the elements in the subsequence important?
    • No, the order is not important; only the sum constraint matters.
  4. What are the constraints on the lengths of nums and queries?
    • 1 <= nums.length, queries.length <= 1000
    • 0 <= nums[i], queries[i] <= 10^6

Strategy

Given the constraints, we can utilize sorting and prefix sums to efficiently determine the maximum length of the subsequence for each query.

  1. Sort the nums array: By sorting, we can easily find the smallest elements to include in the subsequence to keep the sum under the given limit.
  2. Compute prefix sums: Compute the prefix sums of the sorted nums array. The prefix sums will help us quickly determine how many elements we can include for a given sum.
  3. Binary search for each query: For each query, use binary search on the prefix sums to determine the maximum number of elements that can form a subsequence within the given sum.

Code

from bisect import bisect_right

def answerQueries(nums, queries):
    # Sort the numbers to make subsequence selection easier
    nums.sort()
    
    # Compute the prefix sums of the sorted nums
    prefix_sums = []
    current_sum = 0
    for num in nums:
        current_sum += num
        prefix_sums.append(current_sum)
    
    answers = []
    for query in queries:
        # Find the maximum length subsequence with sum <= query
        pos = bisect_right(prefix_sums, query)
        answers.append(pos)
    
    return answers

# Example usage
nums = [4, 5, 2, 1]
queries = [3, 10, 21]
print(answerQueries(nums, queries)) # Output: [2, 3, 4]

Time Complexity

Overall time complexity: O(n log n + m log n), where n is the length of nums and m is the length of queries. Given the constraints, this approach will be efficient and suitable.

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