How do you solve Longest Subsequence With Limited Sum on LeetCode?

Longest Subsequence With Limited Sum (LeetCode #2389) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n + m log n) time complexity and O(n) space complexity. Key insight: Sorting the array allows for efficient sum calculations.

What is the brute force solution for Longest Subsequence With Limited Sum?

The brute force approach for Longest Subsequence With Limited Sum is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves generating all possible subsequences of the array and checking their sums against each query. This method is straightforward but inefficient, especially as the size of the input grows. The optimal Optimal Solution approach improves this to O(n log n + m log n).

What algorithm or data structure is used to solve Longest Subsequence With Limited Sum?

Longest Subsequence With Limited Sum uses the following concepts: Array, Binary Search, Greedy, Sorting, Prefix Sum. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Longest Subsequence With Limited Sum?

Always consider sorting and prefix sums for array-related problems. Practice binary search as it's a common technique. Think about the constraints and optimize accordingly.

What are common mistakes when solving Longest Subsequence With Limited Sum?

Not sorting the array before calculating sums. Failing to handle edge cases like empty subsequences.

#2389

Longest Subsequence With Limited Sum

Easy

ArrayBinary SearchGreedySortingPrefix SumHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n + m log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n + m log n)Space O(n)

By sorting the `nums` array and using a prefix sum approach, we can efficiently determine the maximum size of a subsequence for each query. This method leverages the fact that smaller numbers contribute more to the sum, allowing us to maximize the size of the subsequence.

⚙️

Algorithm

3 steps

1Step 1: Sort the `nums` array in ascending order.
2Step 2: Compute the prefix sums of the sorted array.
3Step 3: For each query, use binary search to find the largest index where the prefix sum is less than or equal to the query.

solution.py13 lines

1# Full working Python code
2import bisect
3
4def maxSubsequenceSize(nums, queries):
5    nums.sort()
6    prefix_sum = [0] * (len(nums) + 1)
7    for i in range(1, len(nums) + 1):
8        prefix_sum[i] = prefix_sum[i - 1] + nums[i - 1]
9    results = []
10    for query in queries:
11        max_size = bisect.bisect_right(prefix_sum, query) - 1
12        results.append(max_size)
13    return results

ℹ

Complexity note: The sorting step takes O(n log n), and for each query, we perform a binary search on the prefix sums, which takes O(log n). Thus, the overall complexity is dominated by the sorting step.

1Sorting the array allows for efficient sum calculations.
2Using prefix sums simplifies the problem of finding valid subsequences.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.