Leetcode 2389: Longest Subsequence With Limited Sum
You are given an integer array
nums of length n and an integer array queries of length m. For each query, return the maximum size of a subsequence that can be selected from nums such that the sum of its elements is less than or equal to the query value. A subsequence is derived by deleting some or no elements from the array while keeping the relative order intact.Problem
Approach
Steps
Complexity
Input: The input consists of an integer array `nums` and an integer array `queries`.
Example: Input: nums = [1,3,5,7], queries = [6,12,18]
Constraints:
• n == nums.length
• m == queries.length
• 1 <= n, m <= 1000
• 1 <= nums[i], queries[i] <= 10^6
Output: For each query, return the maximum size of the subsequence such that its sum is less than or equal to the query value.
Example: Output: [2, 3, 4]
Constraints:
Goal: The goal is to find the maximum size of the subsequence for each query where the sum does not exceed the query value.
Steps:
• 1. Sort the array `nums` to easily select the smallest elements first.
• 2. Compute the prefix sum of the sorted array to get the cumulative sum of elements.
• 3. For each query, use binary search to find the maximum number of elements whose sum is less than or equal to the query value.
Goal: Ensure the solution works efficiently within the provided constraints.
Steps:
• The array `nums` can have up to 1000 elements.
• Each element of `nums` and each query value is between 1 and 10^6.
Assumptions:
• Queries are independent and each query needs to be answered individually.
• Input: Input: nums = [1, 3, 5, 7], queries = [6, 12, 18]
• Explanation: For the query 6, the subsequence [1, 3] has a sum of 4, which is the largest subsequence with a sum <= 6, hence the answer is 2. For the query 12, the subsequence [1, 3, 5] has a sum of 9, and for 18, the subsequence [1, 3, 5, 7] has a sum of 16, so the answers are [2, 3, 4].
• Input: Input: nums = [2, 4, 6, 8], queries = [5]
• Explanation: For the query 5, the largest subsequence with a sum <= 5 is [2], so the answer is 1.
Approach: The approach involves sorting the array `nums`, calculating the prefix sums, and using binary search to efficiently answer each query.
Observations:
• We need to find the maximum subsequence for each query where the sum is within a given limit.
• Sorting `nums` allows us to efficiently calculate the largest subsequence by accumulating smaller numbers first.
• Prefix sums can help to efficiently check the sum of any subsequence. Binary search will help find the largest valid subsequence for each query.
Steps:
• 1. Sort the array `nums` in ascending order.
• 2. Calculate the prefix sum of the sorted array.
• 3. For each query, perform binary search on the prefix sum array to find the largest subsequence whose sum is less than or equal to the query.
Empty Inputs:
• If `nums` is empty, the answer for each query is 0.
Large Inputs:
• Ensure the solution is efficient enough to handle the maximum input size of 1000 elements in `nums`.
Special Values:
• Handle cases where the elements of `nums` are very large or queries are much smaller than the elements.
Constraints:
• The binary search must work within O(log n) for each query to ensure overall efficiency.
vector<int> answerQueries(vector<int> A, vector<int>& queries) {
sort(A.begin(), A.end());
vector<int> res;
for (int i = 1; i < A.size(); ++i)
A[i] += A[i - 1];
for (int& q: queries)
res.push_back(upper_bound(A.begin(), A.end(), q) - A.begin());
return res;
}
1 : Function Declaration
vector<int> answerQueries(vector<int> A, vector<int>& queries) {
Function declaration where the input is an array `A` and a reference to an array `queries`, and the return type is a vector of integers.
2 : Sort Array
sort(A.begin(), A.end());
Sort the input array `A` in ascending order to prepare for efficient prefix sum calculation.
3 : Variable Declaration
vector<int> res;
Declare a result vector `res` that will store the answers to the queries.
4 : Prefix Sum Calculation
for (int i = 1; i < A.size(); ++i)
Start a loop from index 1 to calculate the prefix sum by adding each element to its previous one.
5 : Prefix Sum Update
A[i] += A[i - 1];
Update the current element by adding the value of the previous element, thereby creating a running total.
6 : Query Processing
for (int& q: queries)
Iterate over each query in the `queries` array.
7 : Binary Search
res.push_back(upper_bound(A.begin(), A.end(), q) - A.begin());
Use the `upper_bound` function to find the index of the first element greater than the current query value `q` and push it to the result vector.
8 : Return Result
return res;
Return the result vector containing the answers to all the queries.
Best Case: O(n log n) for sorting and O(m log n) for answering queries.
Average Case: O(n log n) for sorting and O(m log n) for answering queries.
Worst Case: O(n log n) for sorting and O(m log n) for answering queries.
Description: Sorting the array `nums` takes O(n log n) time, and each query can be answered in O(log n) using binary search.
Best Case: O(n) for storing the prefix sums and the result array.
Worst Case: O(n) for storing the prefix sums and the result array.
Description: The space complexity is O(n) due to the storage required for the prefix sum and result arrays.
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