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2389. Longest Subsequence With Limited Sum

Description

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] <= 106

Solutions

  • class Solution {
    public int[] answerQueries(int[] nums, int[] queries) {
        Arrays.sort(nums);
        for (int i = 1; i < nums.length; ++i) {
            nums[i] += nums[i - 1];
        }
        int m = queries.length;
        int[] ans = new int[m];
        for (int i = 0; i < m; ++i) {
            ans[i] = search(nums, queries[i]);
        }
        return ans;
    }

    private int search(int[] nums, int x) {
        int l = 0, r = nums.length;
        while (l < r) {
            int mid = (l + r) >> 1;
            if (nums[mid] > x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    }
}

  • class Solution {
public:
    vector<int> answerQueries(vector<int>& nums, vector<int>& queries) {
        sort(nums.begin(), nums.end());
        for (int i = 1; i < nums.size(); i++) {
            nums[i] += nums[i - 1];
        }
        vector<int> ans;
        for (auto& q : queries) {
            ans.push_back(upper_bound(nums.begin(), nums.end(), q) - nums.begin());
        }
        return ans;
    }
};

  • class Solution:
    def answerQueries(self, nums: List[int], queries: List[int]) -> List[int]:
        nums.sort()
        s = list(accumulate(nums))
        return [bisect_right(s, q) for q in queries]


  • func answerQueries(nums []int, queries []int) (ans []int) {
	sort.Ints(nums)
	for i := 1; i < len(nums); i++ {
		nums[i] += nums[i-1]
	}
	for _, q := range queries {
		ans = append(ans, sort.SearchInts(nums, q+1))
	}
	return
}

  • function answerQueries(nums: number[], queries: number[]): number[] {
    nums.sort((a, b) => a - b);
    for (let i = 1; i < nums.length; i++) {
        nums[i] += nums[i - 1];
    }
    const ans: number[] = [];
    const search = (nums: number[], x: number) => {
        let l = 0;
        let r = nums.length;
        while (l < r) {
            const mid = (l + r) >> 1;
            if (nums[mid] > x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    };
    for (const q of queries) {
        ans.push(search(nums, q));
    }
    return ans;
}


  • public class Solution {
    public int[] AnswerQueries(int[] nums, int[] queries) {
        int[] result = new int[queries.Length];
        Array.Sort(nums);
        for (int i = 0; i < queries.Length; i++) {
            result[i] = getSubsequent(nums, queries[i]);
        }
        return result;

    }

    public int getSubsequent(int[] nums,int query) {
        int sum = 0;
        for (int i = 0; i < nums.Length; i++) {
            sum += nums[i];
            if (sum > query) {
                return i;
            }
        }
        return nums.Length;
    }
}


  • impl Solution {
    pub fn answer_queries(mut nums: Vec<i32>, queries: Vec<i32>) -> Vec<i32> {
        let n = nums.len();
        nums.sort();
        queries
            .into_iter()
            .map(|query| {
                let mut sum = 0;
                for i in 0..n {
                    sum += nums[i];
                    if sum > query {
                        return i as i32;
                    }
                }
                n as i32
            })
            .collect()
    }
}

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