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Formatted question description: https://leetcode.ca/all/2389.html

2389. Longest Subsequence With Limited Sum

  • Difficulty: Easy.
  • Related Topics: Array, Binary Search, Greedy, Sorting, Prefix Sum.
  • Similar Questions: How Many Numbers Are Smaller Than the Current Number, Successful Pairs of Spells and Potions.

Problem

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

Solution (Java, C++, Python)

  • class Solution {
    public int[] answerQueries(int[] nums, int[] queries) {
        // we can sort the nums because the order of the subsequence does not matter
        Arrays.sort(nums);
        for (int i = 1; i < nums.length; i++) {
            nums[i] = nums[i] + nums[i - 1];
        }
        for (int i = 0; i < queries.length; i++) {
            int j = Arrays.binarySearch(nums, queries[i]);
            if (j < 0) {
                j = -j - 2;
            }
            queries[i] = j + 1;
        }
        return queries;
    }
}

############

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

############

# 2389. Longest Subsequence With Limited Sum
# https://leetcode.com/problems/longest-subsequence-with-limited-sum/

class Solution:
    def answerQueries(self, nums: List[int], queries: List[int]) -> List[int]:
        n = len(nums)
        nums.sort()
        res = []
        prefix = list(accumulate(nums, initial = 0))

        for q in queries:
            index = bisect_right(prefix, q) - 1
            res.append(index)

        return res
            


  • 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;
    }
};

  • 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;
}


  • 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()
    }
}

  • 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;
    }
}

Explain:

nope.

Complexity:

  • Time complexity : O(n).
  • Space complexity : O(n).

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