Longest Subsequence With Limited Sum

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

Code

注意：subsequence 和 subarray 不一樣！subsequence 不需要是連續的。因為是 subsequence 所以可以直接 sort nums。

有用到 Search Insert Position 去找 query 該被插入的位置（index 即是題目所求之最大覆蓋長度）。

class Solution {
public:
    vector<int> answerQueries(vector<int>& nums, vector<int>& queries) {
        int qSize = queries.size();
        vector<int> answer(qSize, 0);
        
        sort(nums.begin(), nums.end());
        for(int i = 1; i < nums.size(); i++) {
            nums[i] += nums[i-1];
        }
        for(int i = 0; i < queries.size(); i++) {
            answer[i] = find(nums, queries[i]);
        }
        return answer;
    }
 
    int find(vector<int> &nums, int query) {
        int l = 0, r = nums.size();
        while(l < r) {
            int mid = l + (r - l) / 2;
            if (nums[mid] > query) r = mid;
            else l = mid + 1;
        }
        return l;
    } 
};

find 函式可以用 std::upper_bound 來代替（return 大於指定搜尋元素之 index）

class Solution {
public:
    vector<int> answerQueries(vector<int>& nums, vector<int>& queries) {
        int qSize = queries.size();
        vector<int> answer(qSize, 0);
        
        sort(nums.begin(), nums.end());
        for(int i = 1; i < nums.size(); i++) {
            nums[i] += nums[i-1];
        }
        for(int i = 0; i < queries.size(); i++) {
            answer[i] = upper_bound(nums.begin(), nums.end(), queries[i]) - nums.begin();
    
        }
        return answer;
    }
 
};

Sliding Window

比較慢的解法。

class Solution {
public:
    vector<int> answerQueries(vector<int>& nums, vector<int>& queries) {
        sort(nums.begin(), nums.end());
        int n = queries.size();
        vector<int> res(n, 0);
        for(int k = 0; k < queries.size(); k++) {
            int i = 0;
            int curSum = 0;
            for(int j = 0; j < nums.size(); j++) {
                curSum += nums[j];
                while(curSum > queries[k]) {
                    curSum -= nums[i];
                    i++;
                }
                if(curSum <= queries[k]) {
                    res[k] = max(res[k], j - i + 1);
                }
            }
        }
        return res;
    }
};

Link

• Longest Subsequence With Limited Sum