2099

2099. Find Subsequence of Length K With the Largest Sum (Easy)

You are given an integer array nums and an integer k. You want to find a subsequence of nums of length k that has the largest sum.

Return any such subsequence as an integer array of length k.

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 = [2,1,3,3], k = 2
Output: [3,3]
Explanation:
The subsequence has the largest sum of 3 + 3 = 6.

Example 2:

Input: nums = [-1,-2,3,4], k = 3
Output: [-1,3,4]
Explanation: 
The subsequence has the largest sum of -1 + 3 + 4 = 6.

Example 3:

Input: nums = [3,4,3,3], k = 2
Output: [3,4]
Explanation:
The subsequence has the largest sum of 3 + 4 = 7. 
Another possible subsequence is [4, 3].

 

Constraints:

• 1 <= nums.length <= 1000
• -105 <= nums[i] <= 105
• 1 <= k <= nums.length

Similar Questions:

• Kth Largest Element in an Array (Medium)
• Maximize Sum Of Array After K Negations (Easy)
• Sort Integers by The Number of 1 Bits (Easy)

Solution 1. Sorting

// OJ: https://leetcode.com/problems/find-subsequence-of-length-k-with-the-largest-sum/
// Author: github.com/lzl124631x
// Time: O(NlogN)
// Space: O(N)
class Solution {
public:
    vector<int> maxSubsequence(vector<int>& A, int k) {
        vector<int> id(A.size());
        iota(begin(id), end(id), 0); // Index array 0, 1, 2, ...
        sort(begin(id), end(id), [&](int a, int b) { return A[a] > A[b]; }); // Sort the indexes in descending order of their corresponding values in `A`
        id.resize(k); // Only keep the first `k` indexes with the greatest `A` values
        sort(begin(id), end(id)); // Sort indexes in ascending order
        vector<int> ans;
        for (int i : id) ans.push_back(A[i]);
        return ans;
    }
};

Solution 2. Quick Select

// OJ: https://leetcode.com/problems/find-subsequence-of-length-k-with-the-largest-sum/
// Author: github.com/lzl124631x
// Time: O(N + KlogK) on average, O(N^2 + KlogK) in the worst case
// Space: O(N)
class Solution {
public:
    vector<int> maxSubsequence(vector<int>& A, int k) {
        vector<int> id(A.size());
        iota(begin(id), end(id), 0);
        nth_element(begin(id), begin(id) + k, end(id), [&](int a, int b) { return A[a] > A[b]; });
        id.resize(k);
        sort(begin(id), end(id));
        vector<int> ans;
        for (int i : id) ans.push_back(A[i]);
        return ans;
    }
};

Solution 3. Quick Select

// OJ: https://leetcode.com/problems/find-subsequence-of-length-k-with-the-largest-sum/
// Author: github.com/lzl124631x
// Time: O(N) on average, O(N^2) in the worst case
// Space: O(N)
class Solution {
public:
    vector<int> maxSubsequence(vector<int>& A, int k) {
        if (k == A.size()) return A;
        vector<int> v(begin(A), end(A)), ans;
        nth_element(begin(v), begin(v) + k - 1, end(v), greater<>());
        int cnt = count(begin(v), begin(v) + k, v[k - 1]);
        for (int i = 0; i < A.size(); ++i) {
            if (A[i] > v[k - 1] || (A[i] == v[k - 1] && --cnt >= 0)) ans.push_back(A[i]);
        }
        return ans;
    }
};

Discuss

https://leetcode.com/problems/find-subsequence-of-length-k-with-the-largest-sum/discuss/1623427/C%2B%2B-Sorting-or-Quick-Select