Description#
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
Solutions#
Solution 1#
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| class Solution:
def maxSubsequence(self, nums: List[int], k: int) -> List[int]:
idx = list(range(len(nums)))
idx.sort(key=lambda i: nums[i])
return [nums[i] for i in sorted(idx[-k:])]
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| class Solution {
public int[] maxSubsequence(int[] nums, int k) {
int[] ans = new int[k];
List<Integer> idx = new ArrayList<>();
int n = nums.length;
for (int i = 0; i < n; ++i) {
idx.add(i);
}
idx.sort(Comparator.comparingInt(i -> - nums[i]));
int[] t = new int[k];
for (int i = 0; i < k; ++i) {
t[i] = idx.get(i);
}
Arrays.sort(t);
for (int i = 0; i < k; ++i) {
ans[i] = nums[t[i]];
}
return ans;
}
}
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| class Solution {
public:
vector<int> maxSubsequence(vector<int>& nums, int k) {
int n = nums.size();
vector<pair<int, int>> vals;
for (int i = 0; i < n; ++i) vals.push_back({i, nums[i]});
sort(vals.begin(), vals.end(), [&](auto x1, auto x2) {
return x1.second > x2.second;
});
sort(vals.begin(), vals.begin() + k);
vector<int> ans;
for (int i = 0; i < k; ++i) ans.push_back(vals[i].second);
return ans;
}
};
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| func maxSubsequence(nums []int, k int) []int {
idx := make([]int, len(nums))
for i := range idx {
idx[i] = i
}
sort.Slice(idx, func(i, j int) bool { return nums[idx[i]] > nums[idx[j]] })
sort.Ints(idx[:k])
ans := make([]int, k)
for i, j := range idx[:k] {
ans[i] = nums[j]
}
return ans
}
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