2542. Maximum Subsequence Score

Description

You are given two 0-indexed integer arrays nums1 and nums2 of equal length n and a positive integer k. You must choose a subsequence of indices from nums1 of length k.

For chosen indices i0, i1, ..., ik - 1, your score is defined as:

• The sum of the selected elements from nums1 multiplied with the minimum of the selected elements from nums2.
• It can defined simply as: (nums1[i0] + nums1[i1] +...+ nums1[ik - 1]) * min(nums2[i0] , nums2[i1], ... ,nums2[ik - 1]).

Return the maximum possible score.

A subsequence of indices of an array is a set that can be derived from the set {0, 1, ..., n-1} by deleting some or no elements.

Example 1:

Input: nums1 = [1,3,3,2], nums2 = [2,1,3,4], k = 3
Output: 12
Explanation:
The four possible subsequence scores are:
- We choose the indices 0, 1, and 2 with score = (1+3+3) * min(2,1,3) = 7.
- We choose the indices 0, 1, and 3 with score = (1+3+2) * min(2,1,4) = 6.
- We choose the indices 0, 2, and 3 with score = (1+3+2) * min(2,3,4) = 12.
- We choose the indices 1, 2, and 3 with score = (3+3+2) * min(1,3,4) = 8.
Therefore, we return the max score, which is 12.


Example 2:

Input: nums1 = [4,2,3,1,1], nums2 = [7,5,10,9,6], k = 1
Output: 30
Explanation:
Choosing index 2 is optimal: nums1[2] * nums2[2] = 3 * 10 = 30 is the maximum possible score.


Constraints:

• n == nums1.length == nums2.length
• 1 <= n <= 105
• 0 <= nums1[i], nums2[j] <= 105
• 1 <= k <= n

Solutions

Solution 1: Sorting + Priority Queue (Min Heap)

Sort nums2 and nums1 in descending order according to nums2, then traverse from front to back, maintaining a min heap. The heap stores elements from nums1, and the number of elements in the heap does not exceed $k$. At the same time, maintain a variable $s$ representing the sum of the elements in the heap, and continuously update the answer during the traversal process.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array nums1.

• class Solution {
public long maxScore(int[] nums1, int[] nums2, int k) {
int n = nums1.length;
int[][] nums = new int[n][2];
for (int i = 0; i < n; ++i) {
nums[i] = new int[] {nums1[i], nums2[i]};
}
Arrays.sort(nums, (a, b) -> b[1] - a[1]);
long ans = 0, s = 0;
PriorityQueue<Integer> q = new PriorityQueue<>();
for (int i = 0; i < n; ++i) {
s += nums[i][0];
q.offer(nums[i][0]);
if (q.size() == k) {
ans = Math.max(ans, s * nums[i][1]);
s -= q.poll();
}
}
return ans;
}
}

• class Solution {
public:
long long maxScore(vector<int>& nums1, vector<int>& nums2, int k) {
int n = nums1.size();
vector<pair<int, int>> nums(n);
for (int i = 0; i < n; ++i) {
nums[i] = {-nums2[i], nums1[i]};
}
sort(nums.begin(), nums.end());
priority_queue<int, vector<int>, greater<int>> q;
long long ans = 0, s = 0;
for (auto& [a, b] : nums) {
s += b;
q.push(b);
if (q.size() == k) {
ans = max(ans, s * -a);
s -= q.top();
q.pop();
}
}
return ans;
}
};

• class Solution:
def maxScore(self, nums1: List[int], nums2: List[int], k: int) -> int:
nums = sorted(zip(nums2, nums1), reverse=True)
q = []
ans = s = 0
for a, b in nums:
s += b
heappush(q, b)
if len(q) == k:
ans = max(ans, s * a)
s -= heappop(q)
return ans


• func maxScore(nums1 []int, nums2 []int, k int) int64 {
type pair struct{ a, b int }
nums := []pair{}
for i, a := range nums1 {
b := nums2[i]
nums = append(nums, pair{a, b})
}
sort.Slice(nums, func(i, j int) bool { return nums[i].b > nums[j].b })
q := hp{}
var ans, s int
for _, e := range nums {
a, b := e.a, e.b
s += a
heap.Push(&q, a)
if q.Len() == k {
ans = max(ans, s*b)
s -= heap.Pop(&q).(int)
}
}
return int64(ans)
}

type hp struct{ sort.IntSlice }

func (h hp) Less(i, j int) bool { return h.IntSlice[i] < h.IntSlice[j] }
func (h *hp) Push(v any)        { h.IntSlice = append(h.IntSlice, v.(int)) }
func (h *hp) Pop() any {
a := h.IntSlice
v := a[len(a)-1]
h.IntSlice = a[:len(a)-1]
return v
}