LeetCode 2542. Maximum Subsequence Score Solution in Java, C++, Python & Go | Explanation + Code

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 × log n), and the space complexity is O(n). Here, n is the length of the array nums1.

PythonJavaC++Go

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(code-box)

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; } }(code-box)

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; } };(code-box)

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 }(code-box)