LeetCode #2542 — MEDIUM

Maximum Subsequence Score

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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, ..., i_{k - 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[i_{k - 1}]) * min(nums2[i₀] , nums2[i₁], ... ,nums2[i_{k - 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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Greedy

Example 1

[1,3,3,2]
[2,1,3,4]
3

Example 2

[4,2,3,1,1]
[7,5,10,9,6]
1

Core Insight

What unlocks the optimal approach

How can we use sorting here?
Try sorting the two arrays based on second array.
Loop through nums2 and compute the max product given the minimum is nums2[i]. Update the answer accordingly.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2542: Maximum Subsequence Score
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;
    }
}

// Accepted solution for LeetCode #2542: Maximum Subsequence Score
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
}

# Accepted solution for LeetCode #2542: Maximum Subsequence Score
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

// Accepted solution for LeetCode #2542: Maximum Subsequence Score
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2542: Maximum Subsequence Score
// 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;
//     }
// }

// Accepted solution for LeetCode #2542: Maximum Subsequence Score
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #2542: Maximum Subsequence Score
// 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;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.