LeetCode #3584 — MEDIUM

Maximum Product of First and Last Elements of a Subsequence

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

Solve on LeetCode

The Problem

Problem Statement

You are given an integer array nums and an integer m.

Return the maximum product of the first and last elements of any subsequence of nums of size m.

Example 1:

Input: nums = [-1,-9,2,3,-2,-3,1], m = 1

Output: 81

Explanation:

The subsequence [-9] has the largest product of the first and last elements: -9 * -9 = 81. Therefore, the answer is 81.

Example 2:

Input: nums = [1,3,-5,5,6,-4], m = 3

Output: 20

Explanation:

The subsequence [-5, 6, -4] has the largest product of the first and last elements.

Example 3:

Input: nums = [2,-1,2,-6,5,2,-5,7], m = 2

Output: 35

Explanation:

The subsequence [5, 7] has the largest product of the first and last elements.

Constraints:

1 <= nums.length <= 10⁵
-10⁵ <= nums[i] <= 10⁵
1 <= m <= nums.length

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums and an integer m. Return the maximum product of the first and last elements of any subsequence of nums of size m.

Baseline thinking

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

Pattern signal: Array · Two Pointers

Example 1

[-1,-9,2,3,-2,-3,1]
1

Example 2

[1,3,-5,5,6,-4]
3

Example 3

[2,-1,2,-6,5,2,-5,7]
2

Step 02

Core Insight

What unlocks the optimal approach

We can select nums[i] as the first element of the subsequence, and the last one can be any of nums[i + m - 1], nums[i + m], ..., nums[n - 1].
If we select the first element from the largest i, the suffix just gets longer, and we can update the minimum and maximum values dynamically.

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 #3584: Maximum Product of First and Last Elements of a Subsequence
class Solution {
    public long maximumProduct(int[] nums, int m) {
        long ans = Long.MIN_VALUE;
        int mx = Integer.MIN_VALUE;
        int mi = Integer.MAX_VALUE;
        for (int i = m - 1; i < nums.length; ++i) {
            int x = nums[i];
            int y = nums[i - m + 1];
            mi = Math.min(mi, y);
            mx = Math.max(mx, y);
            ans = Math.max(ans, Math.max(1L * x * mi, 1L * x * mx));
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3584: Maximum Product of First and Last Elements of a Subsequence
func maximumProduct(nums []int, m int) int64 {
	ans := int64(math.MinInt64)
	mx := math.MinInt32
	mi := math.MaxInt32

	for i := m - 1; i < len(nums); i++ {
		x := nums[i]
		y := nums[i-m+1]
		mi = min(mi, y)
		mx = max(mx, y)
		ans = max(ans, max(int64(x)*int64(mi), int64(x)*int64(mx)))
	}

	return ans
}

# Accepted solution for LeetCode #3584: Maximum Product of First and Last Elements of a Subsequence
class Solution:
    def maximumProduct(self, nums: List[int], m: int) -> int:
        ans = mx = -inf
        mi = inf
        for i in range(m - 1, len(nums)):
            x = nums[i]
            y = nums[i - m + 1]
            mi = min(mi, y)
            mx = max(mx, y)
            ans = max(ans, x * mi, x * mx)
        return ans

// Accepted solution for LeetCode #3584: Maximum Product of First and Last Elements of a Subsequence
// 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 #3584: Maximum Product of First and Last Elements of a Subsequence
// class Solution {
//     public long maximumProduct(int[] nums, int m) {
//         long ans = Long.MIN_VALUE;
//         int mx = Integer.MIN_VALUE;
//         int mi = Integer.MAX_VALUE;
//         for (int i = m - 1; i < nums.length; ++i) {
//             int x = nums[i];
//             int y = nums[i - m + 1];
//             mi = Math.min(mi, y);
//             mx = Math.max(mx, y);
//             ans = Math.max(ans, Math.max(1L * x * mi, 1L * x * mx));
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3584: Maximum Product of First and Last Elements of a Subsequence
function maximumProduct(nums: number[], m: number): number {
    let ans = Number.MIN_SAFE_INTEGER;
    let mx = Number.MIN_SAFE_INTEGER;
    let mi = Number.MAX_SAFE_INTEGER;

    for (let i = m - 1; i < nums.length; i++) {
        const x = nums[i];
        const y = nums[i - m + 1];
        mi = Math.min(mi, y);
        mx = Math.max(mx, y);
        ans = Math.max(ans, x * mi, x * mx);
    }

    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)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

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.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.