LeetCode #3652 — MEDIUM

Best Time to Buy and Sell Stock using Strategy

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

The Problem

Problem Statement

You are given two integer arrays prices and strategy, where:

prices[i] is the price of a given stock on the i^th day.
strategy[i] represents a trading action on the i^th day, where:
- -1 indicates buying one unit of the stock.
- 0 indicates holding the stock.
- 1 indicates selling one unit of the stock.

You are also given an even integer k, and may perform at most one modification to strategy. A modification consists of:

Selecting exactly k consecutive elements in strategy.
Set the first k / 2 elements to 0 (hold).
Set the last k / 2 elements to 1 (sell).

The profit is defined as the sum of strategy[i] * prices[i] across all days.

Return the maximum possible profit you can achieve.

Note: There are no constraints on budget or stock ownership, so all buy and sell operations are feasible regardless of past actions.

Example 1:

Input: prices = [4,2,8], strategy = [-1,0,1], k = 2

Output: 10

Explanation:

Modification	Strategy	Profit Calculation	Profit
Original	[-1, 0, 1]	(-1 × 4) + (0 × 2) + (1 × 8) = -4 + 0 + 8	4
Modify [0, 1]	[0, 1, 1]	(0 × 4) + (1 × 2) + (1 × 8) = 0 + 2 + 8	10
Modify [1, 2]	[-1, 0, 1]	(-1 × 4) + (0 × 2) + (1 × 8) = -4 + 0 + 8	4

Thus, the maximum possible profit is 10, which is achieved by modifying the subarray [0, 1].

Example 2:

Input: prices = [5,4,3], strategy = [1,1,0], k = 2

Output: 9

Explanation:

Modification	Strategy	Profit Calculation	Profit
Original	[1, 1, 0]	(1 × 5) + (1 × 4) + (0 × 3) = 5 + 4 + 0	9
Modify [0, 1]	[0, 1, 0]	(0 × 5) + (1 × 4) + (0 × 3) = 0 + 4 + 0	4
Modify [1, 2]	[1, 0, 1]	(1 × 5) + (0 × 4) + (1 × 3) = 5 + 0 + 3	8

Thus, the maximum possible profit is 9, which is achieved without any modification.

Constraints:

2 <= prices.length == strategy.length <= 10⁵
1 <= prices[i] <= 10⁵
-1 <= strategy[i] <= 1
2 <= k <= prices.length
k is even

Step 01

Brute Force Baseline

Problem summary: You are given two integer arrays prices and strategy, where: prices[i] is the price of a given stock on the ith day. strategy[i] represents a trading action on the ith day, where: -1 indicates buying one unit of the stock. 0 indicates holding the stock. 1 indicates selling one unit of the stock. You are also given an even integer k, and may perform at most one modification to strategy. A modification consists of: Selecting exactly k consecutive elements in strategy. Set the first k / 2 elements to 0 (hold). Set the last k / 2 elements to 1 (sell). The profit is defined as the sum of strategy[i] * prices[i] across all days. Return the maximum possible profit you can achieve. Note: There are no constraints on budget or stock ownership, so all buy and sell operations are feasible regardless of past actions.

Baseline thinking

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

Pattern signal: Array · Sliding Window

Example 1

[4,2,8]
[-1,0,1]
2

Example 2

[5,4,3]
[1,1,0]
2

Step 02

Core Insight

What unlocks the optimal approach

Use prefix sums to precompute the base profit and to get fast range queries (sums of <code>prices</code> and counts of each <code>strategy</code> value over any interval).
Try every segment of length <code>k</code>: compute the profit delta caused by replacing that segment (using the prefix queries) and take the maximum of <code>base + delta</code>.

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 #3652: Best Time to Buy and Sell Stock using Strategy
class Solution {
    public long maxProfit(int[] prices, int[] strategy, int k) {
        int n = prices.length;
        long[] s = new long[n + 1];
        long[] t = new long[n + 1];
        for (int i = 1; i <= n; i++) {
            int a = prices[i - 1];
            int b = strategy[i - 1];
            s[i] = s[i - 1] + a * b;
            t[i] = t[i - 1] + a;
        }
        long ans = s[n];
        for (int i = k; i <= n; i++) {
            ans = Math.max(ans, s[n] - (s[i] - s[i - k]) + (t[i] - t[i - k / 2]));
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3652: Best Time to Buy and Sell Stock using Strategy
func maxProfit(prices []int, strategy []int, k int) int64 {
	n := len(prices)
	s := make([]int64, n+1)
	t := make([]int64, n+1)

	for i := 1; i <= n; i++ {
		a := prices[i-1]
		b := strategy[i-1]
		s[i] = s[i-1] + int64(a*b)
		t[i] = t[i-1] + int64(a)
	}

	ans := s[n]
	for i := k; i <= n; i++ {
		ans = max(ans, s[n]-(s[i]-s[i-k])+(t[i]-t[i-k/2]))
	}
	return ans
}

# Accepted solution for LeetCode #3652: Best Time to Buy and Sell Stock using Strategy
class Solution:
    def maxProfit(self, prices: List[int], strategy: List[int], k: int) -> int:
        n = len(prices)
        s = [0] * (n + 1)
        t = [0] * (n + 1)
        for i, (a, b) in enumerate(zip(prices, strategy), 1):
            s[i] = s[i - 1] + a * b
            t[i] = t[i - 1] + a
        ans = s[n]
        for i in range(k, n + 1):
            ans = max(ans, s[n] - (s[i] - s[i - k]) + t[i] - t[i - k // 2])
        return ans

// Accepted solution for LeetCode #3652: Best Time to Buy and Sell Stock using Strategy
impl Solution {
    pub fn max_profit(prices: Vec<i32>, strategy: Vec<i32>, k: i32) -> i64 {
        let n: usize = prices.len();
        let k: usize = k as usize;

        let mut s: Vec<i64> = vec![0; n + 1];
        let mut t: Vec<i64> = vec![0; n + 1];

        for i in 1..=n {
            let a: i64 = prices[i - 1] as i64;
            let b: i64 = strategy[i - 1] as i64;
            s[i] = s[i - 1] + a * b;
            t[i] = t[i - 1] + a;
        }

        let mut ans: i64 = s[n];
        for i in k..=n {
            let cur = s[n] - (s[i] - s[i - k]) + (t[i] - t[i - k / 2]);
            if cur > ans {
                ans = cur;
            }
        }

        ans
    }
}

// Accepted solution for LeetCode #3652: Best Time to Buy and Sell Stock using Strategy
function maxProfit(prices: number[], strategy: number[], k: number): number {
    const n = prices.length;
    const s: number[] = Array(n + 1).fill(0);
    const t: number[] = Array(n + 1).fill(0);

    for (let i = 1; i <= n; i++) {
        const a = prices[i - 1];
        const b = strategy[i - 1];
        s[i] = s[i - 1] + a * b;
        t[i] = t[i - 1] + a;
    }

    let ans = s[n];
    for (let i = k; i <= n; i++) {
        const val = s[n] - (s[i] - s[i - k]) + (t[i] - t[i - Math.floor(k / 2)]);
        ans = Math.max(ans, val);
    }
    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(n)

Approach Breakdown

BRUTE FORCE

O(n × k) time

O(1) space

For each starting index, scan the next k elements to compute the window aggregate. There are n−k+1 starting positions, each requiring O(k) work, giving O(n × k) total. No extra space since we recompute from scratch each time.

SLIDING WINDOW

O(n) time

O(k) space

The window expands and contracts as we scan left to right. Each element enters the window at most once and leaves at most once, giving 2n total operations = O(n). Space depends on what we track inside the window (a hash map of at most k distinct elements, or O(1) for a fixed-size window).

Shortcut: Each element enters and exits the window once → O(n) amortized, regardless of window size.

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.

Shrinking the window only once

Wrong move: Using `if` instead of `while` leaves the window invalid for multiple iterations.

Usually fails on: Over-limit windows stay invalid and produce wrong lengths/counts.

Fix: Shrink in a `while` loop until the invariant is valid again.