3652. Best Time to Buy and Sell Stock using Strategy

[mah-shamim]

Topics: Array, Sliding Window, Prefix Sum, Weekly Contest 463

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.

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

Hint:

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

[mah-shamim]

We can solve this problem efficiently using prefix sums to compute the profit change for any modification in O(1) time, after O(n) preprocessing.

Approach:

• Compute Base Profit: Start by calculating the original profit without any modifications.
• Prefix Sums for Efficiency: Precompute prefix sums to quickly calculate sums over any subarray in O(1) time.
• Analyze Modification Impact: For each possible modification window, compute the profit change compared to the original.
• Optimize with Sliding Window: Consider all possible windows of length k and track the maximum improvement.
• Combine Results: Add the maximum possible improvement to the base profit.

Let's implement this solution in PHP: 3652. Best Time to Buy and Sell Stock using Strategy

<?php
/**
 * @param Integer[] $prices
 * @param Integer[] $strategy
 * @param Integer $k
 * @return Integer
 */
function maxProfit($prices, $strategy, $k) {
    $n = count($prices);
    $baseProfit = 0;
    $prefixPrices = array_fill(0, $n + 1, 0);
    $prefixSp = array_fill(0, $n + 1, 0); // prefix of strategy[i] * prices[i]

    // Build prefix arrays and compute base profit
    for ($i = 0; $i < $n; $i++) {
        $baseProfit += $strategy[$i] * $prices[$i];
        $prefixPrices[$i + 1] = $prefixPrices[$i] + $prices[$i];
        $prefixSp[$i + 1] = $prefixSp[$i] + $strategy[$i] * $prices[$i];
    }

    $maxDelta = 0;
    $half = $k / 2; // k is even

    // Try every possible modification window
    for ($l = 0; $l <= $n - $k; $l++) {
        $sumPricesSecondHalf = $prefixPrices[$l + $k] - $prefixPrices[$l + $half];
        $sumOldSegment = $prefixSp[$l + $k] - $prefixSp[$l];
        $delta = $sumPricesSecondHalf - $sumOldSegment;
        if ($delta > $maxDelta) {
            $maxDelta = $delta;
        }
    }

    return $baseProfit + $maxDelta;
}

// Test cases
echo maxProfit([4,2,8], [-1,0,1], 2) . "\n";    // Output: 10
echo maxProfit([5,4,3], [1,1,0], 2) . "\n";     // Output: 9
?>

Explanation:

1. Base Profit Calculation:

   • Calculate the initial profit as sum(strategy[i] * prices[i]) for all days.
   • This represents the profit without any modifications.

2. Prefix Sum Arrays:

   • Create prefixPrices: cumulative sum of prices, allows O(1) range sum queries.
   • Create prefixSp: cumulative sum of strategy[i] * prices[i], allows O(1) queries for original profit in any window.

3. Understanding the Modification:

   • When modifying a window of length k:
     • First k/2 elements become 0 (hold) → their contribution becomes 0.
     • Last k/2 elements become 1 (sell) → their contribution becomes prices[i].
   • Original profit in window: sum(strategy[i] * prices[i]) for i in window.
   • New profit in window: sum(prices[i]) for i in second half of window.
   • Change in profit (delta) = new_profit - old_profit.

4. Efficient Delta Calculation:

   • For window [l, l+k-1]:
     • sumPricesSecondHalf = sum of prices from l+half to l+k-1.
     • sumOldSegment = original profit contribution from entire window.
     • delta = sumPricesSecondHalf - sumOldSegment.

5. Iterate Through All Windows:

   • Try every possible starting position l where window fits (0 ≤ l ≤ n-k).
   • Track maximum delta found.
   • If no positive delta exists, no modification is beneficial.

6. Final Result:

   • Maximum profit = baseProfit + max(0, maxDelta).

Complexity Analysis

• Time Complexity: O(n) - Single pass to build prefix arrays, then O(n) to check all windows.
• Space Complexity: O(n) - For storing prefix sums.

[topugit]

Thanks for solving the problem of "Best Time to Buy and Sell Stock using Strategy"

[mah-shamim]

Thanks