LeetCode in Kotlin

3573. Best Time to Buy and Sell Stock V

Medium

You are given an integer array prices where prices[i] is the price of a stock in dollars on the ith day, and an integer k.

You are allowed to make at most k transactions, where each transaction can be either of the following:

• Normal transaction: Buy on day i, then sell on a later day j where i < j. You profit prices[j] - prices[i].

• Short selling transaction: Sell on day i, then buy back on a later day j where i < j. You profit prices[i] - prices[j].

Note that you must complete each transaction before starting another. Additionally, you can’t buy or sell on the same day you are selling or buying back as part of a previous transaction.

Return the maximum total profit you can earn by making at most k transactions.

Example 1:

Input: prices = [1,7,9,8,2], k = 2

Output: 14

Explanation:

We can make $14 of profit through 2 transactions:

• A normal transaction: buy the stock on day 0 for $1 then sell it on day 2 for $9.
• A short selling transaction: sell the stock on day 3 for $8 then buy back on day 4 for $2.

Example 2:

Input: prices = [12,16,19,19,8,1,19,13,9], k = 3

Output: 36

Explanation:

We can make $36 of profit through 3 transactions:

• A normal transaction: buy the stock on day 0 for $12 then sell it on day 2 for $19.
• A short selling transaction: sell the stock on day 3 for $19 then buy back on day 4 for $8.
• A normal transaction: buy the stock on day 5 for $1 then sell it on day 6 for $19.

Constraints:

• 2 <= prices.length <= 103
• 1 <= prices[i] <= 109
• 1 <= k <= prices.length / 2

Solution

import kotlin.math.max

class Solution {
    fun maximumProfit(prices: IntArray, k: Int): Long {
        val n = prices.size
        var prev = LongArray(n)
        var curr = LongArray(n)
        for (t in 1..k) {
            var bestLong = -prices[0].toLong()
            var bestShort = prices[0].toLong()
            curr[0] = 0
            for (i in 1..<n) {
                var res = curr[i - 1]
                res = max(res, prices[i] + bestLong)
                res = max(res, -prices[i] + bestShort)
                curr[i] = res
                bestLong = max(bestLong, prev[i - 1] - prices[i])
                bestShort = max(bestShort, prev[i - 1] + prices[i])
            }
            val tmp = prev
            prev = curr
            curr = tmp
        }
        return prev[n - 1]
    }
}

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