The stock buy and sell problem is a classic algorithmic problem in which you are given a list of stock prices over time and your task is to find the best times to buy and sell the stock to maximize your profit. The problem assumes that you can only buy and sell the stock once, i.e., you can only hold one share of the stock at a time.
The input to the problem is an array of stock prices, where each element in the array represents the stock price for a given day. The output of the problem is a pair of integers representing the day to buy and the day to sell the stock to maximize your profit.
This problem has practical applications in the financial industry and is often used to make investment decisions.
Here is a step-by-step algorithm for solving the stock buy and sell problem:
- Initialize two variables,
buy
andsell
, to -1. - Loop through the array of stock prices from left to right.
- If
buy
is -1 or the current stock price is less than the stock price at indexbuy
, updatebuy
to the current index. - Otherwise, calculate the potential profit by subtracting the stock price at index
buy
from the current stock price. - If the potential profit is greater than the current profit (which is the difference between the stock prices at indices
buy
andsell
), updatesell
to the current index. - Return a pair of integers
(buy, sell)
.
Here is a Kotlin implementation of the stock buy and sell problem algorithm:
fun findBuySellPrices(prices: IntArray): Pair<Int, Int> {
var buy = -1
var sell = -1
var profit = 0
for (i in prices.indices) {
if (buy == -1 || prices[i] < prices[buy]) {
buy = i
} else {
val currProfit = prices[i] - prices[buy]
if (currProfit > profit) {
sell = i
profit = currProfit
}
}
}
return Pair(buy+1, sell+1)
}
fun main() {
val prices = intArrayOf(7, 1, 5, 3, 6, 4)
val (buy, sell) = findBuySellPrices(prices)
if (buy == -1 || sell == -1) {
println("No opportunity to make a profit.")
} else {
println("Buy on day $buy, sell on day $sell.")
}
}
Here is a detailed explanation of the findBuySellPrices
function step by step:
- The function takes an integer array
prices
as input and returns aPair<Int, Int>
representing the optimal buy and sell days. If there is no opportunity to make a profit, the function returns (-1, -1). - The function initializes three variables:
buy
,sell
, andprofit
.buy
andsell
are initialized to -1 because they haven’t been set yet, andprofit
is initialized to 0 because there is no profit yet. - The function loops through the array of stock prices using the
indices
property of the array. This allows us to loop over the indices of the array, rather than the values themselves. - For each index
i
, the function checks ifbuy
is -1 or if the current stock price atprices[i]
is less than the stock price atprices[buy]
. If so, it updatesbuy
toi
. This ensures thatbuy
always points to the lowest stock price seen so far. - If the current stock price at
prices[i]
is greater than the stock price atprices[buy]
, the function calculates the potential profit by subtracting the stock price atprices[buy]
from the current stock price atprices[i]
. - If the potential profit is greater than the current profit (which is the difference between the stock prices at indices
buy
andsell
), the function updatessell
toi
andprofit
tocurrProfit
. This ensures thatsell
always points to the highest stock price seen so far andprofit
always holds the highest potential profit. - After the loop finishes, the function returns a pair of integers
(buy, sell)
.
The time complexity of this function is O(n), where n
is the length of the input array. This is because the function loops through the array once to find the optimal buy and sell days.
The space complexity of this function is O(1), because the function only uses a constant amount of extra space to store the variables buy
, sell
, and profit
.