Algorithm
Problem Nmae: 123. Best Time to Buy and Sell Stock III
You are given an array prices where prices[i] is the price of a given stock on the ith day.
Find the maximum profit you can achieve. You may complete at most two transactions.
Note: You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).
Example 1:
Input: prices = [3,3,5,0,0,3,1,4] Output: 6 Explanation: Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3. Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 3.
Example 2:
Input: prices = [1,2,3,4,5] Output: 4 Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4. Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are engaging multiple transactions at the same time. You must sell before buying again.
Example 3:
Input: prices = [7,6,4,3,1] Output: 0 Explanation: In this case, no transaction is done, i.e. max profit = 0.
Constraints:
1 <= prices.length <= 1050 <= prices[i] <= 105
Code Examples
#1 Code Example with C Programming
Code -
C Programming
int maxProfit(int* prices, int pricesSize) {
int buy1, buy2; // max profit if I buy
int sell1, sell2; // max profit if I sell
int i;
buy1 = buy2 = 0x80000000; // min of integer
sell1 = sell2 = 0;
for (i = 0; i < pricesSize; i ++) {
// possible profit
buy1 = buy1 > 0 - prices[i] ? buy1 : 0 - prices[i];
sell1 = sell1 > buy1 + prices[i] ? sell1 : buy1 + prices[i];
buy2 = buy2 > sell1 - prices[i] ? buy2 : sell1 - prices[i];
sell2 = sell2 > buy2 + prices[i] ? sell2 : buy2 + prices[i];
}
return sell2;
}
Input
prices = [3,3,5,0,0,3,1,4]
Output
6
#2 Code Example with Javascript Programming
Code -
Javascript Programming
const maxProfit = function(prices) {
let maxProfit1 = 0
let maxProfit2 = 0
let lowestBuyPrice1 = Number.MAX_VALUE
let lowestBuyPrice2 = Number.MAX_VALUE
for (let p of prices) {
maxProfit2 = Math.max(maxProfit2, p - lowestBuyPrice2)
lowestBuyPrice2 = Math.min(lowestBuyPrice2, p - maxProfit1)
maxProfit1 = Math.max(maxProfit1, p - lowestBuyPrice1)
lowestBuyPrice1 = Math.min(lowestBuyPrice1, p)
}
return maxProfit2
}
Input
prices = [3,3,5,0,0,3,1,4]
Output
6
#3 Code Example with Python Programming
Code -
Python Programming
class Solution:
def maxProfit(self, prices):
s1 = s2 = 0
b1 = b2 = -float("inf")
for p in prices:
if -p > b1: b1 = -p
if b1 + p > s1: s1 = b1 + p
if s1 - p > b2: b2 = s1 - p
if b2 + p > s2: s2 = b2 + p
return s2
Input
prices = [1,2,3,4,5]
Output
4
#4 Code Example with C# Programming
Code -
C# Programming
using System;
namespace LeetCode
{
public class _0123_BestTimeToBuyAndSellStock3
{
public int MaxProfit(int[] prices)
{
int buy1 = int.MaxValue, buy2 = int.MaxValue;
int sell1 = 0, sell2 = 0;
for (int i = 0; i < prices.Length; i++)
{
buy1 = Math.Min(buy1, prices[i]);
sell1 = Math.Max(sell1, prices[i] - buy1);
buy2 = Math.Min(buy2, prices[i] - sell1);
sell2 = Math.Max(sell2, prices[i] - buy2);
}
return sell2;
}
}
}
Input
prices = [1,2,3,4,5]