Algorithm
Problem Nmae: 122. Best Time to Buy and Sell Stock II
You are given an integer array prices where prices[i] is the price of a given stock on the ith day.
On each day, you may decide to buy and/or sell the stock. You can only hold at most one share of the stock at any time. However, you can buy it then immediately sell it on the same day.
Find and return the maximum profit you can achieve.
Example 1:
Input: prices = [7,1,5,3,6,4] Output: 7 Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4. Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3. Total profit is 4 + 3 = 7.
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. Total profit is 4.
Example 3:
Input: prices = [7,6,4,3,1] Output: 0 Explanation: There is no way to make a positive profit, so we never buy the stock to achieve the maximum profit of 0.
Constraints:
1 <= prices.length <= 3 * 1040 <= prices[i] <= 104
Code Examples
#1 Code Example with C Programming
Code -
C Programming
int maxProfit(int* prices, int pricesSize) {
int i, d, p;
if (pricesSize < 2) return 0;
p = 0;
for (i = 1; i < pricesSize; i ++) {
d = prices[i] - prices[i - 1];
p = d > 0 ? p + d : p; // get it as long as it is a profit!
}
return p;
}
Input
Output
#2 Code Example with Java Programming
Code -
Java Programming
class Solution {
public int maxProfit(int[] prices) {
int stockBuyPrice = Integer.MAX_VALUE;
int profit = 0;
for (int i = 0; i < prices.length; i++) {
if (prices[i] > stockBuyPrice) {
profit += prices[i] - stockBuyPrice;
stockBuyPrice = prices[i];
} else {
stockBuyPrice = prices[i];
}
}
return profit;
}
}
Input
Output
#3 Code Example with Javascript Programming
Code -
Javascript Programming
const maxProfit = function (prices) {
let p = 0
for (let i = 1; i < prices.length; ++i) {
let delta = prices[i] - prices[i - 1]
if (delta > 0) {
p += delta
}
}
return p
}
Input
Output
#4 Code Example with Python Programming
Code -
Python Programming
class Solution:
def maxProfit(self, prices: List[int]) -> int:
# stock, no stock
dp = [-float('inf'), 0]
for p in prices:
x = max(dp[1] - p, dp[0])
y = max(dp[1], dp[0] + p)
dp = [x, y]
return dp[-1]
Input
#5 Code Example with C# Programming
Code -
C# Programming
using System;
namespace LeetCode
{
public class _0122_BestTimeToBuyAndSellStock2
{
public int MaxProfit(int[] prices)
{
var maxProfit = 0;
for (int i = 1; i < prices.Length; i++)
maxProfit += Math.Max(0, prices[i] - prices[i - 1]);
return maxProfit;
}
}
}
Input