https://github.com/emahtab/best-time-to-buy-and-sell-stock
Find the max profit buying and selling stock with at most one transaction.
https://github.com/emahtab/best-time-to-buy-and-sell-stock
buy-and-sell leetcode problem-solving
Last synced: 3 months ago
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Find the max profit buying and selling stock with at most one transaction.
- Host: GitHub
- URL: https://github.com/emahtab/best-time-to-buy-and-sell-stock
- Owner: eMahtab
- Created: 2020-04-05T13:39:43.000Z (over 5 years ago)
- Default Branch: master
- Last Pushed: 2020-04-05T14:48:16.000Z (over 5 years ago)
- Last Synced: 2025-03-28T01:43:24.915Z (7 months ago)
- Topics: buy-and-sell, leetcode, problem-solving
- Homepage:
- Size: 28.3 KB
- Stars: 1
- Watchers: 1
- Forks: 0
- Open Issues: 0
-
Metadata Files:
- Readme: README.md
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README
# Best time to buy and sell stock
## https://leetcode.com/problems/best-time-to-buy-and-sell-stock
Say you have an array for which the ith element is the price of a given stock on day i.
If you were only permitted to complete at most one transaction (i.e., buy one and sell one share of the stock), design an algorithm to find the maximum profit.
**Note that you cannot sell a stock before you buy one.**
```
Example 1:
Input: [7,1,5,3,6,4]
Output: 5
Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5.
Not 7-1 = 6, as selling price needs to be larger than buying price.
Example 2:
Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.
```
# Implementation 1 : O(n^2)
```java
class Solution {
public int maxProfit(int[] prices) {
if(prices == null || prices.length <= 1)
return 0;
int maxProfit = 0;
for(int i = 0; i < prices.length - 1; i++) {
for(int j = i+1; j < prices.length; j++) {
maxProfit = Math.max(maxProfit, prices[j] - prices[i]);
}
}
return maxProfit;
}
}
```
# Implementation 2 : O(n)
Say the given array is:
[7, 1, 5, 3, 6, 4]
If we plot the numbers of the given array on a graph, we get:
```java
class Solution {
public int maxProfit(int[] prices) {
if(prices == null || prices.length <= 1)
return 0;
int maxProfit = 0;
int minPrice = prices[0];
for(int i = 1; i < prices.length; i++) {
if(prices[i] < minPrice)
minPrice = prices[i];
else
maxProfit = Math.max(maxProfit, prices[i] - minPrice);
}
return maxProfit;
}
}
```
# Tip :
Make a dotted graph representation for better visualization, this helps in solving the problem.
# References :
1. https://leetcode.com/articles/best-time-to-buy-and-sell-stock
2. https://www.youtube.com/watch?v=mj7N8pLCJ6w