Best Time to Buy and Sell Stock

Say you have an array for which the i^th 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.

Algorithm:

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:

The points of interest are the peaks and valleys in the given graph. We need to find the largest peak following the smallest valley. We can maintain two variables – minprice and maxprofit corresponding to the smallest valley and maximum profit (maximum difference between selling price and minprice) obtained so far respectively.

Java Program:

public class BuySellArrayMaxProfit {
	public static void main(String[] args) {
		int[] prices = { 7, 1, 5, 3, 6, 4 };
		System.out.println("maxProfitUsingOneMinVariable :: " + maxProfitUsingOneMinVariable(prices));
		System.out.println("maxProfitUsingTwoMinMaxVariable :: " + maxProfitUsingTwoMinMaxVariable(prices));
	}

	private static int maxProfitUsingOneMinVariable(int[] prices) {
		int min = prices[0];
		int profit = 0;
		for (int i = 1; i < prices.length; i++) {
			if (prices[i] < min) {
				min = prices[i];
			} else if (profit < (prices[i] - min))
				profit = prices[i] - min;
		}
		return profit;
	}

	private static int maxProfitUsingTwoMinMaxVariable(int[] prices) {
		int min = prices[0];
		int max = prices[0];
		for (int i = 1; i < prices.length; i++) {
			if (prices[i] < min) {
				min = prices[i];
				max = prices[i];
			}
			if (max < prices[i])
				max = prices[i];
		}
		return max - min;
	}
}

// Output:
maxProfitUsingOneMinVariable :: 5
maxProfitUsingTwoMinMaxVariable :: 5

Complexity:

• Time complexity : O(n). Only a single pass is needed.
• Space complexity : O(1). Only two variables(min, max or min, profit) are used.