121. Best Time to Buy and Sell Stock

Problem

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.

Problem Summary

Given an array, its i-th element is the price of a given stock on day i. If you are allowed to complete at most one transaction (i.e., buy and sell one share of the stock), design an algorithm to calculate the maximum profit you can obtain. Note that you cannot sell a stock before you buy one.

Solution Ideas

The problem asks us to find the price difference that yields the maximum profit from the stock
There are multiple solutions to this problem as well; you can use DP, or you can use a monotonic stack

Code


package leetcode

// Solution 1: Simulated DP
func maxProfit(prices []int) int {
	if len(prices) < 1 {
		return 0
	}
	min, maxProfit := prices[0], 0
	for i := 1; i < len(prices); i++ {
		if prices[i]-min > maxProfit {
			maxProfit = prices[i] - min
		}
		if prices[i] < min {
			min = prices[i]
		}
	}
	return maxProfit
}

// Solution 2: Monotonic stack
func maxProfit1(prices []int) int {
	if len(prices) == 0 {
		return 0
	}
	stack, res := []int{prices[0]}, 0
	for i := 1; i < len(prices); i++ {
		if prices[i] > stack[len(stack)-1] {
			stack = append(stack, prices[i])
		} else {
			index := len(stack) - 1
			for ; index >= 0; index-- {
				if stack[index] < prices[i] {
					break
				}
			}
			stack = stack[:index+1]
			stack = append(stack, prices[i])
		}
		res = max(res, stack[len(stack)-1]-stack[0])
	}
	return res
}