Description

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 (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Example 1:
Input: [7, 1, 5, 3, 6, 4]
Output: 5

max. difference = 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

In this case, no transaction is done, i.e. max profit = 0.

Code

class Solution(object):
    def maxProfit(self, prices):
        """
        :type prices: List[int]
        :rtype: int
        """
        profit, minPrice = 0, sys.maxint
        for i in prices:
            if i < minPrice:
                minPrice = i
            else:
                profit = max(profit, i - minPrice)
        return profit

Conventional practice, although I never thought of this practice --- The ___________

Extension:

The logic to solve this problem is same as "max subarray problem" using Kadane's Algorithm. Since no body has mentioned this so far, I thought it's a good thing for everybody to know.

All the straight forward solution should work, but if the interviewer twists the question slightly by giving the difference array of prices, Ex: for {1, 7, 4, 11}, if he gives {0, 6, -3, 7}, you might end up being confused.

Here, the logic is to calculate the difference (maxCur += prices[i] - prices[i-1]) of the original array, and find a contiguous subarray giving maximum profit. If the difference falls below 0, reset it to zero.

public int maxProfit(int[] prices) {
    int maxCur = 0, maxSoFar = 0;
    for(int i = 1; i < prices.length; i++) {
        maxCur = Math.max(0, maxCur += prices[i] - prices[i-1]);
        maxSoFar = Math.max(maxCur, maxSoFar);
    }
    return maxSoFar;
}

*maxCur = current maximum value

*maxSoFar = maximum value found so far

Kadane's Algorithm s algorithm is used to solve the problem of maximal self-sequence, that is, finding a continuous sub-sequence in the one-dimensional direction of the sequence to maximize the sum of the sub-sequence.
The algorithm is:

def max_subarray(A):
    max_ending_here = max_so_far = A[0]
    for x in A[1:]:
        max_ending_here = max(x, max_ending_here + x)
        max_so_far = max(max_so_far, max_ending_here)
    return max_so_far

A variant of this problem is that if a sequence contains negative elements, it is allowed to return a subsequence of zero length. This problem can be solved with the following code:

def max_subarray(A):
    max_ending_here = max_so_far = 0
    for x in A:
        max_ending_here = max(0, max_ending_here + x)
        max_so_far = max(max_so_far, max_ending_here)
    return max_so_far

Conclustion

Learning Recording New Algorithms, Recently Major in Dynamic Programming