Friday, October 2, 2015

Zenefits: [OA]Stock Maximize

Problem Statement
Your algorithms have become so good at predicting the market that you now know what the share price of Wooden Orange Toothpicks Inc. (WOT) will be for the next N days.
Each day, you can either buy one share of WOT, sell any number of shares of WOT that you own, or not make any transaction at all. What is the maximum profit you can obtain with an optimum trading strategy?
The first line contains the number of test cases T. T test cases follow:
The first line of each test case contains a number N. The next line contains N integers, denoting the predicted price of WOT shares for the next N days.
Output T lines, containing the maximum profit which can be obtained for the corresponding test case.
1 <= T <= 10 
1 <= N <= 50000
All share prices are between 1 and 100000
Sample Input
5 3 2
1 2 100
1 3 1 2
Sample Output
For the first case, you cannot obtain any profit because the share price never rises. 
For the second case, you can buy one share on the first two days, and sell both of them on the third day. 
For the third case, you can buy one share on day 1, sell one on day 2, buy one share on day 3, and sell one share on day 4.
Understand the problem:
The problem looks very similar to the best time to buy and sell stock. The mainly difference is in the best time to buy and sell stock, you can only hold one share of the stock. In this problem, however, you can hold as many shares as you want. But each day, you can only buy one stock, and sell any you want. 
So the idea for this problem is starting from the last day N, iterate in backward order. Each time mark the highest price so far, denoting, max1,  when we see a higher price, i.e, max2 > max1. Then we perform transactions between the two days. That is, buy all stocks after max2 and sell all on day max1. 

Code (Java):
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {
    public static long stockMaximize(long[] prices) {
        if (prices == null || prices.length <= 1) {
            return 0;
        int n = prices.length;
        long maxPrice = Long.MIN_VALUE;
        long maxProfit = 0;
        long sum = 0;
        int j = n - 1;
        for (int i = n - 1; i >= 0; i--) {
            if (prices[i] > maxPrice) {
                long localMax = (j - i) * maxPrice - sum;
                maxProfit += localMax;
                j = i;
                maxPrice = prices[i];
                sum = prices[i];
            } else {
                sum += prices[i];
        if (j > 0) {
            maxProfit += (j + 1) * maxPrice - sum;
        return maxProfit;
    public static void main(String[] args) {
        /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
        Scanner sc = new Scanner(;
        int numOfTestCases = sc.nextInt();
        for (int i = 0; i < numOfTestCases; i++) {
            int n = sc.nextInt();
            long[] prices = new long[n];
            for (int j = 0; j < n; j++) {
                prices[j] = sc.nextLong();
            long maxProfit = stockMaximize(prices);

No comments:

Post a Comment