121 Best Time to Buy and Sell Stock

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title: Best Time to Buy and Sell Stock
tags:
- best-time-to-buy-and-sell-stock
- No.121
- simple
- divide-conquer
- max-subarray

---

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 (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.

Corner Cases

• decreasing
• single [1]
• binary [1,2]
• negative case [-1, 1, 2]

Solutions

Divide and Conquer

Transfer the price array to difference array, which indicates the pure profit. Divide the array into 2 parts, solve each part independently and the case crossing the middle point.

The time complexity is O(n lg(n)), since the recursive relationship is T(n) = 2 T(\frac{n}{2}) + O(n):

class Solution {
    public int maxProfit(int[] prices) {
        if (prices.length < 2) {return 0;}
        boolean dec  = true;
        int[]   diff = new int[prices.length - 1];
        for (int i=0; i prices[i+1]);
        }
        // transfer to maximum subarray
        return (dec) ? 0 : divide_conquer(diff, 0, diff.length-1);
    }

    private int divide_conquer(int[] diff, int a, int b) {
        if (a  == b) {return diff[a];}
        int c  = (a + b)/2;
        int s1 = divide_conquer(diff, a, c);
        int s2 = divide_conquer(diff, c+1, b);
        int s3 = crossing(diff, a, b, c);
        int m  = (s1 > s2) ? s1 : s2;
        return (m > s3) ? m : s3;
    }

    private int crossing(int[] diff, int a, int b, int c) {
        int sum  = 0;
        int lsum = -2147483648;
        int rsum = -2147483648;

        for (int i=c; a<=i; i--) {
            sum = sum + diff[i];
            lsum = (sum > lsum) ? sum : lsum;
        }
        sum = 0;
        for (int j=c+1; j<=b; j++) {
            sum = sum + diff[j];
            rsum = (sum > rsum) ? sum : rsum;
        }

        return lsum + rsum;
    }
}