Best Time to Buy and Sell Stock

Question

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
Given an example [3,2,3,1,2], return 1

Analysis

最初的想法是问题可以转化一下:计算相邻两天的变化量array,再对这个差值array寻找maximum subarray,得到的maximum也就是对应的maximum profit
另一种想法是:仅记录loop过的历史最小值,再寻找当前profit最大值.
Time complexity: O(n). Only a single pass is needed.
Space complexity: O(1). Only two variables are used.

Solution

// Diff + Maximum Subarray
public class Solution {
/**
* @param prices: Given an integer array
* @return: Maximum profit
*/
public int maxProfit(int[] prices) {
if (prices == null || prices.length == 0) {
return 0;
}
int[] diff = new int[prices.length];
diff[0] = 0;
for (int i = 1; i < prices.length; i++) {
diff[i] = prices[i] - prices[i - 1];
}
return maxSubArray(diff);
}
public int maxSubArray(int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
}
int length = nums.length;
int max = Integer.MIN_VALUE;
int sum = 0;
for (int i = 0; i < length; i++) {
sum = sum + nums[i];
max = Math.max(sum, max);
sum = Math.max(sum, 0);
}
return max;
}
}
One pass - record lowest price O(n) (1ms 99.88% AC)
class Solution {
public int maxProfit(int[] prices) {
int minPrice = Integer.MAX_VALUE;
int maxProfit = 0;
for (int price : prices) {
if (price < minPrice) minPrice = price;
maxProfit = Math.max(maxProfit, price - minPrice);
}
return maxProfit;
}
}
One Pass
// Only remember the lowest price
public class Solution {
public int maxProfit(int[] prices) {
if (prices == null || prices.length == 0) {
return 0;
}
// Only remember the smallest price
int min = Integer.MAX_VALUE;
int profit = 0;
for (int i : prices) {
min = i < min ? i : min;
profit = (i - min) > profit ? i - min : profit;
}
return profit;
}
}

Reference

Best Time to Buy and Sell Stock Series: I, II, III, IV