Say you have an array for which thei_thelement 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.
Solution
class Solution {
public int maxProfit(int[] prices) {
// take an element and calculate difference for all next elements (keep max only) repeate for other elements
// O(n2)
// int max = 0;
// for (int i = 0; i < prices.length; i++) {
// for (int j = i + 1; j < prices.length; j++) {
// max = Math.max(max, prices[j] - prices[i]);
// }
// }
// O(n) [7,1,5,3,6,4]
if (prices.length == 0) return 0;
int min = prices[0];
int max = 0;
for (int i = 1; i < prices.length; i++) {
min = Math.min(min, prices[i]);
max = Math.max(max, prices[i] - min);
}
return max;
}
}