Java Program For Stock Buy Sell To Maximize Profit
The cost of a stock on each day is given in an array, find the max profit that you can make by buying and selling in those days. For example, if the given array is {100, 180, 260, 310, 40, 535, 695}, the maximum profit can be earned by buying on day 0, and selling on day 3. Again buy on day 4 and sell on day 6. If the given array of prices is sorted in decreasing order, then profit cannot be earned at all.
Naive approach: A simple approach is to try buying the stocks and selling them every single day when profitable and keep updating the maximum profit so far.
Below is the implementation of the above approach:
Java
// Java implementation of the approach import java.util.*; class GFG{ // Function to return the maximum profit // that can be made after buying and // selling the given stocks static int maxProfit( int price[], int start, int end) { // If the stocks can't be bought if (end <= start) return 0 ; // Initialise the profit int profit = 0 ; // The day at which the stock // must be bought for ( int i = start; i < end; i++) { // The day at which the // stock must be sold for ( int j = i + 1 ; j <= end; j++) { // If buying the stock at ith day and // selling it at jth day is profitable if (price[j] > price[i]) { // Update the current profit int curr_profit = price[j] - price[i] + maxProfit(price, start, i - 1 ) + maxProfit(price, j + 1 , end); // Update the maximum profit so far profit = Math.max(profit, curr_profit); } } } return profit; } // Driver code public static void main(String[] args) { int price[] = { 100 , 180 , 260 , 310 , 40 , 535 , 695 }; int n = price.length; System.out.print(maxProfit( price, 0 , n - 1 )); } } // This code is contributed by PrinciRaj1992 |
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Time complexity: O(n^2) where n is the size of a given stock array
Auxiliary Space: O(1)
Efficient approach: If we are allowed to buy and sell only once, then we can use following algorithm. Maximum difference between two elements. Here we are allowed to buy and sell multiple times.
Following is the algorithm for this problem.
- Find the local minima and store it as starting index. If not exists, return.
- Find the local maxima. and store it as an ending index. If we reach the end, set the end as the ending index.
- Update the solution (Increment count of buy-sell pairs)
- Repeat the above steps if the end is not reached.
Java
// Program to find best buying and // selling days import java.util.ArrayList; // Solution structure class Interval { int buy, sell; } class StockBuySell { // This function finds the buy sell // schedule for maximum profit void stockBuySell( int price[], int n) { // Prices must be given for at least // two days if (n == 1 ) return ; int count = 0 ; // Solution array ArrayList<Interval> sol = new ArrayList<Interval>(); // Traverse through given price array int i = 0 ; while (i < n - 1 ) { // Find Local Minima. Note that the // limit is (n-2) as we are comparing // present element to the next element. while ((i < n - 1 ) && (price[i + 1 ] <= price[i])) i++; // If we reached the end, break as no // further solution possible if (i == n - 1 ) break ; Interval e = new Interval(); e.buy = i++; // Store the index of minima // Find Local Maxima. Note that the // limit is (n-1) as we are comparing // to previous element while ((i < n) && (price[i] >= price[i - 1 ])) i++; // Store the index of maxima e.sell = i - 1 ; sol.add(e); // Increment number of buy/sell count++; } // print solution if (count == 0 ) System.out.println( "There is no day when buying the stock " + "will make profit" ); else for ( int j = 0 ; j < count; j++) System.out.println( "Buy on day: " + sol.get(j).buy + " " + "Sell on day : " + sol.get(j).sell); return ; } // Driver code public static void main(String args[]) { StockBuySell stock = new StockBuySell(); // stock prices on consecutive days int price[] = { 100 , 180 , 260 , 310 , 40 , 535 , 695 }; int n = price.length; // function call stock.stockBuySell(price, n); } } // This code is contributed by Mayank Jaiswal |
Buy on day: 0 Sell on day : 3 Buy on day: 4 Sell on day : 6
Time Complexity: The outer loop runs till I become n-1. The inner two loops increment value of I in every iteration. So overall time complexity is O(n)
Auxiliary Space: O(1) since using constant variables
Valley Peak Approach:
In this approach, we just need to find the next greater element and subtract it from the current element so that the difference keeps increasing until we reach a minimum. If the sequence is a decreasing sequence so the maximum profit possible is 0.
Java
// Java program for the above approach import java.io.*; class GFG { static int maxProfit( int prices[], int size) { // maxProfit adds up the difference // between adjacent elements if they // are in increasing order int maxProfit = 0 ; // The loop starts from 1 as its // comparing with the previous for ( int i = 1 ; i < size; i++) if (prices[i] > prices[i - 1 ]) maxProfit += prices[i] - prices[i - 1 ]; return maxProfit; } // Driver code public static void main(String[] args) { // stock prices on consecutive days int price[] = { 100 , 180 , 260 , 310 , 40 , 535 , 695 }; int n = price.length; // function call System.out.println(maxProfit(price, n)); } } // This code is contributed by rajsanghavi9. |
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Time Complexity: O(n)
Auxiliary Space: O(1)
This article is compiled by Ashish Anand and reviewed by the w3wiki team.