Best time to buy and Sell stock in java

October 14, 2022

Best time to buy and Sell stock in Java

Here, in this page we will discuss the program to find the best time to buy and sell stock in C  We are given with an array represents 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.

Example :

Input: Prices[ ] = [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. Note that buying on day 2 and selling on day 1 is not allowed because you must buy before you sell.

 
Best time to buy and sell in C

Algorithm :

  • Take the size of the array from the user and store it in variable say n.
  • Now, declare an array of size n and takes n integer value from the user and store them in array.
  • Create an array named max_sp of size equals size of price array and a variable max and initialize it as the minimum value.
  • Start from the last index in price array.
    1. If prices[i] is greater than max
      1. Update max as prices[i] and make max_sp[i] as minimum value
    2. Else if price[i] is not greater than max
      1. Update max_sp[i] = max.
  • After the pre computation, we follow the naive approach and replace the inner nested loop by using the max_sp array that we just created.
  • Print value of max.

Code in Java

Run 
public class Main {
    static int findMaximumProfit(int[] prices, int i, int k, int buy, int[][] v) {
        // If no stock can be chosen
        if (i >= prices.length || k <= 0) return 0;
        if (v[i][buy] != -1) return v[i][buy];
        int nbuy;
        if (buy == 1)
            nbuy = 0;
        else
            nbuy = 1;
        if (buy == 1) {
            return v[i][buy] = Math.max(-prices[i] + findMaximumProfit(prices, i + 1, k, nbuy, v),
                                        findMaximumProfit(prices, i + 1, k, (int)(buy), v));
        }
        // Otherwise
        else {
            // Buy now
            if (buy == 1)
                nbuy = 0;
            else
                nbuy = 1;
            return v[i][buy] =
                       Math.max(prices[i] + findMaximumProfit(prices, i + 1, k - 1, nbuy, v),
                                findMaximumProfit(prices, i + 1, k, buy, v));
        }
    }
    static int maxProfit(int[] prices) {
        int n = prices.length;
        int[][] v = new int[n][2];
        for (int i = 0; i < v.length; i++) {
            v[i][0] = -1;
            v[i][1] = -1;
        }
        return findMaximumProfit(prices, 0, 1, 1, v);
    }
    // Driver Code
    public
    static void main(String[] args) {
        // Given prices
        int[] prices = {7, 1, 5, 3, 6, 4};
        int ans = maxProfit(prices);
        // Print answer
        System.out.println("Best time to buy and Sell stock = "+ans);
    }
}

Output

5

Given an array which consists of only 0, 1 and 2

Move all the negative elements to one side of the array

Find the Union and Intersection of the two sorted arrays

Find Largest sum contiguous Subarray

Minimize the maximum difference between heights 

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