## 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 earned by buying on day 0, 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 on every single day when profitable and keep updating the maximum profit so far.

Below is the implementation of the above approach:

`// C++ implementation of the approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to return the maximum profit` `// that can be made after buying and` `// selling the given stocks` `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 = max(profit, curr_profit);` ` ` `}` ` ` `}` ` ` `}` ` ` `return` `profit;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `price[] = { 100, 180, 260, 310,` ` ` `40, 535, 695 };` ` ` `int` `n = ` `sizeof` `(price) / ` `sizeof` `(price[0]);` ` ` `cout << maxProfit(price, 0, n - 1);` ` ` `return` `0;` `}` |

**Output**

865

**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.

`// C++ Program to find best buying and selling days` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// 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` `;` ` ` `// 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` `;` ` ` `// Store the index of minima` ` ` `int` `buy = i++;` ` ` `// 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` ` ` `int` `sell = i - 1;` ` ` `cout << ` `"Buy on day: "` `<< buy` ` ` `<< ` `"\t Sell on day: "` `<< sell << endl;` ` ` `}` `}` `// Driver code` `int` `main()` `{` ` ` `// Stock prices on consecutive days` ` ` `int` `price[] = { 100, 180, 260, 310, 40, 535, 695 };` ` ` `int` `n = ` `sizeof` `(price) / ` `sizeof` `(price[0]);` ` ` `// Function call` ` ` `stockBuySell(price, n);` ` ` `return` `0;` `}` `// This is code is contributed by rathbhupendra` |

**Output**

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)

**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.

`#include <iostream>` `using` `namespace` `std;` `// Preprocessing helps the code run faster` `#define fl(i, a, b) for (int i = a; i < b; i++)` `// Function that return` `int` `maxProfit(` `int` `* prices, ` `int` `size)` `{` ` ` `// maxProfit adds up the difference between` ` ` `// adjacent elements if they are in increaisng order` ` ` `int` `maxProfit = 0;` ` ` `// The loop starts from 1` ` ` `// as its comparing with the previous` ` ` `fl(i, 1, size) ` `if` `(prices[i] > prices[i - 1]) maxProfit` ` ` `+= prices[i] - prices[i - 1];` ` ` `return` `maxProfit;` `}` `// Driver Function` `int` `main()` `{` ` ` `int` `prices[] = { 100, 180, 260, 310, 40, 535, 695 };` ` ` `int` `N = ` `sizeof` `(prices) / ` `sizeof` `(prices[0]);` ` ` `cout << maxProfit(prices, N) << endl;` ` ` `return` `0;` `}` `// This code is contributed by Kingshuk Deb` |

**Output**

865

**Time Complexity**: O(n)

**Auxiliary** **Space: **O(1)

Last Updated on October 16, 2021 by admin