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:
Javascript
<script>// Javascript program to implement// the above approach// Function to return the maximum profit// that can be made after buying and// selling the given stocksfunction maxProfit(price, start, end){ // If the stocks can't be bought if (end <= start) return 0; // Initialise the profit let profit = 0; // The day at which the stock // must be bought for (let i = start; i < end; i++) { // The day at which the // stock must be sold for (let 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 let 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 codelet price = [100, 180, 260, 310, 40, 535, 695];let n = price.length;document.write(maxProfit( price, 0, n - 1));</script> |
Output:
865
Time Complexity: O(N2)
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.
Javascript
<script>// JavaScript program to implement // the above approach// This function finds the buy sell// schedule for maximum profitfunction stockBuySell(price, n) { // Prices must be given for at // least two days if (n == 1) return; // Traverse through given price array let 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 let 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 let sell = i - 1; document.write(`Buy on day: ${buy} Sell on day: ${sell}<br>`); }}// Driver code// Stock prices on consecutive dayslet price = [100, 180, 260, 310, 40, 535, 695];let n = price.length;// Function callstockBuySell(price, n);// This code is contributed by Potta Lokesh</script> |
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)
Space Complexity: O(1) since using constant variables
Please refer complete article on Stock Buy Sell to Maximize Profit for more details!
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