Best time to Buy & Sell Stock (121 LeetCode)

This problem is similar to finding maximum difference between two numbers in a given array.

Leet Code Problem Description

You are given an array prices where prices[i] is the price of a given stock on the ith day.

You want to maximize your profit by choosing a single day to buy one stock and choosing a different day in the future to sell that stock. 

Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return 0.

Example 1:

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.

Example 2:

Input: prices = [7,6,4,3,1]
Output: 0
Explanation: In this case, no transactions are done and the max profit = 0.

 

Constraints:

  • 1 <= prices.length <= 105
  • 0 <= prices[i] <= 104


For example, if an array is 3, 6, 4, 3, 2, 7, the difference would be 5. 

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#include <iostream>

using namespace std;

int findMaxDifference(int arr[], int size) {
   int maxDiff = 0, difference = 0;
   int numToCmp = arr[0];
   
   for (int i = 1; i < size; i++) {
       if (numToCmp < arr[i]) {
           difference = arr[i] - numToCmp;
           if (difference > maxDiff) {
               maxDiff = difference;
           }
       } else {
           numToCmp = arr[i];
       }      
   }
   return maxDiff;   
}

int main()
{
    int array[] = {3, 6, 4, 3, 2, 7};
    //int array[] = {3, 6, 4, 3, 2, 3};
    int size = sizeof(array)/sizeof(array[0]);
    int maxDiff = findMaxDifference(array, size);
    cout << "Max Difference is " << maxDiff << endl;
    return 0;
}

Output:

1

Max Difference is 5

Click here to deep dive into this implementation. 

Time Complexity of this algorithm would be O(n). 

The below is the Visualisation of this implementation: 


For loop, i = 1


For loop, i = 2


For loop, i = 3



For loop, i = 4


For loop, i = 5



The below is the optimised solution for the problem.

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int maxProfit = 0;
        int buyDay = prices[0];

        for (int i = 1; i < prices.size(); i++) {            
             maxProfit = max(prices[i] - buyDay, maxProfit);
             buyDay = min(buyDay, prices[i]);
        }
        
        return maxProfit;
    }
};

This is one of the exercises in my module 'Analysis & Design of Algorithms'.

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