LeetCode - Best Time to Buy and Sell Stock

Problem statement

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.

Problem statement taken from: https://leetcode.com/problems/best-time-to-buy-and-sell-stock

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 <= 10^5
- 0 <= prices[i] <= 10^4

Explanation

Brute force approach

The naive approach is to use two nested for loops and
get the maximum difference between two numbers.

A C++ snippet of the above approach is as below:

int maxProfit = 0;

for (int i = 0; i < prices.length - 1; i++) {
    for (int j = i + 1; j < prices.length; j++) {
        int profit = prices[j] - prices[i];
        if (profit > maxProfit)
            maxProfit = profit;
    }
}

return maxProfit;

The time complexity of the above program is O(N^2).

One pass approach

If we check the below image of the stock values across days,
we need to consider the maximum and minimum values.

Let's check the algorithm below:

- set maxP = 0
      minP = INT_MAX

- loop for i = 0; i < prices.size(); i++
  - minP = min(minP, prices[i])

  - if prices[i] > minP
    - maxP = max(maxP, prices[i] - minP)

- return maxP

The time complexity of the above approach is O(log(N)) and,
space complexity is O(1).

C++ solution

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int maxP = 0;
        int minP = INT_MAX;

        for(int i = 0; i < prices.size(); i++){
            minP = min(minP, prices[i]);
            if(prices[i] > minP){
                maxP = max(maxP, prices[i] - minP);
            }
        }

        return maxP;
    }
};

Golang solution

const MaxUint = ^uint(0)
const MaxInt = int(MaxUint >> 1)

func maxProfit(prices []int) int {
    maxP := 0
    minP := MaxInt

    for i := 0; i < len(prices); i++ {
        minP = int(math.Min(float64(minP), float64(prices[i])))

        if prices[i] > minP {
            maxP = int(math.Max(float64(maxP), float64(prices[i] - minP)))
        }
    }

    return maxP
}

Javascript solution

var maxProfit = function(prices) {
    let maxP = 0;
    let minP = Number.MAX_VALUE;

    for( let i = 0; i < prices.length; i++ ) {
        minP = Math.min(minP, prices[i]);

        if( prices[i] > minP ) {
           maxP = Math.max(maxP, prices[i] - minP);
        }
    }

    return maxP;
};

Let's dry-run our algorithm to see how the solution works.

Input: prices = [7, 1, 5, 3, 6, 4]

Step 1: maxP = 0
        minP = INT_MAX

Step 2: loop for i = 0; i < prices.size()
        0 < 6
        true

        minP = min(minP, prices[i]);
             = min(INT_MAX, prices[0])
             = min(INT_MAX, 7)
             = 7

        if prices[i] > minP
           prices[0] > 7
           7 > 7
           false

        i++
        i = 1

Step 3: loop for i < prices.size()
        1 < 6
        true

        minP = min(minP, prices[i]);
             = min(7, prices[1])
             = min(7, 1)
             = 1

        if prices[i] > minP
           prices[1] > 1
           1 > 1
           false

        i++
        i = 2

Step 4: loop for i < prices.size()
        2 < 6
        true

        minP = min(minP, prices[i]);
             = min(1, prices[2])
             = min(1, 5)
             = 1

        if prices[i] > minP
           prices[2] > 1
           5 > 1
           true

           maxP = max(maxP, prices[i] - minP)
                = max(0, 5 - 1)
                = max(0, 4)
                = 4

        i++
        i = 3

Step 5: loop for i < prices.size()
        3 < 6
        true

        minP = min(minP, prices[i]);
             = min(1, prices[3])
             = min(1, 3)
             = 1

        if prices[i] > minP
           prices[3] > 1
           3 > 1
           true

           maxP = max(maxP, prices[i] - minP)
                = max(4, 3 - 1)
                = max(4, 2)
                = 4

        i++
        i = 4

Step 6: loop for i < prices.size()
        4 < 6
        true

        minP = min(minP, prices[i]);
             = min(1, prices[4])
             = min(1, 6)
             = 1

        if prices[i] > minP
           prices[4] > 1
           6 > 1
           true

           maxP = max(maxP, prices[i] - minP)
                = max(4, 6 - 1)
                = max(4, 5)
                = 5

        i++
        i = 5

Step 7: loop for i < prices.size()
        5 < 6
        true

        minP = min(minP, prices[i]);
             = min(1, prices[5])
             = min(1, 4)
             = 1

        if prices[i] > minP
           prices[5] > 1
           4 > 1
           true

           maxP = max(maxP, prices[i] - minP)
                = max(5, 4 - 1)
                = max(5, 3)
                = 5

        i++
        i = 6

Step 8: loop for i < prices.size()
        6 < 6
        false

Step 9: return maxP

So we return the answer as 5.