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Best Time to Buy and Sell Stock – LeetCode – Solution

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121. Best Time to Buy and Sell Stock – Solution

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 (pr ice = 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.
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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

Solution

we need to find max(prices[j] - prices[i]), for every i and j such that j > i

Approach 1: Brute Force

public class Solution {
    public int maxProfit(int prices[]) {
      	int n = prices.length;
        int maxProfit = 0;
        for (int i = 0; i < n - 1; i++) {
            for (int j = i + 1; j < n; j++) {
                maxProfit = Math.max(maxProfit, prices[j] - prices[i]);
            }
        }
        return maxProfit;
    }
}

Time complexity: O(n2) Space complexity: O(1)

Approach 2: Precalulate max array from right

class Solution {
    public int maxProfit(int[] prices) {
        int n = prices.length;
        int maxProfit = 0;
        int[] maxArray = new int[n];
        int maxTillNow = Integer.MIN_VALUE;
      
      	// calculate max array from right
        for(int i=n-1; i>=0; i--){
            maxTillNow = Math.max(maxTillNow,prices[i]);
            maxArray[i] = maxTillNow;
        }
        
        for(int i=0; i<n-1; i++)
            maxProfit = Math.max(maxProfit,maxArray[i+1]-prices[i]);
        return maxProfit;
    }
}

Time complexity: O(n) Space complexity: O(n)

Approach 3: One pass solution without extra space

class Solution {
    
    public int maxProfit(int[] prices) {
        int n = prices.length;
        int minSoFar = Integer.MAX_VALUE;
        int maxProfit = 0;
        for(int i=0;i< n;i++){
            if(minSoFar > prices[i])
                minSoFar = prices[i];
            else
                maxProfit = Math.max(prices[i] - minSoFar, maxProfit);
        }
        return maxProfit;
     }
}

Time complexity: O(n) Space complexity: O(1)

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