This is part of the Easy way to write algorithms in ABAP: Series 01. For more algorithms, please check the main blog-post.

Problem

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

Solution

Time Complexity: O(n)

Space Complexity: O(1)

ABAP

CLASS zcl_stock DEFINITION
  PUBLIC
  FINAL
  CREATE PUBLIC .

  PUBLIC SECTION.

* Mandatory declaration
    INTERFACES if_oo_adt_classrun.

  PROTECTED SECTION.
  PRIVATE SECTION.

    TYPES ty_prices TYPE STANDARD TABLE OF i WITH DEFAULT KEY.

    METHODS StockBuySell
      IMPORTING lt_prices            TYPE ty_prices
      RETURNING VALUE(rv_max_profit) TYPE i.

ENDCLASS.


CLASS zcl_stock IMPLEMENTATION.

  METHOD if_oo_adt_classrun~main.
    out->write( |{ StockBuySell( VALUE #( ( 7 ) ( 1 ) ( 5 ) ( 3 ) ( 6 ) ( 4 ) ) ) }| ).
  ENDMETHOD.

  METHOD StockBuySell.

    DATA(lv_min_price) = lt_prices[ 1 ].

    LOOP AT lt_prices ASSIGNING FIELD-SYMBOL(<lfs_wa>).
      lv_min_price = nmin( val1 = lv_min_price
                           val2 = <lfs_wa> ).
      rv_max_profit = nmax( val1 = rv_max_profit
                            val2 = ( <lfs_wa> - lv_min_price ) ).
    ENDLOOP.

    UNASSIGN <lfs_wa>.
    FREE lv_min_price.

  ENDMETHOD.

ENDCLASS.

JavaScript

var maxProfit = function(prices) {

    var lv_min_price = prices[0],
        lv_max_profit = 0;

    for(let i = 0; i < prices.length; i++){
        lv_min_price = Math.min(lv_min_price, prices[i]);
        lv_max_profit = Math.max(lv_max_profit, (prices[i] - lv_min_price));
    }

    return lv_max_profit;
    
};

Python

class Solution:
    def maxProfit(self, prices: List[int]) -> int:

        lv_min_price = prices[0]
        lv_max_profit = 0

        for i in range(0, len(prices)):
            lv_min_price = min(lv_min_price, prices[i])
            lv_max_profit = max(lv_max_profit, (prices[i] - lv_min_price))

        return lv_max_profit

N.B: For ABAP, I am using SAP BTP ABAP Environment 2211 Release.