maxProfit.js

// An array A consisting of N integers is given. It contains daily prices of a stock share for a period of N consecutive days. If a single share was bought on day P and sold on day Q, where 0 ≤ P ≤ Q < N, then the profit of such transaction is equal to A[Q] − A[P], provided that A[Q] ≥ A[P]. Otherwise, the transaction brings loss of A[P] − A[Q].

// For example, consider the following array A consisting of six elements such that:

//   A[0] = 23171
//   A[1] = 21011
//   A[2] = 21123
//   A[3] = 21366
//   A[4] = 21013
//   A[5] = 21367
// If a share was bought on day 0 and sold on day 2, a loss of 2048 would occur because A[2] − A[0] = 21123 − 23171 = −2048. If a share was bought on day 4 and sold on day 5, a profit of 354 would occur because A[5] − A[4] = 21367 − 21013 = 354. Maximum possible profit was 356. It would occur if a share was bought on day 1 and sold on day 5.

// Write a function,

// function solution(A);

// that, given an array A consisting of N integers containing daily prices of a stock share for a period of N consecutive days, returns the maximum possible profit from one transaction during this period. The function should return 0 if it was impossible to gain any profit.

// For example, given array A consisting of six elements such that:

//   A[0] = 23171
//   A[1] = 21011
//   A[2] = 21123
//   A[3] = 21366
//   A[4] = 21013
//   A[5] = 21367
// the function should return 356, as explained above.

// Write an efficient algorithm for the following assumptions:

// N is an integer within the range [0..400,000];
// each element of array A is an integer within the range [0..200,000].

console.log(solution([23171, 21011, 21123, 21366, 21013, 21367]));
function solution(A){
    let maxProfit = 0
    let min = 200000
     
    for (let i = 0; i < A.length; i++) {
        min = Math.min(min, A[i])
        maxProfit = Math.max(maxProfit, A[i]-min)
    }
     
    return maxProfit
}