A zero-indexed 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:

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,

class Solution { public int solution(int[] A); }

that, given a zero-indexed 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:

the function should return 356, as explained above.

Assume that:

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

Complexity:

• expected worst-case time complexity is O(N);
• expected worst-case space complexity is O(1), beyond input storage (not counting the storage required for input arguments).

Elements of input arrays can be modified.

// you can also use imports, for example:// import java.util.*;// you can write to stdout for debugging purposes, e.g.// System.out.println("this is a debug message");import java.lang.Math;class Solution {    public int solution(int[] A) {        // write your code in Java SE 8        if(A.length == 0)            return 0;        int fmin = A[0], profit = 0;        for(int i = 1; i < A.length; i++) {            fmin = Math.min(fmin,A[i-1]);            profit = Math.max(profit,A[i] - fmin);        }        return profit;    }}


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