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:
A[0] = 23171 A[1] = 21011 A[2] = 21123 A[3] = 21366 A[4] = 21013 A[5] = 21367If 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:
A[0] = 23171 A[1] = 21011 A[2] = 21123 A[3] = 21366 A[4] = 21013 A[5] = 21367the 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;
}
}
https://codility.com/demo/results/trainingX49Y72-U4Z/