Problem:
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Maximum Profit

You can obtain profits from foreign exchange margin transactions. For example, if you buy 1000 dollar at a rate of 100 yen per dollar, and sell them at a rate of 108 yen per dollar, you can obtain (108 - 100) × 1000 = 8000 yen.

Write a program which reads values of a currency $R_t$ at a certain time $t$ ($t = 0, 1, 2, ... n-1$), and reports the maximum value of $R_j - R_i$ where $j > i$ .

Input

The first line contains an integer $n$. In the following $n$ lines, $R_t$ ($t = 0, 1, 2, ... n-1$) are given in order.

Output

Print the maximum value in a line.

Constraints

$2 \leq n \leq 200,000$

$1 \leq R_t \leq 10^9$

Sample Input 1

6
5
3
1
3
4
3

Sample Output 1

3

Sample Input 2

3
4
3
2

Sample Output 2

-1

Solution in JAVA (judge verdict: Accepted)

import java.util.Scanner;

class Main {
	public static void main(String args[]) {
		Scanner in = new Scanner(System.in);

		int n = in.nextInt();
		int[] minsL = new int[n+2];
		int[] maxsL = new int[n+2];
		int[] maxsR = new int[n+2];
		int[] minsR = new int[n+2];
		int[] kawase = new int[n+2];

		minsL[0] = Integer.MAX_VALUE;
		maxsL[0] = 0;
		for(int i=1;i<=n;i++) {
			kawase[i] = in.nextInt();
			minsL[i] = Math.min(minsL[i-1], kawase[i]);
			maxsL[i] = Math.max(maxsL[i-1], kawase[i]);
		}
		minsR[n+1] = Integer.MAX_VALUE;
		maxsR[n+1] = 0;
		for(int i=n;i>=1;i--) {
			minsR[i] = Math.min(minsR[i+1], kawase[i]);
			maxsR[i] = Math.max(maxsR[i+1], kawase[i]);
		}

		int ans = Integer.MIN_VALUE;
		for(int i=1;i<n;i++) {
			ans = Math.max(ans, maxsR[i+1] - minsL[i]);
//			ans = Math.max(ans, Math.abs(maxsL[i] - minsR[i+1]));
		}
		System.out.println(ans);
	}
}