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 C (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

int max(int a, int b)
{
    return a > b ? a : b;
}

int min(int a, int b)
{
    return a < b ? a : b;
}

/* O(n) */
int maximum_profit(int *A, int n)
{
    int maxv = -1000000000;
    int minv = A[0];

    for (int j = 1; j < n ; j++)
    {
        maxv = max(maxv, A[j] - minv);
        minv = min(minv, A[j]);
    }
    return maxv;
}

/* 効率悪い (O(n^2)) */
// int maximum_profit(int *A, int n)
// {
//     int maxv = -1000000;

//     for (int j = 1; j < n; j++)
//     {
//         for (int i = 0; i < j; i++)
//         {
//             if (A[j] - A[i] > maxv)
//             {
//                 maxv = A[j] - A[i];
//             }
//         }
//     }
//     return maxv;
// }

int main(void)
{
    int n;
    int A[200000];

    scanf("%d", &n);
    for (int i = 0; i < n; i++)
    {
        scanf("%d", &A[i]);
    }

    printf("%d\n", maximum_profit(A, n));

    return 0;
}