Maximum difference between two elements s.t larger element appears after the smaller number

Find maximum profit if you can buy & sell share just once.

Given an array of integers, find out the difference between any two elements such that larger element appears after the smaller number.

Examples: If array is [2, 3, 10, 6, 4, 8, 1] then returned value should be 8 (Diff between 10 and 2). If array is [ 7, 9, 5, 6, 3, 2 ] then returned value should be 2 (Diff between 7 and 9).
This difference can be the largest profit if shares are allowed to buy and sell once.

so in simple words 
we need to find A[i] and A[j] such that A[i] < A[j] and i< j

Think what can be done to achieve that.

[2, 3, 10, 6, 4, 8, 1] in this wee nedd to find 2 and 10. 
[2, 3, 10, 1, 4, 8, 11] thn in this it should be 1 and 11.
so as we read array from index 0 to N, we can find max_number later.
but we need to keep track of minimum number.

We can keep track of 
1) minimum number , min_num found so far
2) maximum difference, max_dif  found so far.

now if for next element in array , current diff ( A [ j ] - min_num ) is larger than existing then update max diff. 
and if A[ j ] is smaller then min_num update it.

code 

#include<stdio.h>
 
int maxDiff(int a[], int length)
{
 int max_diff = a[1] - a[0];
 int min_element = a[0];
 for( int i = 1 ; i < length ; i++ )
 { 
 if(a[i] - min_num > max_diff) 
 max_diff = a[i] - min_num;
 if(a[i] < min_num)
 min_num = a[i]; 
 }
 return max_diff;
} 

int main()
{
 int a[] = {1, 2, 6, 80, 100};
 int len = sizeof(a)/sizeof(a[0]);
 printf("Maximum difference is %d", maxDiff(a, len));
 return 0;
}