Thanks for pointing out the mistake.Though my code will correctly calculate
the max_profit but I must keep track of the buying_day.I have made some
modification in my code.Hope it works fine now.
int min = a[0]; // initialized to first element
int max_profit = 0; //when you buy and sell on same day
int buying_day = a[0];
for ( int i = 1; i < n; i++ ){
if ( max_profit < (a[i] - min ) ){
max_profit = a[i] - min;
buying_day = min;
}
if ( a[i] < min )
min = a[i];
}
Finally. I'll have buying_day and max_profit, so if you need to find the
selling day you can easily calculate :
Selling day = buying_day+max_profit;
Correct me if I'm wrong.
On Sun, Aug 5, 2012 at 5:43 PM, Arun Kindra <arunkin...@gmail.com> wrote:
> @harsha : yes, the problem is if u r finding only min and max value, it
> might happen that u sell the stock before buying. Ex- int a [ ] = { 5, 10,
> 4, 6, 7 }; the min value is 4 and max is 10 and 10 comes before 4, means u
> sell the stock before buying.
> and i think the sol given by mukul does the same mistake.we need to keep
> track this case also whether the day(array index) i m buying is not more
> than the day(array index) we are selling the stock.
>
> *correct me if m wrong*.....
>
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