Guys i have read that we cant define function in another function in c
Then why this followung program running fine on gcc
#include
void abc()
{
printf("bac");
void abf()
{
printf("bas");
getchar();
}
}
int main()
{
abc();
getch
@umer : how no. of comparison are reduced to half by moving both
sidesyou have 2 if condition inside, so you are making 2
comparisons at each iteration + n/2 comparison for while loop so
number of comparisons are n+n/2
On 10/2/12, Umer Farooq wrote:
> why don't we try it from both ends ... so
@Umer: I say that you forgot to increment i and decrement j in the loop.
But you don't need any loop-counting comparisons at all.
Dave
On Tuesday, October 2, 2012 3:36:58 AM UTC-5, Umer Farooq wrote:
> why don't we try it from both ends ... something like this:
>
> int i = 0; j = size-1;
>
>
The following link might help you:
http://www.geeksforgeeks.org/archives/767
On Tue, Oct 2, 2012 at 1:03 AM, rahul sharma wrote:
>
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why don't we try it from both ends ... something like this:
int i = 0; j = size-1;
while (i != j)
{
if (arr[i] == elem)
return arr[i];
if (arr[j] == elem)
return arr[j];
}
this way, we have eliminated half of the comparisons in for loop? What do
you guys say?
On Tue
@Rafi: Try this:
int arrExsits( int* arr, int size, int elem)
{
int temp = arr[0];
arr[0] = elem;
while( arr[--size] != elem );
arr[0] = temp;
return a[size] == elem ? size : -1;
}
n+1 comparisons, and it leaves the array unaltered.
Dave
On Sunday, September 30, 2012 4:27:2
Vikas is right!
while (n) is equal to (while n!=0)
you have 2n compares!
בתאריך יום שני, 1 באוקטובר 2012 12:12:21 UTC+2, מאת vikas:
>
> still there is no improvement, compiler will generate the code to compare
> with zero here. what you have accomplished is , hide it from human eyes
>
> On Monday
