Technically linear!
On Mon, Nov 26, 2012 at 9:47 PM, shady sinv...@gmail.com wrote:
what is the time complexity of this?
str_reverse(str){
if(isempty(str)) return str;
else if(length(str) = even) then split str into str_1 and str_2; (of
equal length) (Calculate mid =O(1), then
Yes, my bad. I din notice the recursion at all! Thot it to be a flat
mid-split followed by a reverse followed by a concat. Thanks.
On Mon, Nov 26, 2012 at 11:18 PM, atul anand atul.87fri...@gmail.comwrote:
considering '+' , here will take Cn time . Here '+' is for concatenate ,
now this
Reversing a string should be linear time.
void reverse(char *str)
{
int i = 0;
int j = strlen(str)-1;
while(i j)
swap(str[i++], str[--j]);
}
On Nov 27, 12:47 am, shady sinv...@gmail.com wrote:
what is the time complexity of this?
str_reverse(str){
if(isempty(str)) return
i guess it should be swap(str[i++], str[j--]); because we
already subtracted 1
On Tue, Nov 27, 2012 at 8:58 PM, Don dondod...@gmail.com wrote:
swap(str[i++], str[--j]);
--
Define the deficit of a subarray S to be k*size(S)-sum(S), or how much
its sum falls short of averaging k.
If the deficit of the entire array is negative, the result is the
entire array.
Otherwise, the result is the subarray created by removing the smallest
possible subarray from the front and
Define the deficit of a subarray S to be k*size(S)-sum(S), or how much
its sum falls short of averaging k.
If the deficit of the entire array is negative, the result is the
entire array.
Otherwise, the result is the subarray created by removing the smallest
possible subarray from the front and
Define the deficit of a subarray S to be k*size(S)-sum(S), or how much
its sum falls short of averaging k.
If the deficit of the entire array is negative, the result is the
entire array.
Otherwise, the result is the subarray created by removing the smallest
possible subarray from the front and
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