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hey! i am also unable to explain but one thing i got
if we use %d in place of %f,we will get the correct output.
we must use %d as x is of int data type.so thq code is itself
wrong.
or
u can change x to float data type
--
Ashutosh Tripathi
UG Student
Computer
ya ...
but you have to modify the structure ..
keep an extra pointer for max and min in each node ..
that will point to min and max upto that node ..
that means when you enter smthing you have to compare it with previous min
and previous max . and set the pointer accordin to
try that .. its safe .
n*ceil(log_2n)
On Sep 24, 9:23 am, Krunal Modi wrote:
> But, how was it derived ? :(
> What is the intuition behind ? :( :(
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Here is a C++ version of the algorithm to solve this problem.
#include
#define TRUE 1
#define FALSE 0
typedef struct Binarr{
int iNumber;
};
class binaryarr{
Binarr * arr;
public:
int iHead;
int iSize;
int iZeroes;
int iOnes;
int iTail;
binaryarr(void);
~bi
Dave, check my pseudo code. Both the tasks of checking if the number of 1s
and 0s is equal and sorting the array seems to to work in a single pass.
2010/8/19 Dave
> It seems easy enough to put all of the zeros in the even positions and
> all of the ones in the odd positions in one pass. The dif
element at the tail by going down the tail.
Regards
Ashutosh
Pseudo code of the algorithm
head = 0;
tail = sizeOfArray;
zeros = 0;
ones = 0;
for (head = 0; head < sizeOfArray; head++) {
if head is odd
{
if array[head] is 0
{
zeros = zeros + 1
if tail = head; tail = tail
Hi Amit
Am I allowed to keep counters to count the number of 1's and 0s right?
The condition of not using extra memory is for the array!?
Regards
Ashutosh
2010/8/18 amit
> you are given an array of integers containing only 0s and 1s. you have
> to place all the 0s in even positio
..
Regards
Ashutosh
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For more opti
try it by longest common sequence.
then interchange values .
then delete and insert or watever..
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