tring(), Search.class);
>> *}*
>>
>>
>> *but its not working any idea ?*
>>
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>> On Thu, May 30, 2013 at 9:40 AM, Ankit Agarwal wrote:
&g
that highest reminder can be got if I divide the number by (⌊(n/2)
>>>>> ⌋+1) .Can anyone give me pointers ?
>>>>>
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I think it's the only way as you need to traverse the entire binary
tree to do it.
On Oct 31, 9:45 pm, Ankuj Gupta wrote:
> How to convert a Binary tree to BST ? Naive way is to create each node
> of Binary tree one by one and keep on creating the BST.
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diagonal and find the middle element, that will be the
median.
Thanks
Ankit Agarwal
On Nov 5, 1:29 am, Gene wrote:
> Here's an idea. Say we pick any element P in the 2D array A and use
> it to fill in an N element array X as follows.
>
> j = N;
> for i = 1 to N do
>
2 question
numbers are (a1+1)*a2*a3... an = a1*a2*a3...an + a2*a3...an
the first term is same...
for second term is (a1*a2...an)/(a1)
now we have to find max of ( ((a1*a2..an)/a1), (a1*a2...an)/a2)
so the question of max becomes min of( a1, a2, a3... an)
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let x = 2^j + 2 ^i
new number after swapping the digits is x XOR n
eg n = 1101
j = 6 i = 2
x = 0100 0100
new number = x XOR n = 0100 1001
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Department of Electronics & Computer Engineering
Indian Institute of Te
void function(int start, int end){
if (start > end) return;
int x= a[end]-a[start];
a[start] = a[start] + x;
a[end] = a[end] - x;
function(start+1, end-1);
}
main(){
scanf("%s", &a[0]);
int len = strlen(a);
function(0,len-1);
printf("%s\
>
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>> > Deoki Nandan Vishwakarma
>> > IITR MCA
>> > *
>> > *
>>
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This problem can also be done by using Merging function as in the merge
sort.
1. Copy the sorted elements of the first half in one array (arr L) and
second half in another (arr R). Original array N.
2. count vary from 1 to n.
if (L[i] < R[j] ) { N[count] = L[i], i++}
else { N[count] = R[j]
less quantity with 0.08
> i.e 'if ' condition is true and print it HELLO
>
> On Sun, May 29, 2011 at 9:21 AM, Ankit Agarwal wrote:
>
>> #include
>>
>> int main(void)
>> {
>> float a=0.08;
>> if(a<0.08)
>> printf("
#include
int main(void)
{
float a=0.08;
if(a<0.08)
printf("Hello\n");
else
printf("Hii\n");
return 0;
}
The o/p is: *Hello * why
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&quo
Thanq ol...
On Mon, Feb 7, 2011 at 8:56 PM, Dave wrote:
> @Ankit: Yes. The bug is that the constant in the comparison should
> have been written 11.202e0.
>
> Dave
>
> On Feb 7, 9:05 am, ankit agarwal wrote:
> > But is the bug in the given program
> >
>
uot;Hello!!!\n");
> return 0;
> }
> you will get :
> Hello!!!
>
> On Mon, Feb 7, 2011 at 8:22 PM, ankit agarwal wrote:
>
>> #include
>>
>> int main()
>> {
>> float a=11.202;
>> if(a<11.202)
>> printf(&qu
#include
int main()
{
float a=11.202;
if(a<11.202)
printf("Hiii!!!\n");
else
printf("Hello!!!\n");
return 0;
}
output: Hiii!!!
why does this output comes???
Ankit Agarwal
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>
>
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> IIT Madras.
>
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> You received this message because you are subscr
it is (p(a)p(b)+p(b)p(c)+p(c)p(a))/3
On Jan 12, 1:51 am, SVIX wrote:
> anuragh
>
> assume each can shoot the target everytime...
> P(A) = 1
> P(B) = 1
> P(C) = 1
>
> per your logic, the probability that the target will be hit is 3
> actually, it should have only been 2 as we're going to p
as all prime no. greater than 3 are of the form 6n+1 or 6n-1
so start checking for all these numbers and if they both are prime
then they will make pair
count the pair no. as well as u move on
Ankit
On Jan 11, 9:23 pm, snehal jain wrote:
> you will be given an input no. n u have to output nth tw
@mohit,ruturaj
We dont need to know the range only if there are +ve no's.
and then also we need only m/2 space.
if(a[i]>m)
skip
else
if(a[i]>m/2)
if(hash[m-a[i]]==1))
return true;
else
hash[m-a[i]]++;
else
if(hash[a[i]]==1))
return true;
else
hash[a[i]]+
Hi,
What will happen if the numbers are 1,1,10001,10002 etc..
then you will lot of memory.
On Nov 4, 11:41 am, Naveen Kumar wrote:
> Hi,
>
> If we don't have to preserve the order than,
>
> a[1...N] can be used and if we have input like 1,4,3,6
>
> Insertion can be done as
> a[1]++, a[4]++,
Do level order traversal using two queues.
On Oct 23, 8:19 pm, "juver++" wrote:
> When visiting appropriate vertex v, increment counter +
> +levels[current_depth] and go further.
> You may done this using DFS or BFS.
>
> On 23 окт, 17:31, Harshal wrote:
>
>
>
>
>
>
>
> > hi, i need to find the
case 1: L+R=N+1
arrange Nth bloack at L position and from N-1 to N-L on its left in
decreasing order.
N-L-1 to 1
on its right in decreasing order.
eg 3 4 5 6 2 1
for n=6
l=4
r=3
L+R wrote:
> You are given N blocks of height 1…N
when you press enter after entering a character.
scanf reads the character first time from the input buffer and the
next time it reads only enter from input buffer.
Solution to the problem is using "fflush(stdin)" after scanf command.
On Oct 12, 2:25 am, carry wrote:
> int main()
> {
> int i;
> c
Your Solution considers only those windows which contain continuous
elements of query array in input array.
It won't work for:
i/p array : 12345
query array: 1235
also it won't work for:
i/p array: 2
query array: 1112
Ankit Agarwal
On Oct 10, 9:04 pm, Mridul Malpani wrote:
&
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