@luckyzoner
can post the c program of what u ave said above..
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In this method overflow will be there..if number is just bigger...so by
doing XOR we can get missing number and repeated number .
take xor of all element of array and take this xor with array[1...n]
So we get xor of two numbers.
now get set bit of this xor and proceed.
On Thu, Sep 2, 2010 at
I think it will be 1x
On Wed, Sep 1, 2010 at 10:53 PM, Yan Wang wangyanadam1...@gmail.com wrote:
Maybe you misunderstand the question.
The question is how to compute 2^X where 0 = X = 9?
How?
On Wed, Sep 1, 2010 at 10:48 PM, Ruturaj rutura...@gmail.com wrote:
a 5 digit number is
Suppose the number of shifts be x.
Also let the integer be represented by 16 bits on that machine.
Now take int n= (int)(x/16 + 0.5), to take the upper cap on result :) .
SO having 2^x will be same as doing 2(x-1) so essentially if we represent
the resultant number in a linked list of nodes, where
nice...
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@ashish: cud u plzz explain a bit more...
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can post the c program of what u ave said above
On 2 September 2010 16:06, vikash jain vikash.ro...@gmail.com wrote:
nice...
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this question can be sovled very easily
1.jus sum the given array...x
2.sum the squares of the given array..y
3.now use the AP.n(n+1)/2..for n=100
4.similarly compute n(n+1)(2n+1)/6 for n =100..
Now solve these eqns ...u get the missing and the dupicate..
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trie will be the best choice for this..
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For
@Dhritiman, It's good algo man!!!The only thing is we are destroying
the array but also that's mandatory as only o(n) complexity we are
interested in.
As Somebody wanted the code, here I am attaching below: -
int a[SIZE_A] = {0,2,1,4,0};
int i = 0, dup = 0, pos = 0, t =0;
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