*for i=1 to sumofXY/2*
*if (xorofXY XOR i XOR (sumofXY-i)) is zero*
*then x is i, y is (sumofXY-i)*
*endfor*
Channa, is this logic based on the assumption/fact that no other other pair
of numbers (m, n) can have the same XOR and SUMMATION as that of the (x, y)
pair? If so, could you please
@Dave,
*Otherwise, set the even position to 0 and the odd position to 1.*
I think your solution might be inserting 0's and 1's into the array from
nowhere (thus filling the whole array with alternating 0's and 1's up to the
given size !). The question is to re-arrange existing elements in the
switch them. Switching a 1 and a 0 simply means storing 0
where the 1 was and storing 1 where the 0 was; there is no need to use
the usual code to interchange two values.
Dave
On Nov 13, 8:37 am, Vikram Venkatesan [EMAIL PROTECTED]
wrote:
@Dave,
*Otherwise, set the even position to 0
Hi,
I think the question is not to FORM A MATRIX, but to PRINT the elements
in the form of spiral matrix, where the problem of SEQUENTIAL OUTPUT
comes...
-Vikram
On 12/6/06, Gene [EMAIL PROTECTED] wrote:
hijkl wrote:
this question was asked by Google..
Write a program of spiral
Hi Ksitami,
In the first algorithm, what if the CYLCE OF REPLACEMENTS sorts the
whole array(I mean, all the elements are visited during the 1st iteration
itself, thus creating a COMPLETE CYCLE of the array, putting it in the right
order), and the array gets to the FINAL position (i.e.
Hi, A similar but straightfoward O(n^2) solution ;-) For each element a[i], traverse on either side of the array with a[i] as the pivot until you reach elements a[j]a[i] and a[k]a[i](Two sides). Keep track of the current summations from a[i] (Probably have an array of sub-array
Hi,The problem with that PAIRING solution is that, at each step, some information is lostSay, the array is 2,5,2,7,2,8,2,1While taking elements pair by pair, first, since 2 !=5 , we skip it... (At this step itself, the information about the occurence of '2' is lost) The same happens in the
correct me if i am wrongRegards,VikramOn 6/21/06,
Pradeep Muthukrishnan [EMAIL PROTECTED] wrote:
Please read the question properly...The number can be a majorityelement only if it repeats more than n/2 timesIn your example itdoesnt...On 6/20/06, Vikram Venkatesan
[EMAIL PROTECTED] wrote: Hi
Hi,
It can be derived as follows...
Let f(n) denote the no. of ways in which 'n' steps can be climbed...
So, for climbing 'n' steps, the possible combinations are,
1. Climb one step initially.. so, n-1 steps left.. they can be climbed
in f(n-1) ways, OR
2. Climb 2 steps initally... so, n-2
I agree with phoenixinter... whats the problem with that?? its the
obvious solution, provided O(n) precomputation time... Abhi, can you
explain whether there is any problem with that solution??
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