Re: [algogeeks] Re: Repeated values
yes, that was an implementation mistake but what I meant to say was- Adding extra check of indirect xor'ing could have a pitfall too. Try the case: [0 1 1 1 4 4] http://ideone.com/3sreLZ On 4 November 2012 10:13, Vikram Pradhan vpradha...@gmail.com wrote: It should have caught in the first loop where i checked that condition if the first value (which is 3 in this case) is repeated or not ...but unfortunately i implemented that wrong ...so now i've corrected that mistake ..i hope now it'll work fine .. http://ideone.com/MQ44Rt the whole idea of using xor is that we don't have to modify the array as i've seen this same question somewhere else where the condition was that the array is a read only array ..otherwise Don's method will work fine . Vikram Pradhan | B.Tech| Computer Science Engineering | NIT Jalandhar | 9740186063 | -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Repeated values
@Don :yeah sorry i misinterpreted the if condition ..it'll work fine .
I've modified my previous sol. and in this sol we do not need to modify the
array . Time complexity O(n) and constant space.
http://ideone.com/s2kR24
On Fri, Nov 2, 2012 at 1:14 AM, Don dondod...@gmail.com wrote:
It won't enter an infinite loop in that case. In fact, it would
immediately return.
Don
On Oct 31, 2:39 pm, Vikram Pradhan vpradha...@gmail.com wrote:
@Don It will be an infinite loop for some cases ...like try this i=1,
and
a[1] = 5 , a[5] = 5
*Solution:*
As the numbers are from 0 to N-1 so we can xor the value with its index
in
a loop . if the result is 0 then there is no repetition else there is
some
repetition.
*int result = 0;*
*for(int i=0;iN;i++)*
*{*
*result ^= i ^ array[i];*
*}*
*
*
*if(result==0)*
*There is no repetition.*
*else*
*There is some repetition.*
Time complexity O(N)
Space Complexity : constant
check this :http://ideone.com/RXyynB
As the indexes are from 0 to N-1 and numbers are also from 0 to N-1 in
random order. So if there is no repetition then there is exactly two
copies
of same number in set of (values and index) and when we xor all the
indexes
to all the numbers the result will be zero because xor of same no. is
zero.
--
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|
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Re: [algogeeks] Re: Repeated values
@Vikram - your approach fails for [4 1 1 1 1]
On 1 November 2012 00:09, Vikram Pradhan vpradha...@gmail.com wrote:
@Don It will be an infinite loop for some cases ...like try this i=1, and
a[1] = 5 , a[5] = 5
*Solution:*
As the numbers are from 0 to N-1 so we can xor the value with its index in
a loop . if the result is 0 then there is no repetition else there is some
repetition.
*int result = 0;*
*for(int i=0;iN;i++)*
*{*
*result ^= i ^ array[i];*
*}*
*
*
*if(result==0)*
*There is no repetition.*
*else*
*There is some repetition.*
Time complexity O(N)
Space Complexity : constant
check this : http://ideone.com/RXyynB
As the indexes are from 0 to N-1 and numbers are also from 0 to N-1 in
random order. So if there is no repetition then there is exactly two copies
of same number in set of (values and index) and when we xor all the indexes
to all the numbers the result will be zero because xor of same no. is zero.
--
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9740186063 |
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[algogeeks] Re: Repeated values
It won't enter an infinite loop in that case. In fact, it would
immediately return.
Don
On Oct 31, 2:39 pm, Vikram Pradhan vpradha...@gmail.com wrote:
@Don It will be an infinite loop for some cases ...like try this i=1, and
a[1] = 5 , a[5] = 5
*Solution:*
As the numbers are from 0 to N-1 so we can xor the value with its index in
a loop . if the result is 0 then there is no repetition else there is some
repetition.
*int result = 0;*
*for(int i=0;iN;i++)*
*{*
*result ^= i ^ array[i];*
*}*
*
*
*if(result==0)*
*There is no repetition.*
*else*
*There is some repetition.*
Time complexity O(N)
Space Complexity : constant
check this :http://ideone.com/RXyynB
As the indexes are from 0 to N-1 and numbers are also from 0 to N-1 in
random order. So if there is no repetition then there is exactly two copies
of same number in set of (values and index) and when we xor all the indexes
to all the numbers the result will be zero because xor of same no. is zero.
--
Vikram Pradhan | B.Tech| Computer Science Engineering | NIT Jalandhar |
9740186063 |
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Re: [algogeeks] Re: Repeated values
@Don : Shift all the numbers to the corresponding positions in the first
iteration
And in the second loop check whether the number and the position matches or
not.
O(N) space and O(1) time
constraint : the number's range must be from 0 to N-1
Cheers,
On Wed, Oct 31, 2012 at 3:16 AM, Don dondod...@gmail.com wrote:
In addition, inputs such as {2,2,0} will fail because when you try to
make zero negative, it doesn't work so well.
Don
On Oct 30, 5:36 pm, Don dondod...@gmail.com wrote:
That solution is using the sign bit as extra storage, which is clever,
but if you have an array of unsigned integers and N is more than half
of the largest unsigned integer, it won't work. There is a way to do
it without using the sign bit as extra storage.
On Oct 30, 5:16 pm, rahul sharma rahul23111...@gmail.com wrote:
i thnik this is the solution...
http://www.geeksforgeeks.org/archives/9755
On Wed, Oct 31, 2012 at 2:20 AM, Don dondod...@gmail.com wrote:
We can modify the array. The algorithm should work even if we use
unsigned integers and N is the largest unsigned integer.
Don
On Oct 30, 4:42 pm, rahul sharma rahul23111...@gmail.com wrote:
Can we modify the array???we can make index we visit as negative
and then
if any one already containing -ve..then its repeating
On Wed, Oct 31, 2012 at 1:40 AM, Don dondod...@gmail.com wrote:
Does your algorithm work if N=4 and the array is {1,1,2,2}.
Don
On Oct 30, 2:32 pm, arumuga abinesh arumugaabin...@gmail.com
wrote:
if sum of all elements = n(n-1)/2 , no elements are repeated
else some numbers are repeated
On Tue, Oct 30, 2012 at 11:57 PM, Don dondod...@gmail.com
wrote:
Given an array of N integers in the range 0..N-1, determine
if any
number is repeated in the array.
Solution should execute in O(n) time and use constant space.
Don
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[algogeeks] Re: Repeated values
I think this is the answer - http://stackoverflow.com/questions/5739024/finding-duplicates-in-on-time-and-o1-space On Tuesday, October 30, 2012 2:27:29 PM UTC-4, Don wrote: Given an array of N integers in the range 0..N-1, determine if any number is repeated in the array. Solution should execute in O(n) time and use constant space. Don -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To view this discussion on the web visit https://groups.google.com/d/msg/algogeeks/-/XPyEFSy1Gh0J. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Repeated values
A solution is to sort the array taking advantage of the fact that if
the contents are in fact a permutation of the values 0..n-1, value x
always goes in position A[x]. Therefore you can immediately swap value
x into the correct position. If that position already contains value
x, you have identified a duplicate.
bool hasDuplicate(int *a, int n)
{
for(int i = 0; i n; ++i)
{
while(a[i] != i)
{
int j = a[i];
if (a[j] == j) return true;
a[i] = a[j];
a[j] = j;
}
}
return false;
}
At first it may appear to be O(n^2) but notice that each pass through
the inner loop puts one value in its correct place. Thus the inner
loop executes at most n times. Unlike a similar solution above, if it
finds a duplicate it immediately returns, so in some cases it doesn't
require n iterations.
This solution is also faster than the solution using the sign bit,
based on a benchmark test I ran.
Don
On Oct 30, 2:27 pm, Don dondod...@gmail.com wrote:
Given an array of N integers in the range 0..N-1, determine if any
number is repeated in the array.
Solution should execute in O(n) time and use constant space.
Don
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[algogeeks] Re: Repeated values
Does your algorithm work if N=4 and the array is {1,1,2,2}.
Don
On Oct 30, 2:32 pm, arumuga abinesh arumugaabin...@gmail.com wrote:
if sum of all elements = n(n-1)/2 , no elements are repeated
else some numbers are repeated
On Tue, Oct 30, 2012 at 11:57 PM, Don dondod...@gmail.com wrote:
Given an array of N integers in the range 0..N-1, determine if any
number is repeated in the array.
Solution should execute in O(n) time and use constant space.
Don
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Re: [algogeeks] Re: Repeated values
Can we modify the array???we can make index we visit as negative and then
if any one already containing -ve..then its repeating
On Wed, Oct 31, 2012 at 1:40 AM, Don dondod...@gmail.com wrote:
Does your algorithm work if N=4 and the array is {1,1,2,2}.
Don
On Oct 30, 2:32 pm, arumuga abinesh arumugaabin...@gmail.com wrote:
if sum of all elements = n(n-1)/2 , no elements are repeated
else some numbers are repeated
On Tue, Oct 30, 2012 at 11:57 PM, Don dondod...@gmail.com wrote:
Given an array of N integers in the range 0..N-1, determine if any
number is repeated in the array.
Solution should execute in O(n) time and use constant space.
Don
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[algogeeks] Re: Repeated values
We can modify the array. The algorithm should work even if we use
unsigned integers and N is the largest unsigned integer.
Don
On Oct 30, 4:42 pm, rahul sharma rahul23111...@gmail.com wrote:
Can we modify the array???we can make index we visit as negative and then
if any one already containing -ve..then its repeating
On Wed, Oct 31, 2012 at 1:40 AM, Don dondod...@gmail.com wrote:
Does your algorithm work if N=4 and the array is {1,1,2,2}.
Don
On Oct 30, 2:32 pm, arumuga abinesh arumugaabin...@gmail.com wrote:
if sum of all elements = n(n-1)/2 , no elements are repeated
else some numbers are repeated
On Tue, Oct 30, 2012 at 11:57 PM, Don dondod...@gmail.com wrote:
Given an array of N integers in the range 0..N-1, determine if any
number is repeated in the array.
Solution should execute in O(n) time and use constant space.
Don
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Re: [algogeeks] Re: Repeated values
i thnik this is the solution...
http://www.geeksforgeeks.org/archives/9755
On Wed, Oct 31, 2012 at 2:20 AM, Don dondod...@gmail.com wrote:
We can modify the array. The algorithm should work even if we use
unsigned integers and N is the largest unsigned integer.
Don
On Oct 30, 4:42 pm, rahul sharma rahul23111...@gmail.com wrote:
Can we modify the array???we can make index we visit as negative and then
if any one already containing -ve..then its repeating
On Wed, Oct 31, 2012 at 1:40 AM, Don dondod...@gmail.com wrote:
Does your algorithm work if N=4 and the array is {1,1,2,2}.
Don
On Oct 30, 2:32 pm, arumuga abinesh arumugaabin...@gmail.com wrote:
if sum of all elements = n(n-1)/2 , no elements are repeated
else some numbers are repeated
On Tue, Oct 30, 2012 at 11:57 PM, Don dondod...@gmail.com wrote:
Given an array of N integers in the range 0..N-1, determine if any
number is repeated in the array.
Solution should execute in O(n) time and use constant space.
Don
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[algogeeks] Re: Repeated values
That solution is using the sign bit as extra storage, which is clever,
but if you have an array of unsigned integers and N is more than have
of the largest unsigned integer, it won't work. There is a way to do
it without using the sign bit as extra storage.
Don
On Oct 30, 5:16 pm, rahul sharma rahul23111...@gmail.com wrote:
i thnik this is the solution...http://www.geeksforgeeks.org/archives/9755
On Wed, Oct 31, 2012 at 2:20 AM, Don dondod...@gmail.com wrote:
We can modify the array. The algorithm should work even if we use
unsigned integers and N is the largest unsigned integer.
Don
On Oct 30, 4:42 pm, rahul sharma rahul23111...@gmail.com wrote:
Can we modify the array???we can make index we visit as negative and then
if any one already containing -ve..then its repeating
On Wed, Oct 31, 2012 at 1:40 AM, Don dondod...@gmail.com wrote:
Does your algorithm work if N=4 and the array is {1,1,2,2}.
Don
On Oct 30, 2:32 pm, arumuga abinesh arumugaabin...@gmail.com wrote:
if sum of all elements = n(n-1)/2 , no elements are repeated
else some numbers are repeated
On Tue, Oct 30, 2012 at 11:57 PM, Don dondod...@gmail.com wrote:
Given an array of N integers in the range 0..N-1, determine if any
number is repeated in the array.
Solution should execute in O(n) time and use constant space.
Don
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[algogeeks] Re: Repeated values
That solution is using the sign bit as extra storage, which is clever,
but if you have an array of unsigned integers and N is more than half
of the largest unsigned integer, it won't work. There is a way to do
it without using the sign bit as extra storage.
On Oct 30, 5:16 pm, rahul sharma rahul23111...@gmail.com wrote:
i thnik this is the solution...http://www.geeksforgeeks.org/archives/9755
On Wed, Oct 31, 2012 at 2:20 AM, Don dondod...@gmail.com wrote:
We can modify the array. The algorithm should work even if we use
unsigned integers and N is the largest unsigned integer.
Don
On Oct 30, 4:42 pm, rahul sharma rahul23111...@gmail.com wrote:
Can we modify the array???we can make index we visit as negative and then
if any one already containing -ve..then its repeating
On Wed, Oct 31, 2012 at 1:40 AM, Don dondod...@gmail.com wrote:
Does your algorithm work if N=4 and the array is {1,1,2,2}.
Don
On Oct 30, 2:32 pm, arumuga abinesh arumugaabin...@gmail.com wrote:
if sum of all elements = n(n-1)/2 , no elements are repeated
else some numbers are repeated
On Tue, Oct 30, 2012 at 11:57 PM, Don dondod...@gmail.com wrote:
Given an array of N integers in the range 0..N-1, determine if any
number is repeated in the array.
Solution should execute in O(n) time and use constant space.
Don
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[algogeeks] Re: Repeated values
In addition, inputs such as {2,2,0} will fail because when you try to
make zero negative, it doesn't work so well.
Don
On Oct 30, 5:36 pm, Don dondod...@gmail.com wrote:
That solution is using the sign bit as extra storage, which is clever,
but if you have an array of unsigned integers and N is more than half
of the largest unsigned integer, it won't work. There is a way to do
it without using the sign bit as extra storage.
On Oct 30, 5:16 pm, rahul sharma rahul23111...@gmail.com wrote:
i thnik this is the solution...http://www.geeksforgeeks.org/archives/9755
On Wed, Oct 31, 2012 at 2:20 AM, Don dondod...@gmail.com wrote:
We can modify the array. The algorithm should work even if we use
unsigned integers and N is the largest unsigned integer.
Don
On Oct 30, 4:42 pm, rahul sharma rahul23111...@gmail.com wrote:
Can we modify the array???we can make index we visit as negative and
then
if any one already containing -ve..then its repeating
On Wed, Oct 31, 2012 at 1:40 AM, Don dondod...@gmail.com wrote:
Does your algorithm work if N=4 and the array is {1,1,2,2}.
Don
On Oct 30, 2:32 pm, arumuga abinesh arumugaabin...@gmail.com wrote:
if sum of all elements = n(n-1)/2 , no elements are repeated
else some numbers are repeated
On Tue, Oct 30, 2012 at 11:57 PM, Don dondod...@gmail.com wrote:
Given an array of N integers in the range 0..N-1, determine if any
number is repeated in the array.
Solution should execute in O(n) time and use constant space.
Don
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