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Given an array of size N in which every number is between 1
and N, determine if there are any duplicates in it. You are
allowed to destroy the array if you like.
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Answer Posted / faykarta
I'm nearly 100% sure that the only way to do this without
allocating more space is to loop through the entire array
checking elements against each other. Because its an
iterative method it can be optimised for linear memory access.
consider:
for(i = 0; i < N; ++i)
{
for(j = i + 1; j < N; ++j)
{
if(array[i] == array[j])
{
//index look up avoided through compiler optimisation???
return true;
}
}
}
return false;
*(Cache Integrity)The optimisation could be made using
pointer arithmetic or iterator style traversal in C/C++ anyway.
You could perform a custom sort operation that throws out
but i think it will be about the same algorithmic complexity
with less linear memory access and more complex code....
consider: (havent tested this one)
for(i = 0; i < N; ++i)
{
j = array[i] - 1;
while(j != i)
{
//check
if(array[i] == array[j])
{
return true;
}
//swap
array[i] ^= array[j] ^= array[i] ^= array[j];
j = array[i] - 1;
}
}
return false;
Any thoughts on these?
I would stick with brute force method.
| Is This Answer Correct ? | 1 Yes | 1 No |
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