How to tell if an array is a permutation in O(n)? - arrays

Question

Input: A read-only array of N elements containing integer values from 1 to N (some integer values can appear more than once!). And a memory zone of a fixed size (10, 100, 1000 etc - not depending on N).
How to tell in O(n) if the array represents a permutation?
--What I achieved so far (an answer proved that this was not good):--
I use the limited memory area to store the sum and the product of the array.
I compare the sum with N*(N+1)/2 and the product with N!
I know that if condition (2) is true I might have a permutation. I'm wondering if there's a way to prove that condition (2) is sufficient to tell if I have a permutation. So far I haven't figured this out ...

Answer 1

I'm very slightly skeptical that there is a solution. Your problem seems to be very close to one posed several years ago in the mathematical literature, with a summary given here ("The Duplicate Detection Problem", S. Kamal Abdali, 2003) that uses cycle-detection -- the idea being the following:
If there is a duplicate, there exists a number j between 1 and N such that the following would lead to an infinite loop:
x := j;
do
{
x := a[x];
}
while (x != j);
because a permutation consists of one or more subsets S of distinct elements s0, s1, ... sk-1 where sj = a[sj-1] for all j between 1 and k-1, and s0 = a[sk-1], so all elements are involved in cycles -- one of the duplicates would not be part of such a subset.
e.g. if the array = [2, 1, 4, 6, 8, 7, 9, 3, 8]
then the element in bold at position 5 is a duplicate because all the other elements form cycles: { 2 -> 1, 4 -> 6 -> 7 -> 9 -> 8 -> 3}. Whereas the arrays [2, 1, 4, 6, 5, 7, 9, 3, 8] and [2, 1, 4, 6, 3, 7, 9, 5, 8] are valid permutations (with cycles { 2 -> 1, 4 -> 6 -> 7 -> 9 -> 8 -> 3, 5 } and { 2 -> 1, 4 -> 6 -> 7 -> 9 -> 8 -> 5 -> 3 } respectively).
Abdali goes into a way of finding duplicates. Basically the following algorithm (using Floyd's cycle-finding algorithm) works if you happen across one of the duplicates in question:
function is_duplicate(a, N, j)
{
/* assume we've already scanned the array to make sure all elements
are integers between 1 and N */
x1 := j;
x2 := j;
do
{
x1 := a[x1];
x2 := a[x2];
x2 := a[x2];
} while (x1 != x2);
/* stops when it finds a cycle; x2 has gone around it twice,
x1 has gone around it once.
If j is part of that cycle, both will be equal to j. */
return (x1 != j);
}
The difficulty is I'm not sure your problem as stated matches the one in his paper, and I'm also not sure if the method he describes runs in O(N) or uses a fixed amount of space. A potential counterexample is the following array:
[3, 4, 5, 6, 7, 8, 9, 10, ... N-10, N-9, N-8, N-7, N-2, N-5, N-5, N-3, N-5, N-1, N, 1, 2]
which is basically the identity permutation shifted by 2, with the elements [N-6, N-4, and N-2] replaced by [N-2, N-5, N-5]. This has the correct sum (not the correct product, but I reject taking the product as a possible detection method since the space requirements for computing N! with arbitrary precision arithmetic are O(N) which violates the spirit of the "fixed memory space" requirement), and if you try to find cycles, you will get cycles { 3 -> 5 -> 7 -> 9 -> ... N-7 -> N-5 -> N-1 } and { 4 -> 6 -> 8 -> ... N-10 -> N-8 -> N-2 -> N -> 2}. The problem is that there could be up to N cycles, (identity permutation has N cycles) each taking up to O(N) to find a duplicate, and you have to keep track somehow of which cycles have been traced and which have not. I'm skeptical that it is possible to do this in a fixed amount of space. But maybe it is.
This is a heavy enough problem that it's worth asking on mathoverflow.net (despite the fact that most of the time mathoverflow.net is cited on stackoverflow it's for problems which are too easy)
edit: I did ask on mathoverflow, there's some interesting discussion there.

Answer 2

This is impossible to do in O(1) space, at least with a single-scan algorithm.
Proof
Suppose you have processed N/2 of the N elements. Assuming the sequence is a permutation then, given the state of the algorithm, you should be able to figure out the set of N/2 remaining elements. If you can't figure out the remaining elements, then the algorithm can be fooled by repeating some of the old elements.
There are N choose N/2 possible remaining sets. Each of them must be represented by a distinct internal state of the algorithm, because otherwise you couldn't figure out the remaining elements. However, it takes logarithmic space to store X states, so it takes BigTheta(log(N choose N/2)) space to store N choose N/2 states. That values grows with N, and therefore the algorithm's internal state can not fit in O(1) space.
More Formal Proof
You want to create a program P which, given the final N/2 elements and the internal state of the linear-time-constant-space algorithm after it has processed N/2 elements, determines if the entire sequence is a permutation of 1..N. There is no time or space bound on this secondary program.
Assuming P exists we can create a program Q, taking only the internal state of the linear-time-constant-space algorithm, which determines the necessary final N/2 elements of the sequence (if it was a permutation). Q works by passing P every possible final N/2 elements and returning the set for which P returns true.
However, because Q has N choose N/2 possible outputs, it must have at least N choose N/2 possible inputs. That means the internal state of the original algorithm must store at least N choose N/2 states, requiring BigTheta(log N choose N/2), which is greater than constant size.
Therefore the original algorithm, which does have time and space bounds, also can't work correctly if it has constant-size internal state.
[I think this idea can be generalized, but thinking isn't proving.]
Consequences
BigTheta(log(N choose N/2)) is equal to BigTheta(N). Therefore just using a boolean array and ticking values as you encounter them is (probably) space-optimal, and time-optimal too since it takes linear time.

Answer 3

I doubt you would be able to prove that ;)
(1, 2, 4, 4, 4, 5, 7, 9, 9)
I think that more generally, this problem isn't solvable by processing the numbers in order. Suppose you are processing the elements in order and you are halfway the array. Now the state of your program has to somehow reflect which numbers you've encountered so far. This requires at least O(n) bits to store.

Answer 4

This isn't going to work due to the complexity being given as a function of N rather than M, implying that N >> M
This was my shot at it, but for a bloom filter to be useful, you need a big M, at which point you may as well use simple bit toggling for something like integers
http://en.wikipedia.org/wiki/Bloom_filter
For each element in the array
Run the k hash functions
Check for inclusion in the bloom filter
If it is there, there is a probability you've seen the element before
If it isn't, add it
When you are done, you may as well compare it to the results of a 1..N array in order, as that'll only cost you another N.
Now if I haven't put enough caveats in. It isn't 100%, or even close since you specified complexity in N, which implies that N >> M, so fundamentally it won't work as you have specified it.
BTW, the false positive rate for an individual item should be
e = 2^(-m/(n*sqrt(2)))
Which monkeying around with will give you an idea how big M would need to be to be acceptable.

Answer 5

I don't know how to do it in O(N), or even if it can be done in O(N). I know that it can be done in O(N log N) if you (use an appropriate) sort and compare.
That being said, there are many O(N) techniques that can be done to show that one is NOT a permutation of the other.
Check the length. If unequal, obviously not a permutation.
Create an XOR fingerprint. If the value of all the elements XOR'ed together does not match, then it can not be a permutation. A match would however be inconclusive.
Find the sum of all elements. Although the result may overflow, that should not be a worry when matching this 'fingerprint'. If however, you did a checksum that involved multiplying then overflow would be an issue.
Hope this helps.

Answer 6

You might be able to do this in randomized O(n) time and constant space by computing sum(x_i) and product(x_i) modulo a bunch of different randomly chosen constants C of size O(n). This basically gets you around the problem that product(x_i) gets too large.
There's still a lot of open questions, though, like if sum(x_i)=N(N+1)/2 and product(x_i)=N! are sufficient conditions to guarantee a permutation, and what is the chance that a non-permutation generates a false positive (I would hope ~1/C for each C you try, but maybe not).

Answer 7

it's a permutation if and only if there are no duplicate values in the array, should be easy to check that in O(N)

Answer 8

Depending on how much space you have, relative to N, you might try using hashing and buckets.
That is, iterate over the entire list, hash each element, and store it in a bucket. You'll need to find a way to reduce bucket collisions from the hashes, but that is a solved problem.
If an element tries to go into a bucket with an item identical to it, it is a permutation.
This type of solution would be O(N) as you touch each element only once.
However, the problem with this is whether space M is larger than N or not. If M > N, this solution will be fine, but if M < N, then you will not be able to solve the problem with 100% accuracy.

Answer 9

First, an information theoretic reason why this may be possible. We can trivially check that the numbers in the array are in bounds in O(N) time and O(1) space. To specify any such array of in-bounds numbers requires N log N bits of information. But to specify a permutation requires approximately (N log N) - N bits of information (Stirling's approximation). Thus, if we could acquire N bits of information during testing, we might be able to know the answer. This is trivial to do in N time (in fact, with M static space we can pretty easily acquire log M information per step, and under special circumstances we can acquire log N information).
On the other hand, we only get to store something like M log N bits of information in our static storage space, which is presumably much less than N, so it depends greatly what the shape of the decision surface is between "permutation" and "not".
I think that this is almost possible but not quite given the problem setup. I think one is "supposed" to use the cycling trick (as in the link that Iulian mentioned), but the key assumption of having a tail in hand fails here because you can index the last element of the array with a permutation.

Answer 10

The sum and the product will not guarantee the correct answer, since these hashes are subject to collisions, i.e. different inputs might potentially produce identical results. If you want a perfect hash, a single-number result that actually fully describes the numerical composition of the array, it might be the following.
Imagine that for any number i in [1, N] range you can produce a unique prime number P(i) (for example, P(i) is the i-th prime number). Now all you need to do is calculate the product of all P(i) for all numbers in your array. The product will fully and unambiguously describe the composition of your array, disregarding the ordering of values in it. All you need to do is to precalculate the "perfect" value (for a permutation) and compare it with the result for a given input :)
Of course, the algorithm like this does not immediately satisfy the posted requirements. But at the same time it is intuitively too generic: it allows you to detect a permutation of absolutely any numerical combination in an array. In your case you need to detect a permutation of a specific combination 1, 2, ..., N. Maybe this can somehow be used to simplify things... Probably not.

Answer 11

Alright, this is different, but it appears to work!
I ran this test program (C#):
static void Main(string[] args) {
for (int j = 3; j < 100; j++) {
int x = 0;
for (int i = 1; i <= j; i++) {
x ^= i;
}
Console.WriteLine("j: " + j + "\tx: " + x + "\tj%4: " + (j % 4));
}
}
Short explanation: x is the result of all the XORs for a single list, i is the element in a particular list, and j is the size of the list. Since all I'm doing is XOR, the order of the elements don't matter. But I'm looking at what correct permutations look like when this is applied.
If you look at j%4, you can do a switch on that value and get something like this:
bool IsPermutation = false;
switch (j % 4) {
case 0:
IsPermutation = (x == j);
break;
case 1:
IsPermutation = (x == 1);
break;
case 2:
IsPermutation = (x == j + 1);
break;
case 3:
IsPermutation = (x == 0);
break;
}
Now I acknowledge that this probably requires some fine tuning. It's not 100%, but it's a good easy way to get started. Maybe with some small checks running throughout the XOR loop, this could be perfected. Try starting somewhere around there.

Answer 12

it looks like asking to find duplicate in array with stack machine.
it sounds impossible to know the full history of the stack , while you extract each number and have limited knowledge of the numbers that were taken out.

Answer 13

Here's proof it can't be done:
Suppose by some artifice you have detected no duplicates in all but the last cell. Then the problem reduces to checking if that last cell contains a duplicate.
If you have no structured representation of the problem state so far, then you are reduced to performing a linear search over the entire previous input, for EACH cell. It's easy to see how this leaves you with a quadratic-time algorithm.
Now, suppose through some clever data structure that you actually know which number you expect to see last. Then certainly that knowledge takes at least enough bits to store the number you seek -- perhaps one memory cell? But there is a second-to-last number and a second-to-last sub-problem: then you must similarly represent a set of two possible numbers yet-to-be-seen. This certainly requires more storage than encoding only for one remaining number. By a progression of similar arguments, the size of the state must grow with the size of the problem, unless you're willing to accept a quadratic-time worst-case.
This is the time-space trade-off. You can have quadratic time and constant space, or linear time and linear space. You cannot have linear time and constant space.

Answer 14

Check out the following solution. It uses O(1) additional space.
It alters the array during the checking process, but returns it back to its initial state at the end.
The idea is:
Check if any of the elements is out of the range [1, n] => O(n).
Go over the numbers in order (all of them are now assured to be in the range [1, n]), and for each number x (e.g. 3):
go to the x'th cell (e.g. a[3]), if it's negative, then someone already visited it before you => Not permutation. Otherwise (a[3] is positive), multiply it by -1.
=> O(n).
Go over the array and negate all negative numbers.
This way, we know for sure that all elements are in the range [1, n], and that there are no duplicates => The array is a permutation.
int is_permutation_linear(int a[], int n) {
int i, is_permutation = 1;
// Step 1.
for (i = 0; i < n; ++i) {
if (a[i] < 1 || a[i] > n) {
return 0;
}
}
// Step 2.
for (i = 0; i < n; ++i) {
if (a[abs(a[i]) - 1] < 0) {
is_permutation = 0;
break;
}
a[i] *= -1;
}
// Step 3.
for (i = 0; i < n; ++i) {
if (a[i] < 0) {
a[i] *= -1;
}
}
return is_permutation;
}
Here is the complete program that tests it:
/*
* is_permutation_linear.c
*
* Created on: Dec 27, 2011
* Author: Anis
*/
#include <stdio.h>
int abs(int x) {
return x >= 0 ? x : -x;
}
int is_permutation_linear(int a[], int n) {
int i, is_permutation = 1;
for (i = 0; i < n; ++i) {
if (a[i] < 1 || a[i] > n) {
return 0;
}
}
for (i = 0; i < n; ++i) {
if (a[abs(a[i]) - 1] < 0) {
is_permutation = 0;
break;
}
a[abs(a[i]) - 1] *= -1;
}
for (i = 0; i < n; ++i) {
if (a[i] < 0) {
a[i] *= -1;
}
}
return is_permutation;
}
void print_array(int a[], int n) {
int i;
for (i = 0; i < n; i++) {
printf("%2d ", a[i]);
}
}
int main() {
int arrays[9][8] = { { 1, 2, 3, 4, 5, 6, 7, 8 },
{ 8, 6, 7, 2, 5, 4, 1, 3 },
{ 0, 1, 2, 3, 4, 5, 6, 7 },
{ 1, 1, 2, 3, 4, 5, 6, 7 },
{ 8, 7, 6, 5, 4, 3, 2, 1 },
{ 3, 5, 1, 6, 8, 4, 7, 2 },
{ 8, 3, 2, 1, 4, 5, 6, 7 },
{ 1, 1, 1, 1, 1, 1, 1, 1 },
{ 1, 8, 4, 2, 1, 3, 5, 6 } };
int i;
for (i = 0; i < 9; i++) {
printf("array: ");
print_array(arrays[i], 8);
printf("is %spermutation.\n",
is_permutation_linear(arrays[i], 8) ? "" : "not ");
printf("after: ");
print_array(arrays[i], 8);
printf("\n\n");
}
return 0;
}
And its output:
array: 1 2 3 4 5 6 7 8 is permutation.
after: 1 2 3 4 5 6 7 8
array: 8 6 7 2 5 4 1 3 is permutation.
after: 8 6 7 2 5 4 1 3
array: 0 1 2 3 4 5 6 7 is not permutation.
after: 0 1 2 3 4 5 6 7
array: 1 1 2 3 4 5 6 7 is not permutation.
after: 1 1 2 3 4 5 6 7
array: 8 7 6 5 4 3 2 1 is permutation.
after: 8 7 6 5 4 3 2 1
array: 3 5 1 6 8 4 7 2 is permutation.
after: 3 5 1 6 8 4 7 2
array: 8 3 2 1 4 5 6 7 is permutation.
after: 8 3 2 1 4 5 6 7
array: 1 1 1 1 1 1 1 1 is not permutation.
after: 1 1 1 1 1 1 1 1
array: 1 8 4 2 1 3 5 6 is not permutation.
after: 1 8 4 2 1 3 5 6

Answer 15

Java solution below answers question partly. Time complexity I believe is O(n). (This belief based on the fact that solution doesn't contains nested loops.) About memory -- not sure. Question appears first on relevant requests in google, so it probably can be useful for somebody.
public static boolean isPermutation(int[] array) {
boolean result = true;
array = removeDuplicates(array);
int startValue = 1;
for (int i = 0; i < array.length; i++) {
if (startValue + i != array[i]){
return false;
}
}
return result;
}
public static int[] removeDuplicates(int[] input){
Arrays.sort(input);
List<Integer> result = new ArrayList<Integer>();
int current = input[0];
boolean found = false;
for (int i = 0; i < input.length; i++) {
if (current == input[i] && !found) {
found = true;
} else if (current != input[i]) {
result.add(current);
current = input[i];
found = false;
}
}
result.add(current);
int[] array = new int[result.size()];
for (int i = 0; i < array.length ; i ++){
array[i] = result.get(i);
}
return array;
}
public static void main (String ... args){
int[] input = new int[] { 4,2,3,4,1};
System.out.println(isPermutation(input));
//output true
input = new int[] { 4,2,4,1};
System.out.println(isPermutation(input));
//output false
}

Answer 16

int solution(int A[], int N) {
int i,j,count=0, d=0, temp=0,max;
for(i=0;i<N-1;i++) {
for(j=0;j<N-i-1;j++) {
if(A[j]>A[j+1]) {
temp = A[j+1];
A[j+1] = A[j];
A[j] = temp;
}
}
}
max = A[N-1];
for(i=N-1;i>=0;i--) {
if(A[i]==max) {
count++;
}
else {
d++;
}
max = max-1;
}
if(d!=0) {
return 0;
}
else {
return 1;
}
}