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

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

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:

1. Check if any of the elements is out of the range [1, n] => O(n).
2. 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).
3. 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;
}

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, ... s_k-1 where s_j = a[s_j-1] for all j between 1 and k-1, and s₀ = a[s_k-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).

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