CareerCup Find if an array is a sequence

Given an int array which might contain duplicates, find if it is a sequence.

Eg. {45,50,47,46,49,48}

is a sequence 45, 46,47,48,49,50

Sorting is an obvious solution. Can this be done in O(n) time and O(1) space

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Here's a in-place algorithm with O(n) time and O(1) extra-space

Given array 'A' of size 'n', the goal is to reorder elements in the given array so that they are in their correct positions i.e. A[i]-min(A) is at A[0] when A[i]==min(A)

and A[j]-min(A) is at A[1] if A[j] = min(A)+1

and A[k]-min(A) is at A[2] if A[k] = min(A)+2 ..... so on....

[code]1. Pass1: Find max(A) and min(A)
   if ( max(A)-min(A) > n ) then return false //i.e. you cannot have a sequence greater than n
2. Pass2: For every element 'i', 
              a. if A[i] == A[A[i]-min(A)] //already at the right position
                    if i != arr[i]-minArr then set A[i]=-Inf //this is a duplicate
                    else continue next iteration
              b. else //swap to move it to the right position
                     swap A[i] with A[A[i]-min(A)]
                     after swapping if A[i] != min(A)+i repeat from step 'a'
3. Pass3: For every element 'i', check if next element == A[i]+1,
              if not then return false.

An example with duplicates: {45,50,47,45,50,46,49,48,49}

Pass1: max(A) = 50, min(A) = 45

Pass2: modified Array:

[45,50,47,45,50,46,49,48,49] //45 already at A[A[0]-min(A)]

[45,46,47,45,50,50,49,48,49] //swap 50 & 46

[45,46,47,45,50,50,49,48,49] //47 already at A[47-45]

[45,46,47,-Inf,50,50,49,48,49] //A[3] = -Inf since it is a duplicate

[45,46,47,-Inf,-Inf,50,49,48,49]

[45,46,47,-Inf,-Inf,50,49,48,49]

[45,46,47,-Inf,49,50,-Inf,48,49]

[45,46,47,48,49,50,-Inf,-Inf,49]

[45,46,47,48,49,50,-Inf,-Inf,-Inf]

Pass3: return true

//Note : instead of -Inf you can use some other marker such as min(A)-2

ok, since many of you are disagreeing, I've posted the code below