Given an array of integers, find the first missing positive integer in linear time and constant space

Assuming the array can be modified,

1. We divide the array into 2 parts such that the first part consists of only positive numbers. Say we have the starting index as 0 and the ending index as end(exclusive).

2. We traverse the array from index 0 to end. We take the absolute value of the element at that index - say the value is x.

   1. If x > end we do nothing.
   2. If not, we make the sign of the element at index x-1 negative. (Clarification: We do not toggle the sign. If the value is positive, it becomes negative. If it is negative, it remains negative. In pseudo code, this would be something like if (arr[x-1] > 0) arr[x-1] = -arr[x-1] and not arr[x-1] = -arr[x-1].)

3. Finally, we traverse the array once more from index 0 to end. In case we encounter a positive element at some index, we output index + 1. This is the answer. However, if we do not encounter any positive element, it means that integers 1 to end occur in the array. We output end + 1.

It can also be the case that all the numbers are non-positive making end = 0. The output end + 1 = 1 remains correct.

All the steps can be done in O(n) time and using O(1) space.

Example:

Initial Array:            1 -1 -5 -3 3 4 2 8
Step 1 partition:         1 8 2 4 3 | -3 -5 -1, end = 5

In step 2 we change the signs of the positive numbers to keep track of which integers have already occurred. For example, here array[2] = -2 < 0, it suggests that 2 + 1 = 3 has already occurred in the array. Basically, we change the value of the element having index i to negative if i+1 is in the array.

Step 2 Array changes to: -1 -8 -2 -4 3 | -3 -5 -1

In step 3, if some value array[index] is positive, it means that we did not find any integer of value index + 1 in step 2.

Step 3: Traversing from index 0 to end, we find array[4] = 3 > 0
        The answer is 4 + 1 = 5