Daily Coding Problem Solution 4 This problem was asked by Stripe.

Given an array of integers, find the first missing positive integer in linear time and constant space. In other words, find the lowest positive integer that does not exist in the array. The array can contain duplicates and negative numbers as well.

For example, the input [3, 4, -1, 1] should give 2. The input [1, 2, 0] should give 3.

You can modify the input array in-place.

### Attempt 1

``````def lowest_positive(arr):
num = None
arr = sorted(list(set(arr)))
# arr = [i for i in arr if i > 0]
for i, n in enumerate(arr):
if n > 0: # if uncomment above, comment here
if i+1 < len(arr):
if n+1 == arr[i+1]:
continue
else:
num = n+1
break
if num is None:
num = arr[-1] + 1
return num

assert lowest_positive([3, 4, -1, 1]) == 2, 'fail'
assert lowest_positive([1, 2, 0]) == 3, 'fail'

``````

nLog(n)

### Attempt 2

``````def lowest_positive2(arr):
# 2nd solution
minimum = None
nums = []

for i, n in enumerate(arr):
if n > 0:
nums.append(n)
if minimum is None:
minimum = n
if n < minimum:
minimum = n

i = 1
while 1:
if minimum+i in nums:
i+=1
continue
else:
return minimum+i

``````

O(n^2) worst case

### Attempt 3

``````def lowest_positive3(arr):
# 3rd solution
minimum = None
maximum = None
nums = []

for i, n in enumerate(arr):
if n > -1:
nums.append(n)
if minimum is None:
minimum = n
if n < minimum:
minimum = n

if maximum is None:
maximum = n
if n > maximum:
maximum = n
i = 1
while 1:
if minimum+i == nums[i-1]:

i+=1
continue
else:
return (minimum+i)

return maximum+1

``````

O(n)