给定一个包含 N 个整数的数组 A，返回在 O(n) 时间复杂度内不会出现在 A 中的最小正 integer（大于 0）（Python Sol）

Question

For anyone confused on the Codility Sample Test

Write a function:

def solution(A)

that, given an array A of N integers, returns the smallest positive integer (greater than 0) that does not occur in A.

For example, given A = [1, 3, 6, 4, 1, 2], the function should return 5.

Given A = [1, 2, 3], the function should return 4.

Given A = [−1, −3], the function should return 1.

Write an efficient algorithm for the following assumptions:

N is an integer within the range [1..100,000]; each element of array A is an integer within the range [−1,000,000..1,000,000].

# you can write to stdout for debugging purposes, e.g.
# print("this is a debug message")

def solution(A):
    # write your code in Python 3.6
    pass

    #check that if maximum number in array is less than 1 then the smallest positive integer that doesn't occur in A will be 1
    if max(A) < 1:
        return 1
    
    #sorted set of A
    hashset = set(A)

    #smallest possible positive integer
    result = 1

    #iterate through array seeing if the integer is in the hashset
    while result in hashset:
        
        #increment result
        result += 1
    
    #return smallest positive integer (greater than 0) that does not occur in A
    return result

Answer 1

I think the fastest solution is define an external vector "n" with all possible int from 1 to 100'000

N = np.arange(100_000)
def solution(A):
    return n[~np.isin(N,A)][0]

I hope this is useful.