Solution:

This java solution to codility problem PermMissingElem scored a 100%

Logic: Iterate over the array and add the elements to a set on which you can do a contains operation. Iterate over the hashset for the first n+1 numbers and if when a specific number is not found in the hashset then that is the missing element.

import java.util.HashSet;
import java.util.Set;
class PermMissingElement{
    public int solution(int[] A) {
        int n = A.length + 1;
        int missingNumber = 0;
        Set<Integer> values = new HashSet<>();
        for (int i=0; i<A.length; i++) {
            values.add(A[i]);
        }
        for (int i=1; i<=n; i++) {
            if (!values.contains(i)) {
                missingNumber = i;
            }
        }
        return missingNumber;
    }
}

Solution 2:

This java solution scored 80% as it failed with larger length (~100,000)

Logic: Calculate the sum of the first n+1 natural numbers. Then iterate over the array and deduct the value from the sum. Whats left of the sum is the missing number. But aha fails with bigger ranges..

    
class PermMissingElement{
   public int solution1(int[] A) {

        int n = A.length + 1;
        long sumOfFirstN = n * (n+1)/2;
        for (int i=0; i<A.length; i++) {
            sumOfFirstN = sumOfFirstN - A[i];
        }
        return (int)(sumOfFirstN);
    }
}