Java:

Using bitwise operator will be time complexity of O(n) without requiring addition memory space.

when we use xor between same number,

7^7 = (0111)^(0111) -> 0^0 = 0, 1^1 = 0

therefore,

7^7 = 0000 -> its decimal value becomes zero

conclusion: xor with same number provide zero.

when we use xor between any value and zero,

7^0 = (0111)^(0000) -> 0^0 = 0, 1^0 = 1

therefore,

7^0 = 0111 -> its decimal value becomes 7 (same number)

conclusion: xor with a number provide the same number. No effect when we xor a value with zero

Also, 7^4^7 is same as 7^7^4

7^7^4 -> (7^7)^4 -> 0^4 (xor between same value is zero)

0^4 -> 4 (xor a value with zero provides the same value)

Now with more values,

7^4^5^7^5 -> (7^7)^(5^5)^4 ->0^0^4 -> 4

Since hackerrank gurantees only one unique element always there, so we can just use xor across the list and return the final output.

    // Write your code here
        int value = a.get(0);
        for(int i=1; i<a.size(); i++){
            value ^= a.get(i); 
        }
         
        return value;
    }

Perl:

sub lonelyinteger {
    my $arr = shift;
    
    my %h;
    foreach (@$arr) {
        $h{$_} += 1;        
    }
    
    my ($tmp) = sort { $h{$a} <=> $h{$b} } keys %h;

    return $tmp;
}

Here is my c++ solution, you can watch the vidéo explanation here : https://youtu.be/E4PVCUOdeXM

    int result = 0;
    for(int x: a) result ^= x;
    return result;
}

hi yall