int n = 10;
int n1 = ~n;

System.out.println(Integer.toBinaryString(n)); // 1010 but, maybe 00000000 00000000 00000000 00001010
System.out.println(n); // 10
System.out.println(Integer.toBinaryString(n1)); // 11111111111111111111111111110101
System.out.println(n1); // -11

How does 11111111111111111111111111110101 output as -11?

I couldn’t find a way to convert the complement back to decimal

>Solution :

The two’s complement is used for negative numbers

-n = ~n + 1

• -0 (-0…0000) = 1…1111 + 1 = 0…0000
• -1 (-0…0001) = 1…1110 + 1 = 1…1111
• -2 (-0…0010) = 1…1101 + 1 = 1…1110
• -10 (-0…1010) = 1…0101 + 1 = 1…0110

Your miscomprehension is that the ~ operator makes a bit "negative." It just is the one’s complement, bitwise negate.

1...0110 (-10)
0...1010 (+10)
-------- +
      :0
     C0       C=carry
    C0
...C0
--------
0...0000 (0)
========

As ~n = -n -1, ~10 = -10 – 1 = -11.
Just as ~0 = -0 – 1 = -1.

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Array.binarySearch(array, searchKey) uses the one’s complement, ~. It gives a non-negative index of the key when found, and the ones complement (hence negative) for the insert position when the key was not found. A result of -1 means insert position 0, before the first element in the array.