Sort the array, now let <code>x <= y</code> which means <code>|x - y| <= min(x, y)</code> can now be written as <code>y - x <= x</code> or in other words, <code>y <= 2 * x</code>.

If <code>x</code> and <code>y</code> have the same number of bits, try making<code>y</code>’s bits different from x if possible for each bit starting from the second most significant bit.

If <code>y</code> has 1 more bit than <code>x</code> and <code>y <= 2 * x</code> use the idea about Digit DP to make <code>y</code>’s prefix smaller than <code>2 * x + 1</code> as well as trying to make each bit different from <code>x</code> using a Hashmap.

Alternatively, use Trie data structure to find the pair with maximum <code>XOR</code>.