The Maximum Possible Number by Binary Concatenation interview question is a bit manipulation and permutation problem. You are given an array of three integers. You need to convert these integers into their binary representations (without leading zeros), concatenate them in any of the 6 possible orders, and find the maximum decimal value that can be formed from these concatenations.
For example, if the numbers are 1, 2, and 3, their binary forms are "1", "10", and "11". Concatenating them as "11" + "10" + "1" gives "11101", which is 29.
This Maximum Possible Number by Binary Concatenation coding problem is used by companies like Amazon and Google to test basic bitwise operations and enumeration. It evaluates whether a candidate can correctly convert numbers to binary, handle string concatenations, and convert back to decimal. Since there are only 3 numbers, the problem also tests if you can recognize that trying all permutations (enumeration) is a perfectly valid and efficient strategy.
The Array, Enumeration, Bit Manipulation interview pattern is the go-to here.
Numbers: [2, 8, 10]
Permutation Example: "1010" + "1000" + "10" = "1010100010" Decimal value: 512 + 128 + 32 + 2 = 674. Another Permutation: "10" + "1000" + "1010" = "1010001010" Decimal value: 512 + 128 + 10 = 650.
The algorithm would check all 6 and find the maximum.
In the Maximum Possible Number by Binary Concatenation coding problem, a common mistake is including leading zeros in the binary representation, which the problem usually forbids. Another error is not trying all permutations, perhaps assuming a greedy sort would work (like in "Largest Number"), but with only 3 elements, brute force is safer and easier. Some candidates might also struggle with the base conversion logic if they don't use built-in language functions.
Get comfortable with your language's built-in functions for binary conversion (like bin() in Python or Integer.toBinaryString() in Java). Also, remember that for a small number of elements (like 3 or 4), explicitly listing permutations or using a simple recursive permutation generator is often better than trying to find a clever mathematical shortcut.