LeetCode #3309 — MEDIUM

Maximum Possible Number by Binary Concatenation

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given an array of integers nums of size 3.

Return the maximum possible number whose binary representation can be formed by concatenating the binary representation of all elements in nums in some order.

Note that the binary representation of any number does not contain leading zeros.

Example 1:

Input: nums = [1,2,3]

Output: 30

Explanation:

Concatenate the numbers in the order [3, 1, 2] to get the result "11110", which is the binary representation of 30.

Example 2:

Input: nums = [2,8,16]

Output: 1296

Explanation:

Concatenate the numbers in the order [2, 8, 16] to get the result "10100010000", which is the binary representation of 1296.

Constraints:

nums.length == 3
1 <= nums[i] <= 127

Step 01

Brute Force Baseline

Problem summary: You are given an array of integers nums of size 3. Return the maximum possible number whose binary representation can be formed by concatenating the binary representation of all elements in nums in some order. Note that the binary representation of any number does not contain leading zeros.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Bit Manipulation

Example 1

[1,2,3]

Example 2

[2,8,16]

Core Insight

What unlocks the optimal approach

How many possible concatenation orders are there?

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3309: Maximum Possible Number by Binary Concatenation
class Solution {
    private int[] nums;

    public int maxGoodNumber(int[] nums) {
        this.nums = nums;
        int ans = f(0, 1, 2);
        ans = Math.max(ans, f(0, 2, 1));
        ans = Math.max(ans, f(1, 0, 2));
        ans = Math.max(ans, f(1, 2, 0));
        ans = Math.max(ans, f(2, 0, 1));
        ans = Math.max(ans, f(2, 1, 0));
        return ans;
    }

    private int f(int i, int j, int k) {
        String a = Integer.toBinaryString(nums[i]);
        String b = Integer.toBinaryString(nums[j]);
        String c = Integer.toBinaryString(nums[k]);
        return Integer.parseInt(a + b + c, 2);
    }
}

// Accepted solution for LeetCode #3309: Maximum Possible Number by Binary Concatenation
func maxGoodNumber(nums []int) int {
	f := func(i, j, k int) int {
		a := strconv.FormatInt(int64(nums[i]), 2)
		b := strconv.FormatInt(int64(nums[j]), 2)
		c := strconv.FormatInt(int64(nums[k]), 2)
		res, _ := strconv.ParseInt(a+b+c, 2, 64)
		return int(res)
	}
	ans := f(0, 1, 2)
	ans = max(ans, f(0, 2, 1))
	ans = max(ans, f(1, 0, 2))
	ans = max(ans, f(1, 2, 0))
	ans = max(ans, f(2, 0, 1))
	ans = max(ans, f(2, 1, 0))
	return ans
}

# Accepted solution for LeetCode #3309: Maximum Possible Number by Binary Concatenation
class Solution:
    def maxGoodNumber(self, nums: List[int]) -> int:
        ans = 0
        for arr in permutations(nums):
            num = int("".join(bin(i)[2:] for i in arr), 2)
            ans = max(ans, num)
        return ans

// Accepted solution for LeetCode #3309: Maximum Possible Number by Binary Concatenation
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3309: Maximum Possible Number by Binary Concatenation
// class Solution {
//     private int[] nums;
// 
//     public int maxGoodNumber(int[] nums) {
//         this.nums = nums;
//         int ans = f(0, 1, 2);
//         ans = Math.max(ans, f(0, 2, 1));
//         ans = Math.max(ans, f(1, 0, 2));
//         ans = Math.max(ans, f(1, 2, 0));
//         ans = Math.max(ans, f(2, 0, 1));
//         ans = Math.max(ans, f(2, 1, 0));
//         return ans;
//     }
// 
//     private int f(int i, int j, int k) {
//         String a = Integer.toBinaryString(nums[i]);
//         String b = Integer.toBinaryString(nums[j]);
//         String c = Integer.toBinaryString(nums[k]);
//         return Integer.parseInt(a + b + c, 2);
//     }
// }

// Accepted solution for LeetCode #3309: Maximum Possible Number by Binary Concatenation
function maxGoodNumber(nums: number[]): number {
    const f = (i: number, j: number, k: number): number => {
        const a = nums[i].toString(2);
        const b = nums[j].toString(2);
        const c = nums[k].toString(2);
        const res = parseInt(a + b + c, 2);
        return res;
    };

    let ans = f(0, 1, 2);
    ans = Math.max(ans, f(0, 2, 1));
    ans = Math.max(ans, f(1, 0, 2));
    ans = Math.max(ans, f(1, 2, 0));
    ans = Math.max(ans, f(2, 0, 1));
    ans = Math.max(ans, f(2, 1, 0));

    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

SORT + SCAN

O(n log n) time

O(n) space

Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.

BIT MANIPULATION

O(n) time

O(1) space

Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.

Shortcut: Bit operations are O(1). XOR cancels duplicates. Single pass → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.