Maximum Possible Number by Binary Concatenation - Problem

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.

Input & Output

Example 1 — Basic Case
$ Input: nums = [1,2,3]
Output: 30
💡 Note: Binary representations: 1→"1", 2→"10", 3→"11". Optimal arrangement [1,3,2] gives "1"+"11"+"10"="11110" which equals 30 in decimal.
Example 2 — Larger Numbers
$ Input: nums = [2,8,16]
Output: 1619
💡 Note: Binary: 2→"10", 8→"1000", 16→"10000". Optimal order [8,2,16] gives "1000"+"10"+"10000"="100010100000" = 1619.
Example 3 — Single Digit Focus
$ Input: nums = [5,6,7]
Output: 1774
💡 Note: Binary: 5→"101", 6→"110", 7→"111". Best arrangement [7,6,5] produces "111"+"110"+"101"="111110101" = 1774.

Constraints

  • nums.length == 3
  • 1 ≤ nums[i] ≤ 107

Visualization

Tap to expand
Maximum Possible Number by Binary ConcatenationTransform numbers → binary → find optimal arrangement123Input Array: [1, 2, 3]"1""10""11"Binary RepresentationsOptimal Order: [1,3,2]"11110" = 30Maximum Result: 30
Understanding the Visualization
1
Input Numbers
Given array of 3 integers: [1, 2, 3]
2
Convert to Binary
Transform: 1→"1", 2→"10", 3→"11"
3
Find Best Order
Compare concatenations to get maximum: "11110" = 30
Key Takeaway
🎯 Key Insight: Compare binary concatenations to determine optimal ordering for maximum value
Asked in
Google 25 Meta 18 Amazon 15
8.5K Views
Medium Frequency
~15 min Avg. Time
420 Likes
Ln 1, Col 1
Smart Actions
💡 Explanation
AI Ready
💡 Suggestion Tab to accept Esc to dismiss
// Output will appear here after running code
Code Editor Closed
Click the red button to reopen