Maximum Possible Number by Binary Concatenation

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: 2576

💡 Note: Binary: 2→"10", 8→"1000", 16→"10000". Optimal order [2,8,16] gives "10"+"1000"+"10000"="101000010000" = 2576.

Example 3 — Single Digit Focus

$ Input: nums = [5,6,7]

› Output: 501

💡 Note: Binary: 5→"101", 6→"110", 7→"111". Best arrangement [7,5,6] produces "111"+"101"+"110"="111101110" = 494.

Constraints

nums.length == 3
1 ≤ nums[i] ≤ 10⁷

Visualization

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Asked in

G Google 25 f Meta 18 a Amazon 15

The key insight is to arrange numbers so their binary concatenation produces the maximum possible value. Best approach uses custom sorting by comparing binary concatenations. Time: O(1), Space: O(1).

Common Approaches

✓ Brute Force - Try All Permutations

⏱️ Time: O(1) Space: O(1)

Since we have exactly 3 numbers, we can try all 6 possible arrangements (3! = 6). For each permutation, convert numbers to binary, concatenate them, then convert back to decimal to find the maximum.

Optimized Sorting Approach

⏱️ Time: O(1) Space: O(1)

Instead of trying all permutations, we can sort the numbers using a custom comparator. For any two numbers a and b, we compare the binary concatenations of 'a+b' vs 'b+a' and arrange them to get the maximum result.

Brute Force - Try All Permutations — Algorithm Steps

Step 1: Generate all 6 permutations of the 3 numbers
Step 2: For each permutation, convert numbers to binary and concatenate
Step 3: Convert the concatenated binary string back to decimal
Step 4: Return the maximum value found

Visualization

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Step-by-Step Walkthrough

Generate Permutations

Create all 6 possible orders of the 3 numbers

Convert & Concatenate

For each order, convert to binary and join

Find Maximum

Convert back to decimal and track the largest value

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int toBinary(int num, char* binary) {
    if (num == 0) {
        strcpy(binary, "0");
        return 1;
    }
    
    int len = 0;
    int temp = num;
    while (temp > 0) {
        temp /= 2;
        len++;
    }
    
    binary[len] = '\0';
    for (int i = len - 1; i >= 0; i--) {
        binary[i] = (num % 2) + '0';
        num /= 2;
    }
    return len;
}

long long binaryToDecimal(char* binary) {
    long long result = 0;
    for (int i = 0; binary[i]; i++) {
        result = result * 2 + (binary[i] - '0');
    }
    return result;
}

void swap(int* a, int* b) {
    int temp = *a;
    *a = *b;
    *b = temp;
}

long long solution(int* nums) {
    long long maxVal = 0;
    int perms[6][3] = {{0,1,2}, {0,2,1}, {1,0,2}, {1,2,0}, {2,0,1}, {2,1,0}};
    
    for (int i = 0; i < 6; i++) {
        char binaryConcat[100] = "";
        for (int j = 0; j < 3; j++) {
            char binary[32];
            toBinary(nums[perms[i][j]], binary);
            strcat(binaryConcat, binary);
        }
        
        long long decimalVal = binaryToDecimal(binaryConcat);
        if (decimalVal > maxVal) {
            maxVal = decimalVal;
        }
    }
    
    return maxVal;
}

int main() {
    char line[100];
    fgets(line, sizeof(line), stdin);
    
    // Parse [a,b,c] format
    int nums[3];
    sscanf(line, "[%d,%d,%d]", &nums[0], &nums[1], &nums[2]);
    
    printf("%lld\n", solution(nums));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(1)

Fixed 6 permutations with constant work per permutation

✓ Linear Growth

Space Complexity

O(1)

Only using variables for current permutation and max result

✓ Linear Space

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Input & Output

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Common Approaches

Brute Force - Try All Permutations — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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