Maximum Odd Binary Number

Maximum Odd Binary Number - Problem

You are given a binary string s that contains at least one '1'.

You have to rearrange the bits in such a way that the resulting binary number is the maximum odd binary number that can be created from this combination.

Return a string representing the maximum odd binary number that can be created from the given combination.

Note that the resulting string can have leading zeros.

Input & Output

Example 1 — Basic Case

$ Input: s = "010"

› Output: "001"

💡 Note: We have one '1' and two '0s. To maximize and keep odd: place 0 '1's at front (since 1-1=0), two '0's in middle, and one '1' at end → "001"

Example 2 — Multiple 1s

$ Input: s = "101"

› Output: "101"

💡 Note: We have two '1's and one '0'. Place 1 '1' at front, one '0' in middle, one '1' at end → "101". This is already optimal.

Example 3 — Many 1s

$ Input: s = "1111"

› Output: "1111"

💡 Note: All bits are '1'. Place 3 '1's at front, no '0's in middle, one '1' at end → "1111". Already maximum odd number.

Constraints

1 ≤ s.length ≤ 100
s consists only of '0' and '1'
s contains at least one '1'

Visualization

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Asked in

G Google 25 a Amazon 20 M Microsoft 15

The key insight is that for maximum odd number, we need all 1s at the front (for maximum value) except one 1 at the end (to ensure odd). Best approach is greedy construction: count bits and arrange optimally. Time: O(n), Space: O(1)

Common Approaches

✓ Generate All Permutations

⏱️ Time: O(n!) Space: O(n!)

Generate all permutations of the binary string, check which ones are odd (end with '1'), and find the maximum among them. This works but is extremely inefficient.

Greedy Construction

⏱️ Time: O(n) Space: O(1)

Count the number of 1s and 0s. To maximize the number, place all 1s at the beginning except one 1 at the end to ensure the number is odd. Fill the middle with all 0s.

Generate All Permutations — Algorithm Steps

Generate all permutations of the binary string
Filter permutations that end with '1' (odd numbers)
Convert each to integer and find maximum
Return the maximum as string

Visualization

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Step-by-Step Walkthrough

Generate All

Create all possible permutations

Filter Odd

Keep only those ending with '1'

Find Max

Select lexicographically largest

Code -

solution.c — C

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

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

void permute(char* str, int left, int right, char* maxOdd) {
    if (left == right) {
        if (str[right] == '1') {
            if (strlen(maxOdd) == 0 || strcmp(str, maxOdd) > 0) {
                strcpy(maxOdd, str);
            }
        }
    } else {
        for (int i = left; i <= right; i++) {
            swap(&str[left], &str[i]);
            permute(str, left + 1, right, maxOdd);
            swap(&str[left], &str[i]);
        }
    }
}

char* solution(char* s) {
    int n = strlen(s);
    char* result = (char*)malloc((n + 1) * sizeof(char));
    result[0] = '\0';
    
    char* temp = (char*)malloc((n + 1) * sizeof(char));
    strcpy(temp, s);
    
    permute(temp, 0, n - 1, result);
    
    free(temp);
    return result;
}

int main() {
    char s[1001];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    
    char* result = solution(s);
    printf("%s\n", result);
    
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n!)

Generating all permutations takes factorial time

✓ Linear Growth

Space Complexity

O(n!)

Storing all permutations requires factorial space

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Generate All Permutations — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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