LeetCode 2165: Smallest Value of the Rearranged Number Solution

Master LeetCode problem 2165 (Smallest Value of the Rearranged Number), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2165. Smallest Value of the Rearranged Number

Medium

Math Sorting

Problem Explanation

Explanation

To solve this problem, we need to rearrange the digits of the given number to form the smallest possible number without any leading zeros. We can achieve this by sorting the digits of the number in non-decreasing order, while ensuring that the most significant digit is not zero. If the original number is negative, we need to consider the sign while rearranging the digits.

Here is the high-level algorithm:

Convert the absolute value of the input number to a list of digits.
Sort the list of digits in non-decreasing order.
If the original number was negative, swap the first digit (excluding 0 if present) with the last digit to maintain the sign.
Convert the sorted list of digits back to a number.

The time complexity of this approach is O(n log n) where n is the number of digits in the input number, due to the sorting step. The space complexity is O(n) to store the list of digits.

Solution Code

class Solution {
    public int rearrangeNumber(int num) {
        if (num == 0) {
            return 0;
        }
        
        boolean isNegative = num < 0;
        num = Math.abs(num);
        
        List<Integer> digits = new ArrayList<>();
        while (num > 0) {
            digits.add(num % 10);
            num /= 10;
        }
        
        Collections.sort(digits);
        
        if (isNegative) {
            for (int i = 0; i < digits.size(); i++) {
                if (digits.get(i) != 0) {
                    int temp = digits.get(i);
                    digits.set(i, digits.get(digits.size() - 1));
                    digits.set(digits.size() - 1, temp);
                    break;
                }
            }
        }
        
        int result = 0;
        for (int digit : digits) {
            result = result * 10 + digit;
        }
        
        return isNegative ? -result : result;
    }
}

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Frequently Asked Questions

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