How do you solve Smallest Value of the Rearranged Number on LeetCode?

Smallest Value of the Rearranged Number (LeetCode #2165) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Sorting digits is a powerful technique for minimizing or maximizing numerical values.

What is the brute force solution for Smallest Value of the Rearranged Number?

The brute force approach for Smallest Value of the Rearranged Number is Brute Force, with O(n!) time complexity and O(n!) space complexity. The brute-force approach involves generating all possible permutations of the digits of the number and then filtering out those that have leading zeros. Finally, we select the smallest valid arrangement. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Smallest Value of the Rearranged Number?

Smallest Value of the Rearranged Number uses the following concepts: Math, Sorting. The recommended patterns to study are: Sorting, Greedy Algorithms.

What are the interview tips for Smallest Value of the Rearranged Number?

Clearly explain your thought process and why you choose a specific approach. Consider edge cases, such as single-digit numbers or numbers with all identical digits. Practice coding the solution without looking at references to build confidence.

What are common mistakes when solving Smallest Value of the Rearranged Number?

Not considering leading zeros when rearranging digits. Overlooking the sign of the number when processing negative values.

#2165

Smallest Value of the Rearranged Number

Medium

MathSortingSortingGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n!)	O(n log n)
Space	O(n!)	O(n)

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Intuition

Time O(n log n)Space O(n)

The optimal solution focuses on sorting the digits of the number. For positive numbers, we ensure the smallest non-zero digit is at the front, while for negative numbers, we sort the digits in descending order to maximize the value. This is efficient and avoids unnecessary permutations.

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Algorithm

5 steps

1Step 1: Convert the number to a string to access its digits.
2Step 2: Separate the digits and sort them.
3Step 3: For positive numbers, find the smallest non-zero digit and place it at the front.
4Step 4: For negative numbers, simply reverse the sorted digits.
5Step 5: Convert the sorted array back to an integer.

solution.py19 lines

1# Full working Python code
2
3def smallest_rearranged_number(num):
4    str_num = str(num)
5    is_negative = str_num[0] == '-'
6    digits = sorted(str_num[1:] if is_negative else str_num)
7    if not is_negative:
8        # Move the smallest non-zero digit to the front
9        for i in range(len(digits)):
10            if digits[i] != '0':
11                digits[0], digits[i] = digits[i], digits[0]
12                break
13        return int(''.join(digits))
14    else:
15        return -int(''.join(digits[::-1]))
16
17# Example usage
18print(smallest_rearranged_number(310))  # Output: 103
19print(smallest_rearranged_number(-7605))  # Output: -7650

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step, which is efficient compared to generating permutations. The space complexity is O(n) for storing the digits.

1Sorting digits is a powerful technique for minimizing or maximizing numerical values.
2Handling leading zeros is crucial for valid number formation.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.