What is the time complexity of Smallest Value of the Rearranged Number?

The optimal solution has a time complexity of O(n log n).

What data structures are used to solve Smallest Value of the Rearranged Number?

This problem uses Math, Sorting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Smallest Value of the Rearranged Number?

Insight 1: The sign of the number is crucial; positive and negative numbers are handled differently. Insight 2: Sorting the digits is the core operation. Ascending order for positive and descending for negative.

Which companies ask LeetCode 2165. Smallest Value of the Rearranged Number in interviews?

This problem is frequently asked at Cognizant.

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Smallest Value of the Rearranged Number - LeetCode 2165 Solution

Smallest Value of the Rearranged Number - Complete Solution Guide

Smallest Value of the Rearranged Number is LeetCode problem 2165, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given an integer num. Rearrange the digits of num such that its value is minimized and it does not contain any leading zeros. Return the rearranged number with minimal value . Note that the sign of the number does not change after rearranging the digits. Example 1: Input: num = 310 Output: 103 Explanation: The possible arrangements for the digits of 310 are 013, 031, 103, 130, 301, 310. The arrangement with the smallest value that does not contain any leading zeros is 103. Example 2: Inp

Detailed Explanation

The problem asks us to rearrange the digits of a given integer `num` to obtain the smallest possible value while ensuring there are no leading zeros. The sign of the number must remain the same after rearrangement. Essentially, we need to sort the digits in ascending order for positive numbers and descending order for negative numbers, handling the leading zero constraint in the positive case.

Solution Approach

The solution involves converting the integer to a string, sorting the digits based on whether the number is positive or negative, and then handling the leading zero case for positive numbers. For negative numbers, the digits are sorted in descending order to achieve the smallest magnitude. For positive numbers, they are sorted in ascending order, and a check is done to swap the first non-zero digit with the first digit (if the first digit is zero) to prevent leading zeros.

Step-by-Step Algorithm

Step 1: Determine if the input number `num` is positive, negative, or zero. If zero, return zero.
Step 2: Convert the absolute value of `num` to a string `s_num`.
Step 3: If `num` is negative, sort the digits in `s_num` in descending order.
Step 4: If `num` is positive, sort the digits in `s_num` in ascending order.
Step 5: If `num` is positive and the first digit of the sorted string is '0', find the first non-zero digit and swap it with the first digit.
Step 6: Convert the sorted string back to an integer. If `num` was negative, negate the result before returning.

Key Insights

Insight 1: The sign of the number is crucial; positive and negative numbers are handled differently.
Insight 2: Sorting the digits is the core operation. Ascending order for positive and descending for negative.
Insight 3: Handling leading zeros in positive numbers is a special case that requires swapping the first non-zero digit with the first digit.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Math, Sorting.

Companies

Asked at: Cognizant.

Smallest Value of the Rearranged Number | LeetCode 2165

Medium

Math Sorting

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

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public long smallestNumber(long num) {
        if (num == 0) {
            return 0;
        }

        boolean isNegative = num < 0;
        String sNum = String.valueOf(Math.abs(num));

        if (isNegative) {
            char[] digits = sNum.toCharArray();
            Arrays.sort(digits);
            StringBuilder sb = new StringBuilder();
            for (int i = digits.length - 1; i >= 0; i--) {
                sb.append(digits[i]);
            }
            String resultStr = sb.toString();
            return -Long.parseLong(resultStr);
        } else {
            char[] digits = sNum.toCharArray();
            Arrays.sort(digits);

            if (digits[0] == '0') {
                int firstNonZeroIndex = 0;
                for (int i = 0; i < digits.length; i++) {
                    if (digits[i] != '0') {
                        firstNonZeroIndex = i;
                        break;
                    }
                }

                char temp = digits[0];
                digits[0] = digits[firstNonZeroIndex];
                digits[firstNonZeroIndex] = temp;
            }

            String resultStr = new String(digits);
            return Long.parseLong(resultStr);
        }
    }
}

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Problem Info

Description

Problem 2165: Smallest Value of the Rearranged Number - Detailed explanation and solution approaches.

Topics

Math Sorting

Companies

Cognizant

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2165

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Determine if the input number `num` is positive, negative, or zero. If zero, return zero.
2
Step 2: Convert the absolute value of `num` to a string `s_num`.
3
Step 3: If `num` is negative, sort the digits in `s_num` in descending order.
4
Step 4: If `num` is positive, sort the digits in `s_num` in ascending order.
5
Step 5: If `num` is positive and the first digit of the sorted string is '0', find the first non-zero digit and swap it with the first digit.
6
Step 6: Convert the sorted string back to an integer. If `num` was negative, negate the result before returning.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is determined by the sorting algorithm used. Most standard sorting algorithms (like `Arrays.sort` in Java, `sort` in C++, and `sorted` in Python) have a time complexity of O(n log n). The space complexity is O(n) because we store the digits of the number in a string or a character array, whose size is proportional to the number of digits.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Counting Sort

Use counting sort since the range of digits is known (0-9).

Time: O(n + k) = O(n) since k is fixed at 10Space: O(k) = O(1) as the count array size is fixed

Trade-offs: Counting sort is more efficient in terms of time complexity but might be slightly more complex to implement compared to general sorting algorithms. It's preferred when performance is critical and the range of input values is small and known in advance. Not easily extendable to other sorting problems.

✨

Key Insights

1Insight 1: The sign of the number is crucial; positive and negative numbers are handled differently.
2Insight 2: Sorting the digits is the core operation. Ascending order for positive and descending for negative.
3Insight 3: Handling leading zeros in positive numbers is a special case that requires swapping the first non-zero digit with the first digit.

📚

Key Terms Explained

For Students

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Input is 0. The solution correctly returns 0.
Edge case 2: Positive number with leading zeros after sorting (e.g., 100 becomes 001 after initial sorting). The solution correctly swaps the first non-zero digit to avoid leading zeros.
Edge case 3: Negative number with all zeros except for one digit (e.g., -10000). The solution sorts in descending order so it would become -10000, the smallest possible value.
Edge case 4: Very large numbers nearing the upper and lower limits of long data type. The solution uses `long long` in C++ and `long` in Java to handle these large numbers.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to handle leading zeros in positive numbers. Beginners often forget to swap the first non-zero digit with the first digit.

✗Common mistake 2: Incorrectly sorting negative numbers in ascending order. Negative numbers need to be sorted in descending order to minimize their value.

✗Common mistake 3: Not handling the case where the input is zero. It's a simple but necessary check.

✗Common mistake 4: Using `int` instead of `long` or `long long` for the data type, which can lead to integer overflow issues for large input values.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Optimizing numerical data storage. In databases or data analysis, rearranging digits to minimize storage space can improve efficiency. (Though highly theoretical and not a primary method)

🏦Real-world scenario 2: Generating minimal representations of numbers for specific constraints. This pattern is applicable to tasks where numerical representations need to adhere to size or order restrictions.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 6, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 7

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 30

❓

Frequently Asked Questions

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Smallest Value of the Rearranged Number"?

For this medium-level Math problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether sorting techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?