What is the time complexity of Reformat Phone Number?

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

What data structures are used to solve Reformat Phone Number?

This problem uses String. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Reformat Phone Number?

Insight 1: The core logic involves iterating through the digits and grouping them according to the specified rules. Handling the different cases for the remaining digits (2, 3, or 4) is crucial. Insight 2: String manipulation is key. Efficiently extracting digits, creating blocks, and joining them with hyphens are essential operations.

Which companies ask LeetCode 1694. Reformat Phone Number in interviews?

This problem is frequently asked at Activision.

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Reformat Phone Number - LeetCode 1694 Solution

Reformat Phone Number - Complete Solution Guide

Reformat Phone Number is LeetCode problem 1694, a Easy 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 a phone number as a string number . number consists of digits, spaces ' ' , and/or dashes '-' . You would like to reformat the phone number in a certain manner. Firstly, remove all spaces and dashes. Then, group the digits from left to right into blocks of length 3 until there are 4 or fewer digits. The final digits are then grouped as follows: 2 digits: A single block of length 2. 3 digits: A single block of length 3. 4 digits: Two blocks of length 2 each. The blocks are then join

Detailed Explanation

The problem asks you to reformat a phone number string. The input string may contain spaces and hyphens along with digits. The goal is to remove all non-digit characters, then group the digits into blocks of three, with the last group being either two, three, or two blocks of two digits depending on the remaining number of digits. These blocks are then joined with hyphens. For example, "1-23-45 6" becomes "123-456", and "123 4-567" becomes "123-45-67".

Solution Approach

The provided solutions all follow a similar approach: first, they filter out non-digit characters. Then, they iterate through the digits, creating blocks of three until fewer than four digits remain. The remaining digits are handled based on the number of digits (2, 3, or 4). Finally, the blocks are joined using hyphens to form the reformatted number. This is done using iterative string manipulation.

Step-by-Step Algorithm

Step 1: Extract only the digits from the input string. This involves iterating through the string and appending only digits to a new string or StringBuilder.
Step 2: Initialize an empty result string or StringBuilder.
Step 3: Iterate through the digits, creating blocks of 3 digits. Append each block to the result string, followed by a hyphen.
Step 4: Handle the remaining digits (less than 4): If 4 digits remain, create two blocks of 2; if 3, one block of 3; if 2, one block of 2.
Step 5: Append the last block(s) to the result string. Remove the trailing hyphen if it exists.
Step 6: Return the reformatted number string.

Key Insights

Insight 1: The core logic involves iterating through the digits and grouping them according to the specified rules. Handling the different cases for the remaining digits (2, 3, or 4) is crucial.
Insight 2: String manipulation is key. Efficiently extracting digits, creating blocks, and joining them with hyphens are essential operations.
Insight 3: Edge cases involve strings with fewer than 2 digits (which are invalid according to the constraints) or strings where the number of digits is a multiple of 3.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: String.

Companies

Asked at: Activision.

Reformat Phone Number | LeetCode 1694

Easy

String

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

Solution Code

Available in:

class Solution {
    public String reformatNumber(String number) {
        StringBuilder digits = new StringBuilder();
        for (char c : number.toCharArray()) {
            if (Character.isDigit(c)) {
                digits.append(c);
            }
        }

        int n = digits.length();
        StringBuilder result = new StringBuilder();
        int i = 0;

        while (n - i > 4) {
            result.append(digits.substring(i, i + 3)).append("-");
            i += 3;
        }

        if (n - i == 4) {
            result.append(digits.substring(i, i + 2)).append("-").append(digits.substring(i + 2, i + 4));
        } else if (n - i == 3) {
            result.append(digits.substring(i, i + 3));
        } else if (n - i == 2) {
            result.append(digits.substring(i, i + 2));
        }

        return result.toString();
    }
}

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Problem Info

Description

Problem 1694: Reformat Phone Number - Detailed explanation and solution approaches.

Topics

String

Companies

Activision

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:1694

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: Extract only the digits from the input string. This involves iterating through the string and appending only digits to a new string or StringBuilder.
2
Step 2: Initialize an empty result string or StringBuilder.
3
Step 3: Iterate through the digits, creating blocks of 3 digits. Append each block to the result string, followed by a hyphen.
4
Step 4: Handle the remaining digits (less than 4): If 4 digits remain, create two blocks of 2; if 3, one block of 3; if 2, one block of 2.
5
Step 5: Append the last block(s) to the result string. Remove the trailing hyphen if it exists.
6
Step 6: Return the reformatted number string.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The string extraction and reformatting operations have linear time complexity. The space required for intermediate strings is also proportional to the input length.

📖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

Recursive Approach

A recursive solution could also work. The base cases would be 2, 3, and 4 digits remaining. The recursive step would handle grouping blocks of 3.

Time: O(n)Space: O(n) due to recursive call stack

Trade-offs: While conceptually elegant, a recursive solution might be less efficient due to function call overhead compared to an iterative approach. For this problem, the iterative approach is generally preferred.

✨

Key Insights

1Insight 1: The core logic involves iterating through the digits and grouping them according to the specified rules. Handling the different cases for the remaining digits (2, 3, or 4) is crucial.
2Insight 2: String manipulation is key. Efficiently extracting digits, creating blocks, and joining them with hyphens are essential operations.
3Insight 3: Edge cases involve strings with fewer than 2 digits (which are invalid according to the constraints) or strings where the number of digits is a multiple of 3.

📚

Key Terms Explained

For Students

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: Input strings with less than two digits. The problem statement specifies this is not allowed, so the solution should ideally handle this (though the provided solutions implicitly rely on the constraint).
Edge case 2: Input strings where the number of digits is a multiple of 3. This is straightforward to handle within the main loop.
Edge case 3: An empty string as input. This is not explicitly covered in the constraints and would require an additional check in a robust solution.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the edge cases of 2, 3, and 4 remaining digits. Proper conditional logic is needed to correctly group these final digits.

✗Common mistake 2: Incorrectly handling the trailing hyphen. It needs to be removed if present after the last block is appended.

✗Common mistake 3: Inefficiently processing the input string. Using `StringBuilder` (Java) or similar mechanisms is recommended for efficient string manipulation rather than repeated string concatenation.

✅

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: Phone number formatting in user interfaces or databases. This algorithm can ensure consistent formatting regardless of how the user inputs the number.

🏦Real-world scenario 2: Data cleaning and preprocessing in data analysis. Removing non-numeric characters and reformatting data into a standardized format is a common task.

💡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

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

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: August 14, 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

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

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

@DevExpert92Jul 7

❓

Frequently Asked Questions

How do I handle string manipulation efficiently?

In languages with immutable strings (Java, Python, C#), use StringBuilder/list to avoid O(n) concatenation cost. Consider using character arrays for in-place modifications. For pattern matching, consider KMP or Rabin-Karp for O(n+m) complexity.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Reformat Phone Number"?

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

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