What is the time complexity of Letter Case Permutation?

The optimal solution has a time complexity of O(2^n).

What is the space complexity of Letter Case Permutation?

The space complexity is O(n*2^n).

What data structures are used to solve Letter Case Permutation?

This problem uses String, Backtracking, Bit Manipulation. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Letter Case Permutation?

Insight 1: The core idea is to use backtracking to explore all possible combinations of letter cases. Each letter has two choices: lowercase or uppercase. Insight 2: When encountering a digit, there is only one choice: keep it as it is. This simplifies the recursion.

Which companies ask LeetCode 784. Letter Case Permutation in interviews?

This problem is frequently asked at Yelp.

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Letter Case Permutation - LeetCode 784 Solution

Q: How to solve Letter Case Permutation?

The provided code employs a backtracking algorithm to solve the letter case permutation problem. The algorithm recursively explores all possible combinations of upper and lower case letters in the input string. For each character in the string, if it's a letter, the algorithm branches into two recursive calls: one with the letter in lowercase and the other with the letter in uppercase. If it's a digit, the algorithm proceeds with only one recursive call, keeping the digit unchanged. The base case for the recursion is when the algorithm reaches the end of the string, at which point it adds the current permutation to the result list.

Letter Case Permutation - Complete Solution Guide

Letter Case Permutation is LeetCode problem 784, 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

Given a string s , you can transform every letter individually to be lowercase or uppercase to create another string. Return a list of all possible strings we could create . Return the output in any order . Example 1: Input: s = "a1b2" Output: ["a1b2","a1B2","A1b2","A1B2"] Example 2: Input: s = "3z4" Output: ["3z4","3Z4"] Constraints: 1 <= s.length <= 12 s consists of lowercase English letters, uppercase English letters, and digits.

Detailed Explanation

The problem 'Letter Case Permutation' asks us to generate all possible strings from a given input string `s` by independently converting each letter in `s` to either lowercase or uppercase. Digits in the input string should remain unchanged. The function should return a list containing all these possible permutations in any order. For instance, given the input 'a1b2', the output should be ['a1b2', 'a1B2', 'A1b2', 'A1B2']. The input string `s` contains only letters (both uppercase and lowercase) and digits, and its length is between 1 and 12.

Solution Approach

The provided code employs a backtracking algorithm to solve the letter case permutation problem. The algorithm recursively explores all possible combinations of upper and lower case letters in the input string. For each character in the string, if it's a letter, the algorithm branches into two recursive calls: one with the letter in lowercase and the other with the letter in uppercase. If it's a digit, the algorithm proceeds with only one recursive call, keeping the digit unchanged. The base case for the recursion is when the algorithm reaches the end of the string, at which point it adds the current permutation to the result list.

Step-by-Step Algorithm

Step 1: Initialize an empty list `result` to store the generated permutations.
Step 2: Define a recursive helper function `backtrack(index, path)` where `index` is the current position in the string and `path` is the current permutation being built.
Step 3: The base case for the recursion is when `index` equals the length of the string `s`. In this case, add the current `path` to the `result` list as a string and return.
Step 4: If `s[index]` is a letter (either uppercase or lowercase), create two branches:
Step 4a: Append the lowercase version of `s[index]` to `path` and recursively call `backtrack(index + 1, path)`. After the recursive call returns, remove the last character from `path` to backtrack.
Step 4b: Append the uppercase version of `s[index]` to `path` and recursively call `backtrack(index + 1, path)`. After the recursive call returns, remove the last character from `path` to backtrack.
Step 5: If `s[index]` is a digit, append it to `path` and recursively call `backtrack(index + 1, path)`. After the recursive call returns, remove the last character from `path` to backtrack.
Step 6: Start the backtracking process by calling `backtrack(0, [])`.
Step 7: Return the `result` list containing all generated permutations.

Key Insights

Insight 1: The core idea is to use backtracking to explore all possible combinations of letter cases. Each letter has two choices: lowercase or uppercase.
Insight 2: When encountering a digit, there is only one choice: keep it as it is. This simplifies the recursion.
Insight 3: The problem can be viewed as a tree traversal where each node represents a character in the input string, and the branches represent the different case options for each letter.

Complexity Analysis

Time Complexity: O(2^n)

Space Complexity: O(n*2^n)

Topics

This problem involves: String, Backtracking, Bit Manipulation.

Companies

Asked at: Yelp.

Letter Case Permutation | LeetCode 784

Medium

String Backtracking Bit Manipulation

Time: O(2^n)Space: O(n*2^n)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.List;

class Solution {
    public List<String> letterCasePermutation(String s) {
        List<String> result = new ArrayList<>();
        
        backtrack(0, new StringBuilder(), s, result);
        return result;
    }

    private void backtrack(int index, StringBuilder path, String s, List<String> result) {
        if (index == s.length()) {
            result.add(path.toString());
            return;
        }

        char charAtCurrentIndex = s.charAt(index);

        if (Character.isLetter(charAtCurrentIndex)) {
            path.append(Character.toLowerCase(charAtCurrentIndex));
            backtrack(index + 1, path, s, result);
            path.deleteCharAt(path.length() - 1);

            path.append(Character.toUpperCase(charAtCurrentIndex));
            backtrack(index + 1, path, s, result);
            path.deleteCharAt(path.length() - 1);
        } else {
            path.append(charAtCurrentIndex);
            backtrack(index + 1, path, s, result);
            path.deleteCharAt(path.length() - 1);
        }
    }
}

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Problem Info

Description

Problem 784: Letter Case Permutation - Detailed explanation and solution approaches.

Topics

String Backtracking Bit Manipulation

Companies

Yelp

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:784

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

9 steps

1
Step 1: Initialize an empty list `result` to store the generated permutations.
2
Step 2: Define a recursive helper function `backtrack(index, path)` where `index` is the current position in the string and `path` is the current permutation being built.
3
Step 3: The base case for the recursion is when `index` equals the length of the string `s`. In this case, add the current `path` to the `result` list as a string and return.
4
Step 4: If `s[index]` is a letter (either uppercase or lowercase), create two branches:
5
Step 4a: Append the lowercase version of `s[index]` to `path` and recursively call `backtrack(index + 1, path)`. After the recursive call returns, remove the last character from `path` to backtrack.
6
Step 4b: Append the uppercase version of `s[index]` to `path` and recursively call `backtrack(index + 1, path)`. After the recursive call returns, remove the last character from `path` to backtrack.
7
Step 5: If `s[index]` is a digit, append it to `path` and recursively call `backtrack(index + 1, path)`. After the recursive call returns, remove the last character from `path` to backtrack.
8
Step 6: Start the backtracking process by calling `backtrack(0, [])`.
9
Step 7: Return the `result` list containing all generated permutations.

💡

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(2^n)

🐌 Exponential - Very slow! Often needs optimization.

💾

Space Complexity

O(n*2^n)

📝

Why?

The time complexity stems from the exponential branching of the backtracking algorithm. Each letter doubles the number of paths to explore. The space complexity considers the storage of all generated strings in the result list and the maximum depth of the recursion stack, which corresponds to the length of the input string.

📖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

Iterative Bit Manipulation

An iterative approach can be implemented using bit manipulation. We can represent each permutation as a bitmask where each bit corresponds to a letter in the input string. If the bit is set, we convert the letter to uppercase; otherwise, we convert it to lowercase.

Time: O(2^n)Space: O(n*2^n)

Trade-offs: While the time complexity is the same, the iterative approach might be slightly more efficient due to the avoidance of recursion overhead. However, it can be more difficult to understand and implement compared to the recursive backtracking approach.

✨

Key Insights

1Insight 1: The core idea is to use backtracking to explore all possible combinations of letter cases. Each letter has two choices: lowercase or uppercase.
2Insight 2: When encountering a digit, there is only one choice: keep it as it is. This simplifies the recursion.
3Insight 3: The problem can be viewed as a tree traversal where each node represents a character in the input string, and the branches represent the different case options for each letter.

📚

Key Terms Explained

For Students

String

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

Edge Cases

Edge case 1: Empty input string: The code will return a list containing only an empty string.
Edge case 2: Input string with only digits: The code will return a list containing only the original input string.
Edge case 3: Input string with a mix of letters and digits: The code correctly generates all possible permutations of letter cases while preserving the digits.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to backtrack by removing the last character from the `path` after each recursive call. This leads to incorrect permutations being generated.

✗Common mistake 2: Not handling digits correctly. Digits should be included in the `path` but should not trigger branching.

✗Common mistake 3: Incorrectly converting characters to lowercase or uppercase. Ensure you use the correct methods (e.g., `tolower()` and `toupper()` in C++, `Character.toLowerCase()` and `Character.toUpperCase()` in Java, and `.lower()` and `.upper()` in Python).

✅

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: Generating password variations. This algorithm can be used to create different password combinations by varying the case of letters, potentially improving security.

🏦Real-world scenario 2: Data fuzzing. When testing software, it is often necessary to generate many different input variations to find bugs. This algorithm can be applied to text-based inputs to generate a wide range of test cases.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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 4, 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 13

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

@DevExpert92Jul 6

❓

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.

What is the difference between backtracking and brute force?

Backtracking prunes invalid paths early using constraints, while brute force explores all possibilities. In backtracking, you "choose" an option, "explore" it recursively, then "unchoose" (backtrack) if it doesn't lead to a solution. This eliminates entire branches of the search tree.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

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 "Letter Case Permutation"?

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

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