Letter Case Permutation

🎯

Letter Case Permutation

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Imagine you are designing a username generator that can create all possible variations of a string by toggling the case of its letters, helping users find unique but related usernames.

💡 This problem is about generating all possible variations of a string by changing letter cases. Beginners often struggle because it requires exploring multiple branching choices recursively, which can be confusing without a clear include/exclude mindset.

📋

Problem Statement

Given a string s, you can transform every letter individually to be lowercase or uppercase to create new strings. Return a list of all possible strings we could create. Digits remain unchanged.

1 ≤ s.length ≤ 12s consists of only letters or digits.

💡

Example

Input"a1b2"

Outputa1b2a1B2A1b2A1B2

For each letter, we choose lowercase or uppercase, digits stay the same. So 'a' → 'a' or 'A', 'b' → 'b' or 'B'.

• Empty string → [] or [''] depending on interpretation
• String with no letters (e.g., '1234') → ['1234'] only
• String with all letters (e.g., 'abc') → all 2^3 = 8 permutations
• String with mixed letters and digits (e.g., 'a1b') → permutations only change letters

⚠️

Common Mistakes

Not handling digits correctly

Digits get toggled or skipped incorrectly, leading to wrong outputs

✅ Explicitly check if character is digit and only append without branching

Not backtracking properly (forgetting to revert changes)

Subsequent recursive calls use wrong characters, corrupting results

✅ Always revert character or path changes after recursive calls

Using string concatenation in recursion without care

Leads to high memory usage and slower performance

✅ Use character arrays or StringBuilder to optimize string building

Not considering empty string or all digits input

May return empty list or miss valid outputs

✅ Handle edge cases explicitly and test them

🧠

Brute Force (Pure Recursion / Backtracking)

💡 This approach exists to build intuition on how to explore all possible choices by branching at each letter. It’s the foundation for understanding backtracking.

Intuition

At each character, if it's a letter, we branch into two recursive calls: one with lowercase and one with uppercase. If it's a digit, we only have one choice. We collect all results at the end.

Algorithm

1. Start from index 0 and an empty path string.
2. If current character is a digit, append it and recurse to next index.
3. If current character is a letter, recurse twice: once with lowercase, once with uppercase appended.
4. When index reaches string length, add the built path to results.

💡 The challenge is to remember to branch only on letters and to accumulate the path correctly through recursion.

</>

Code

from typing import List

def letterCasePermutation(s: str) -> List[str]:
    res = []
    def backtrack(i: int, path: List[str]):
        if i == len(s):
            res.append(''.join(path))
            return
        if s[i].isdigit():
            path.append(s[i])
            backtrack(i + 1, path)
            path.pop()
        else:
            path.append(s[i].lower())
            backtrack(i + 1, path)
            path.pop()
            path.append(s[i].upper())
            backtrack(i + 1, path)
            path.pop()
    backtrack(0, [])
    return res

# Driver code
if __name__ == '__main__':
    print(letterCasePermutation("a1b2"))

Line Notes

def letterCasePermutation(s: str) -> List[str]:Defines the main function with input string and output list of strings to return all permutations.

def backtrack(i: int, path: List[str]):Helper recursive function tracking current index and the path built so far to explore all branches.

if i == len(s):Base case: when index reaches end of string, join path and add to results as one valid permutation.

if s[i].isdigit():Digits do not branch; append digit and recurse to next index to maintain correctness.

path.append(s[i].lower())Try lowercase branch for current letter to explore one variation.

path.pop()Backtrack by removing last character to restore state before exploring next branch.

path.append(s[i].upper())Try uppercase branch for current letter to explore the other variation.

import java.util.*;

public class LetterCasePermutation {
    public static List<String> letterCasePermutation(String s) {
        List<String> res = new ArrayList<>();
        backtrack(s.toCharArray(), 0, new StringBuilder(), res);
        return res;
    }

    private static void backtrack(char[] chars, int i, StringBuilder path, List<String> res) {
        if (i == chars.length) {
            res.add(path.toString());
            return;
        }
        if (Character.isDigit(chars[i])) {
            path.append(chars[i]);
            backtrack(chars, i + 1, path, res);
            path.deleteCharAt(path.length() - 1);
        } else {
            path.append(Character.toLowerCase(chars[i]));
            backtrack(chars, i + 1, path, res);
            path.deleteCharAt(path.length() - 1);
            path.append(Character.toUpperCase(chars[i]));
            backtrack(chars, i + 1, path, res);
            path.deleteCharAt(path.length() - 1);
        }
    }

    public static void main(String[] args) {
        System.out.println(letterCasePermutation("a1b2"));
    }
}

Line Notes

public static List<String> letterCasePermutation(String s)Main function signature returning list of all permutations.

backtrack(s.toCharArray(), 0, new StringBuilder(), res);Start recursion with character array, index 0, empty path builder, and result list.

if (i == chars.length)Base case: when index reaches end, add current path string to results.

if (Character.isDigit(chars[i]))Digits do not branch; append digit and recurse to next index.

path.append(Character.toLowerCase(chars[i]));Try lowercase branch for current letter to explore one variation.

path.deleteCharAt(path.length() - 1);Backtrack by removing last appended character to restore state.

path.append(Character.toUpperCase(chars[i]));Try uppercase branch for current letter to explore the other variation.

#include <iostream>
#include <vector>
#include <string>
using namespace std;

class Solution {
public:
    vector<string> letterCasePermutation(string s) {
        vector<string> res;
        string path;
        backtrack(s, 0, path, res);
        return res;
    }

private:
    void backtrack(const string& s, int i, string& path, vector<string>& res) {
        if (i == s.size()) {
            res.push_back(path);
            return;
        }
        if (isdigit(s[i])) {
            path.push_back(s[i]);
            backtrack(s, i + 1, path, res);
            path.pop_back();
        } else {
            path.push_back(tolower(s[i]));
            backtrack(s, i + 1, path, res);
            path.pop_back();
            path.push_back(toupper(s[i]));
            backtrack(s, i + 1, path, res);
            path.pop_back();
        }
    }
};

int main() {
    Solution sol;
    vector<string> res = sol.letterCasePermutation("a1b2");
    for (auto& str : res) cout << str << endl;
    return 0;
}

Line Notes

vector<string> letterCasePermutation(string s)Main function returning vector of all permutations.

void backtrack(const string& s, int i, string& path, vector<string>& res)Recursive helper with current index and path string reference to build permutations.

if (i == s.size())Base case: when index reaches end, add current path to results.

if (isdigit(s[i]))Digits do not branch; append digit and recurse to next index.

path.push_back(tolower(s[i]));Try lowercase branch for current letter to explore one variation.

path.pop_back();Backtrack by removing last character to restore state.

path.push_back(toupper(s[i]));Try uppercase branch for current letter to explore the other variation.

function letterCasePermutation(s) {
    const res = [];
    function backtrack(i, path) {
        if (i === s.length) {
            res.push(path);
            return;
        }
        const c = s[i];
        if (/[0-9]/.test(c)) {
            backtrack(i + 1, path + c);
        } else {
            backtrack(i + 1, path + c.toLowerCase());
            backtrack(i + 1, path + c.toUpperCase());
        }
    }
    backtrack(0, "");
    return res;
}

// Driver code
console.log(letterCasePermutation("a1b2"));

Line Notes

function letterCasePermutation(s)Main function to generate all permutations from input string.

function backtrack(i, path)Recursive helper tracking current index and the path string built so far.

if (i === s.length)Base case: when index reaches end, add current path to results.

if (/[0-9]/.test(c))Digits do not branch; recurse with same character appended.

backtrack(i + 1, path + c.toLowerCase())Try lowercase branch for current letter to explore one variation.

backtrack(i + 1, path + c.toUpperCase())Try uppercase branch for current letter to explore the other variation.

⏱

Complexity

TimeO(2^k * n) where k is number of letters and n is string length

SpaceO(n) recursion stack + O(2^k * n) output space

Each letter doubles the number of permutations, digits do not branch. We build strings of length n for each permutation.

💡 For example, if input has 3 letters, we get 2^3=8 permutations, each of length n, so about 8*n operations.

Interview Verdict: Accepted

This brute force approach is accepted for typical constraints and is the foundation for more optimized solutions.

🧠

Iterative BFS / Queue Approach

💡 This approach uses a queue to iteratively build permutations level by level, which can be easier to understand for those uncomfortable with recursion.

Intuition

Start with an empty string in a queue. For each character, if digit, append it to all strings in queue. If letter, for each string, append lowercase and uppercase versions, doubling queue size.

Algorithm

1. Initialize a queue with an empty string.
2. For each character in input string:
3. If digit, append it to all strings in queue.
4. If letter, for each string in queue, create two new strings with lowercase and uppercase appended.
5. After processing all characters, queue contains all permutations.

💡 The tricky part is managing the queue size doubling when letters appear and ensuring all strings are updated correctly.

</>

Code

from collections import deque
from typing import List

def letterCasePermutation(s: str) -> List[str]:
    queue = deque([''])
    for c in s:
        n = len(queue)
        if c.isdigit():
            for _ in range(n):
                queue.append(queue.popleft() + c)
        else:
            for _ in range(n):
                curr = queue.popleft()
                queue.append(curr + c.lower())
                queue.append(curr + c.upper())
    return list(queue)

# Driver code
if __name__ == '__main__':
    print(letterCasePermutation("a1b2"))

Line Notes

queue = deque([''])Initialize queue with empty string as starting point for building permutations.

for c in s:Process each character in input string sequentially to build permutations stepwise.

if c.isdigit():Digits do not branch; append digit to each string in queue.

for _ in range(n):Process current queue size to avoid infinite loop while expanding queue.

queue.append(curr + c.lower())Append lowercase version for letter to build one branch.

queue.append(curr + c.upper())Append uppercase version for letter to build the other branch.

import java.util.*;

public class LetterCasePermutation {
    public static List<String> letterCasePermutation(String s) {
        LinkedList<String> queue = new LinkedList<>();
        queue.add("");
        for (char c : s.toCharArray()) {
            int n = queue.size();
            if (Character.isDigit(c)) {
                for (int i = 0; i < n; i++) {
                    queue.add(queue.poll() + c);
                }
            } else {
                for (int i = 0; i < n; i++) {
                    String curr = queue.poll();
                    queue.add(curr + Character.toLowerCase(c));
                    queue.add(curr + Character.toUpperCase(c));
                }
            }
        }
        return queue;
    }

    public static void main(String[] args) {
        System.out.println(letterCasePermutation("a1b2"));
    }
}

Line Notes

LinkedList<String> queue = new LinkedList<>();Initialize queue with empty string to start building permutations.

for (char c : s.toCharArray())Iterate over each character in input string to process sequentially.

if (Character.isDigit(c))Digits do not branch; append digit to each string in queue.

for (int i = 0; i < n; i++)Process current queue size to avoid infinite loop while expanding queue.

queue.add(curr + Character.toLowerCase(c));Add lowercase version for letter to build one branch.

queue.add(curr + Character.toUpperCase(c));Add uppercase version for letter to build the other branch.

#include <iostream>
#include <vector>
#include <string>
#include <queue>
using namespace std;

class Solution {
public:
    vector<string> letterCasePermutation(string s) {
        queue<string> q;
        q.push("");
        for (char c : s) {
            int n = q.size();
            if (isdigit(c)) {
                for (int i = 0; i < n; i++) {
                    string curr = q.front(); q.pop();
                    q.push(curr + c);
                }
            } else {
                for (int i = 0; i < n; i++) {
                    string curr = q.front(); q.pop();
                    q.push(curr + (char)tolower(c));
                    q.push(curr + (char)toupper(c));
                }
            }
        }
        vector<string> res;
        while (!q.empty()) {
            res.push_back(q.front());
            q.pop();
        }
        return res;
    }
};

int main() {
    Solution sol;
    vector<string> res = sol.letterCasePermutation("a1b2");
    for (auto& str : res) cout << str << endl;
    return 0;
}

Line Notes

queue<string> q;Initialize queue with empty string to start building permutations.

for (char c : s)Iterate over each character in input string sequentially.

if (isdigit(c))Digits do not branch; append digit to each string in queue.

for (int i = 0; i < n; i++)Process current queue size to avoid infinite loop while expanding queue.

q.push(curr + (char)tolower(c));Add lowercase version for letter to build one branch.

q.push(curr + (char)toupper(c));Add uppercase version for letter to build the other branch.

function letterCasePermutation(s) {
    let queue = [''];
    for (const c of s) {
        const n = queue.length;
        if (/[0-9]/.test(c)) {
            for (let i = 0; i < n; i++) {
                queue[i] = queue[i] + c;
            }
        } else {
            const temp = [];
            for (let i = 0; i < n; i++) {
                temp.push(queue[i] + c.toLowerCase());
                temp.push(queue[i] + c.toUpperCase());
            }
            queue = temp;
        }
    }
    return queue;
}

// Driver code
console.log(letterCasePermutation("a1b2"));

Line Notes

let queue = [''];Initialize queue with empty string to start building permutations.

for (const c of s)Iterate over each character in input string sequentially.

if (/[0-9]/.test(c))Digits do not branch; append digit to each string in queue by replacing strings.

for (let i = 0; i < n; i++)Process all strings in queue to append current character.

temp.push(queue[i] + c.toLowerCase());Add lowercase version for letter to build one branch.

temp.push(queue[i] + c.toUpperCase());Add uppercase version for letter to build the other branch.

queue = temp;Replace queue with expanded list after processing letter to maintain correct permutations.

⏱

Complexity

TimeO(2^k * n) where k is number of letters and n is string length

SpaceO(2^k * n) for queue and output

Each letter doubles the queue size, digits do not branch. We build strings of length n for each permutation.

💡 This iterative approach uses a queue to build permutations stepwise, which can be easier to visualize than recursion.

Interview Verdict: Accepted

This approach is accepted and useful when recursion is not preferred or stack depth is a concern.

🧠

Backtracking with In-Place Character Array Modification

💡 This approach optimizes memory by modifying the input character array in place during recursion, reducing string concatenation overhead.

Intuition

Instead of building new strings at each recursion, we toggle the case of letters in the character array and recurse, then revert changes to backtrack.

Algorithm

1. Convert input string to character array.
2. At each index, if digit, recurse to next index.
3. If letter, recurse twice: once with lowercase, once with uppercase by modifying array in place.
4. When index reaches end, convert array to string and add to results.
5. Backtrack by reverting character case changes.

💡 The key is to remember to revert changes after recursive calls to maintain correct state.

</>

Code

from typing import List

def letterCasePermutation(s: str) -> List[str]:
    res = []
    arr = list(s)
    def backtrack(i: int):
        if i == len(arr):
            res.append(''.join(arr))
            return
        if arr[i].isdigit():
            backtrack(i + 1)
        else:
            arr[i] = arr[i].lower()
            backtrack(i + 1)
            arr[i] = arr[i].upper()
            backtrack(i + 1)
            arr[i] = s[i]  # revert to original
    backtrack(0)
    return res

# Driver code
if __name__ == '__main__':
    print(letterCasePermutation("a1b2"))

Line Notes

arr = list(s)Convert string to list for in-place modification to avoid costly string concatenations.

def backtrack(i: int):Recursive helper function with current index to explore permutations.

if i == len(arr):Base case: when index reaches end, join array and add to results.

if arr[i].isdigit():Digits do not branch; recurse to next index directly.

arr[i] = arr[i].lower()Set current letter to lowercase to explore one branch.

arr[i] = arr[i].upper()Set current letter to uppercase to explore the other branch.

arr[i] = s[i] # revert to originalBacktrack by restoring original character to maintain correct state.

import java.util.*;

public class LetterCasePermutation {
    public static List<String> letterCasePermutation(String s) {
        List<String> res = new ArrayList<>();
        char[] arr = s.toCharArray();
        backtrack(arr, 0, res, s);
        return res;
    }

    private static void backtrack(char[] arr, int i, List<String> res, String original) {
        if (i == arr.length) {
            res.add(new String(arr));
            return;
        }
        if (Character.isDigit(arr[i])) {
            backtrack(arr, i + 1, res, original);
        } else {
            arr[i] = Character.toLowerCase(arr[i]);
            backtrack(arr, i + 1, res, original);
            arr[i] = Character.toUpperCase(arr[i]);
            backtrack(arr, i + 1, res, original);
            arr[i] = original.charAt(i); // revert
        }
    }

    public static void main(String[] args) {
        System.out.println(letterCasePermutation("a1b2"));
    }
}

Line Notes

char[] arr = s.toCharArray();Convert string to char array for efficient in-place modifications.

private static void backtrack(char[] arr, int i, List<String> res, String original)Recursive helper with index and original string to revert changes.

if (i == arr.length)Base case: when index reaches end, add current permutation to results.

if (Character.isDigit(arr[i]))Digits do not branch; recurse to next index directly.

arr[i] = Character.toLowerCase(arr[i]);Set current letter to lowercase to explore one branch.

arr[i] = Character.toUpperCase(arr[i]);Set current letter to uppercase to explore the other branch.

arr[i] = original.charAt(i); // revertBacktrack by restoring original character to maintain correct state.

#include <iostream>
#include <vector>
#include <string>
using namespace std;

class Solution {
public:
    vector<string> letterCasePermutation(string s) {
        vector<string> res;
        backtrack(s, 0, res, s);
        return res;
    }

private:
    void backtrack(string& arr, int i, vector<string>& res, const string& original) {
        if (i == arr.size()) {
            res.push_back(arr);
            return;
        }
        if (isdigit(arr[i])) {
            backtrack(arr, i + 1, res, original);
        } else {
            arr[i] = tolower(arr[i]);
            backtrack(arr, i + 1, res, original);
            arr[i] = toupper(arr[i]);
            backtrack(arr, i + 1, res, original);
            arr[i] = original[i]; // revert
        }
    }
};

int main() {
    Solution sol;
    vector<string> res = sol.letterCasePermutation("a1b2");
    for (auto& str : res) cout << str << endl;
    return 0;
}

Line Notes

void backtrack(string& arr, int i, vector<string>& res, const string& original)Recursive helper with reference to string for in-place edits and original for revert.

if (i == arr.size())Base case: when index reaches end, add current permutation to results.

if (isdigit(arr[i]))Digits do not branch; recurse to next index directly.

arr[i] = tolower(arr[i]);Set current letter to lowercase to explore one branch.

arr[i] = toupper(arr[i]);Set current letter to uppercase to explore the other branch.

arr[i] = original[i]; // revertBacktrack by restoring original character to maintain correct state.

function letterCasePermutation(s) {
    const res = [];
    const arr = s.split('');
    function backtrack(i) {
        if (i === arr.length) {
            res.push(arr.join(''));
            return;
        }
        if (/[0-9]/.test(arr[i])) {
            backtrack(i + 1);
        } else {
            arr[i] = arr[i].toLowerCase();
            backtrack(i + 1);
            arr[i] = arr[i].toUpperCase();
            backtrack(i + 1);
            arr[i] = s[i]; // revert
        }
    }
    backtrack(0);
    return res;
}

// Driver code
console.log(letterCasePermutation("a1b2"));

Line Notes

const arr = s.split('');Convert string to array for efficient in-place modification.

function backtrack(i)Recursive helper with current index to explore permutations.

if (i === arr.length)Base case: when index reaches end, join array and add to results.

if (/[0-9]/.test(arr[i]))Digits do not branch; recurse to next index directly.

arr[i] = arr[i].toLowerCase();Set current letter to lowercase to explore one branch.

arr[i] = arr[i].toUpperCase();Set current letter to uppercase to explore the other branch.

arr[i] = s[i]; // revertBacktrack by restoring original character to maintain correct state.

⏱

Complexity

TimeO(2^k * n) where k is number of letters and n is string length

SpaceO(n) recursion stack + O(2^k * n) output space

We modify the input array in place, avoiding string concatenation overhead, but still explore all permutations.

💡 This approach is more memory efficient than building new strings each recursion, which can matter for large inputs.

Interview Verdict: Accepted

This is a practical optimization of the brute force approach, often preferred in interviews for cleaner code and better performance.

📊

All Approaches - One-Glance Tradeoffs

💡 For most interviews, the in-place backtracking approach is best to code due to clarity and efficiency.

Approach	Time	Space	Stack Risk	Reconstruct	Use In Interview
1. Brute Force Recursion	O(2^k * n)	O(n) recursion stack + O(2^k * n) output	Yes (deep recursion for large k)	Yes	Mention and code if comfortable with recursion
2. Iterative BFS / Queue	O(2^k * n)	O(2^k * n) for queue and output	No	Yes	Good alternative if recursion is risky
3. In-Place Backtracking	O(2^k * n)	O(n) recursion stack + O(2^k * n) output	Yes	Yes	Preferred approach to code for clarity and efficiency

💼

Interview Strategy

💡 Use this guide to understand the problem deeply before interviews. Start by clarifying the problem, then explain brute force, and finally optimize. Practice coding and testing thoroughly.

How to Present

Step 1: Clarify input constraints and output format.Step 2: Describe brute force recursive backtracking approach.Step 3: Discuss iterative BFS approach as alternative.Step 4: Present in-place backtracking optimization.Step 5: Code chosen approach and test with examples.

Time Allocation

Clarify: 2min → Approach: 5min → Code: 10min → Test: 3min. Total ~20min

What the Interviewer Tests

Interviewer tests your understanding of recursion/backtracking, handling branching logic, edge cases, and optimization skills.

Common Follow-ups

• What if input is very large? → Discuss pruning or iterative approach.
• Can you do this without recursion? → Use iterative BFS with queue.
• How to handle Unicode or special characters? → Clarify problem constraints.
• What is the maximum output size? → 2^k where k is number of letters.

💡 These follow-ups test your ability to adapt solutions, optimize, and clarify assumptions.

🔍

Pattern Recognition

When to Use

1) Problem asks for all subsets or permutations with variations; 2) Input contains letters and digits; 3) Need to toggle letter cases; 4) Output all possible strings.

Signature Phrases

toggle caseall possible stringsletter or digit

NOT This Pattern When

Problems that require counting permutations without generating them or greedy selection problems

Similar Problems

Permutations - generates all orderingsSubsets - generates all subsets of a setCombination Sum - explores combinations with backtracking

Practice

1. You need to find all unique combinations of exactly k distinct numbers from 1 to 9 that add up to a target sum n. Which algorithmic approach guarantees finding all valid combinations efficiently by exploring partial solutions and pruning impossible paths early?

easy

A. Greedy algorithm that picks the largest numbers first until the sum is reached or exceeded

B. Backtracking with pruning based on current sum and remaining numbers' possible sums

C. Dynamic programming that counts combinations without generating them explicitly

D. Sorting the numbers and using two pointers to find pairs that sum to n

Solution

1. Step 1: Understand problem constraints

   The problem requires all unique combinations of size k from numbers 1 to 9 summing to n, which suggests exploring subsets.

2. Step 2: Identify suitable algorithm

   Backtracking with pruning efficiently explores partial combinations and stops early when sums cannot reach n, guaranteeing correctness and efficiency.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Backtracking with pruning is standard for combination sum problems [OK]

Hint: Backtracking with pruning finds all valid combos efficiently [OK]

Common Mistakes:

• Thinking greedy or two pointers can find all combinations
• Confusing counting with generating combinations

2. You are given an array of integers and need to find the number of subsets whose bitwise OR equals the maximum possible OR value among all subsets. Which algorithmic approach guarantees an optimal solution for this problem?

easy

A. Greedy approach selecting elements with highest bits set first

B. Iterative bitmask enumeration generating all subsets and computing their OR

C. Dynamic programming based on subset sums

D. Sorting the array and using two pointers to find max OR subsets

Solution

1. Step 1: Understand the problem requires checking all subsets

   Since the maximum bitwise OR depends on any combination of elements, all subsets must be considered.

2. Step 2: Identify approach that enumerates all subsets efficiently

   Iterative bitmask enumeration systematically generates all subsets and computes their OR, ensuring no subset is missed.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Bitmask enumeration covers all subsets -> optimal [OK]

Hint: Bitmask enumeration covers all subsets exhaustively [OK]

Common Mistakes:

• Assuming greedy selection suffices for max OR
• Confusing subset sum DP with bitwise OR problem

3. The following code attempts to solve Combination Sum (Reuse Allowed). Identify the line containing the subtle bug that causes incorrect results or inefficiency:

def combinationSum(candidates, target):
    candidates.sort()
    result = []
    def backtrack(index, path, target):
        if target == 0:
            result.append(path[:])
            return
        if index == len(candidates) or target < 0:
            return
        max_use = target // candidates[index]
        for count in range(max_use + 1):
            path += [candidates[index]] * count
            backtrack(index + 1, path, target - candidates[index] * count)
            # Missing path restoration here
    backtrack(0, [], target)
    return result

medium

A. Line with 'path += [candidates[index]] * count' because path is modified without backtracking (pop).

B. Line with 'candidates.sort()' because sorting is unnecessary.

C. Line with 'if target == 0:' because it should check for target <= 0.

D. Line with 'max_use = target // candidates[index]' because integer division may cause errors.

Solution

1. Step 1: Analyze path modification in loop

   Path is extended by count copies of candidates[index] but never restored after recursive call, causing path to grow incorrectly across iterations.

2. Step 2: Identify missing backtracking step

   Proper backtracking requires removing added elements after recursion to restore path state for next iteration.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Modifying path without restoration leads to incorrect combinations [OK]

Hint: Always restore path after recursion in backtracking [OK]

Common Mistakes:

• Forgetting to pop elements after recursion
• Assuming path copy avoids mutation issues

4. Examine the following buggy code snippet from the trie-based solution. Which line contains the subtle bug that causes incorrect counts for some puzzles?

medium

A. Line where node.count is incremented

B. Line missing filtering words with more than 7 unique letters before insertion

C. Line checking if bit is set in mask during insertion

D. Line checking if puzzle_mask contains child bit during dfs

Solution

1. Step 1: Identify filtering step

   The code inserts all words regardless of unique letter count, causing large trie and slow traversal.

2. Step 2: Understand impact

   Words with >7 unique letters cannot match puzzles (7 letters max), so filtering them out is necessary for correctness and efficiency.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Filtering missing leads to incorrect counts and performance issues [OK]

Hint: Filter words with >7 unique letters before insertion [OK]

Common Mistakes:

• Not filtering large words
• Incorrect bit checks
• Misplaced count increments

5. Suppose the problem is modified so that numbers 1 to 9 can be reused any number of times in combinations summing to n with size k. Which change to the backtracking algorithm correctly handles this variant?

hard

A. Add a visited set to prevent reusing numbers and keep backtrack(i + 1, comb, total + i)

B. Keep the recursive call as backtrack(i + 1, comb, total + i) but allow duplicates in the result

C. Change the recursive call to backtrack(i, comb, total + i) to allow reuse of the same number i

D. Use a dynamic programming approach instead of backtracking to handle reuse efficiently

Solution

1. Step 1: Understand reuse requirement

   Allowing reuse means the same number can be chosen multiple times, so next recursion should start at current number i, not i+1.

2. Step 2: Modify recursive call

   Change backtrack(i + 1, ...) to backtrack(i, ...) to allow repeated picks of i.

3. Step 3: Confirm correctness

   This change correctly explores combinations with repeated numbers without duplicates.

4. Final Answer:

   Option C -> Option C

5. Quick Check:

   Recursing with start=i enables reuse of number i [OK]

Hint: Reuse requires recursive calls with start index unchanged [OK]

Common Mistakes:

• Not changing start index to allow reuse
• Allowing duplicates by mistake
• Switching to DP unnecessarily