UTF-8 Validation

🎯

UTF-8 Validation

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When transmitting text data over the internet, ensuring that the byte sequences conform to UTF-8 encoding is crucial to avoid misinterpretation or corruption.

💡 This problem involves checking if a sequence of integers represents a valid UTF-8 encoded string. Beginners often struggle because UTF-8 encoding rules depend on bit patterns in bytes, which requires understanding binary representations and bitmasking.

📋

Problem Statement

Given an array of integers representing bytes, determine if it is a valid UTF-8 encoding. Each integer represents one byte (only the least significant 8 bits are used). UTF-8 encoding rules specify how many bytes a character uses and how continuation bytes must be formatted.

1 ≤ n ≤ 10^5Each integer is in the range 0 to 255 (only 8 bits used)

💡

Example

Input"[197, 130, 1]"

Outputtrue

197 (11000101) starts a 2-byte character, 130 (10000010) is a valid continuation byte, 1 (00000001) is a valid 1-byte character.

Input"[235, 140, 4]"

Outputfalse

235 (11101011) starts a 3-byte character, 140 (10001100) is a valid continuation byte, but 4 (00000100) is not a valid continuation byte.

• Single byte character only → true
• Continuation byte without a leading byte → false
• Leading byte indicating 4-byte character but fewer continuation bytes → false
• Bytes with leading bits not matching any UTF-8 pattern → false

⚠️

Common Mistakes

Not checking if enough continuation bytes remain after a leading byte

Index out of range error or false positives

✅ Always verify that the number of continuation bytes does not exceed array length

Misinterpreting continuation bytes by not checking the two leading bits are '10'

Invalid UTF-8 sequences accepted as valid

✅ Use bitmask to ensure continuation bytes start with '10'

Incorrectly identifying the number of bytes in a character by wrong bit shifts

Wrong character length leads to validation errors

✅ Use correct bit shifts and masks to identify leading byte patterns

Not verifying that the final continuation byte count is zero at the end

Partial characters at the end accepted as valid

✅ Return false if continuation byte counter is not zero after processing all bytes

Assuming input integers are always 8-bit without masking

Incorrect bit checks if input has higher bits set

✅ Mask input with 0xFF or ensure input is within 0-255

🧠

Brute Force (Check Each Byte Individually with Nested Conditions)

💡 This approach exists to build intuition by explicitly checking each byte and its expected pattern without using bitmask tricks. It helps beginners understand the UTF-8 rules before optimizing.

Intuition

Check each byte one by one, determine how many bytes the current character should have based on the leading bits, then verify the following bytes are valid continuation bytes.

Algorithm

1. Initialize an index to 0 to iterate over the data array.
2. For each byte, determine the number of bytes in the current UTF-8 character by checking leading bits.
3. Check if the following bytes are valid continuation bytes (start with '10').
4. If any check fails, return false; if all pass, return true.

💡 The challenge is correctly identifying the number of bytes per character and validating continuation bytes, which requires careful bit checks.

</>

Code

def validUtf8(data):
    n = len(data)
    i = 0
    while i < n:
        byte = data[i]
        # Determine number of bytes
        if (byte >> 7) == 0:
            num_bytes = 1
        elif (byte >> 5) == 0b110:
            num_bytes = 2
        elif (byte >> 4) == 0b1110:
            num_bytes = 3
        elif (byte >> 3) == 0b11110:
            num_bytes = 4
        else:
            return False
        # Check continuation bytes
        if i + num_bytes > n:
            return False
        for j in range(i + 1, i + num_bytes):
            if (data[j] >> 6) != 0b10:
                return False
        i += num_bytes
    return True

# Driver code
if __name__ == '__main__':
    print(validUtf8([197, 130, 1]))  # True
    print(validUtf8([235, 140, 4]))  # False

Line Notes

while i < n:Loop over all bytes in the data array

if (byte >> 7) == 0:Check if the first bit is 0, meaning 1-byte character

elif (byte >> 5) == 0b110:Check if first three bits are 110, meaning 2-byte character

for j in range(i + 1, i + num_bytes):Verify each continuation byte starts with '10'

public class Solution {
    public boolean validUtf8(int[] data) {
        int n = data.length;
        int i = 0;
        while (i < n) {
            int byteVal = data[i];
            int numBytes;
            if ((byteVal >> 7) == 0) {
                numBytes = 1;
            } else if ((byteVal >> 5) == 0b110) {
                numBytes = 2;
            } else if ((byteVal >> 4) == 0b1110) {
                numBytes = 3;
            } else if ((byteVal >> 3) == 0b11110) {
                numBytes = 4;
            } else {
                return false;
            }
            if (i + numBytes > n) return false;
            for (int j = i + 1; j < i + numBytes; j++) {
                if ((data[j] >> 6) != 0b10) return false;
            }
            i += numBytes;
        }
        return true;
    }

    public static void main(String[] args) {
        Solution sol = new Solution();
        System.out.println(sol.validUtf8(new int[]{197, 130, 1})); // true
        System.out.println(sol.validUtf8(new int[]{235, 140, 4})); // false
    }
}

Line Notes

while (i < n) {Iterate through each byte in the input array

if ((byteVal >> 7) == 0) {Check if the byte is a single-byte character

for (int j = i + 1; j < i + numBytes; j++) {Check all continuation bytes for correct pattern

return false;Return false immediately if any validation fails

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

class Solution {
public:
    bool validUtf8(vector<int>& data) {
        int n = data.size();
        int i = 0;
        while (i < n) {
            int byteVal = data[i];
            int numBytes;
            if ((byteVal >> 7) == 0) {
                numBytes = 1;
            } else if ((byteVal >> 5) == 0b110) {
                numBytes = 2;
            } else if ((byteVal >> 4) == 0b1110) {
                numBytes = 3;
            } else if ((byteVal >> 3) == 0b11110) {
                numBytes = 4;
            } else {
                return false;
            }
            if (i + numBytes > n) return false;
            for (int j = i + 1; j < i + numBytes; j++) {
                if ((data[j] >> 6) != 0b10) return false;
            }
            i += numBytes;
        }
        return true;
    }
};

int main() {
    Solution sol;
    vector<int> data1 = {197, 130, 1};
    vector<int> data2 = {235, 140, 4};
    cout << boolalpha << sol.validUtf8(data1) << endl; // true
    cout << boolalpha << sol.validUtf8(data2) << endl; // false
    return 0;
}

Line Notes

while (i < n) {Loop through all bytes in the vector

if ((byteVal >> 7) == 0) {Check if the byte is a single-byte UTF-8 character

for (int j = i + 1; j < i + numBytes; j++) {Validate continuation bytes start with '10'

return false;Immediately return false if any byte fails validation

var validUtf8 = function(data) {
    let n = data.length;
    let i = 0;
    while (i < n) {
        let byte = data[i];
        let numBytes;
        if ((byte >> 7) === 0) {
            numBytes = 1;
        } else if ((byte >> 5) === 0b110) {
            numBytes = 2;
        } else if ((byte >> 4) === 0b1110) {
            numBytes = 3;
        } else if ((byte >> 3) === 0b11110) {
            numBytes = 4;
        } else {
            return false;
        }
        if (i + numBytes > n) return false;
        for (let j = i + 1; j < i + numBytes; j++) {
            if ((data[j] >> 6) !== 0b10) return false;
        }
        i += numBytes;
    }
    return true;
};

// Test cases
console.log(validUtf8([197, 130, 1])); // true
console.log(validUtf8([235, 140, 4])); // false

Line Notes

while (i < n) {Iterate over each byte in the input array

if ((byte >> 7) === 0) {Check if the byte is a single-byte UTF-8 character

for (let j = i + 1; j < i + numBytes; j++) {Check that continuation bytes start with '10'

return false;Return false immediately if any byte does not match UTF-8 pattern

⏱

Complexity

TimeO(n)

SpaceO(1)

We scan the array once, checking each byte and its continuation bytes in constant time per character.

💡 For n=100,000 bytes, this means roughly 100,000 operations, which is efficient for interview constraints.

Interview Verdict: Accepted

This approach is straightforward and efficient enough to pass all tests, making it a solid baseline solution.

🧠

Bitmasking with Predefined Masks for Leading Bytes

💡 This approach uses bitmasks to cleanly identify the number of bytes in a character, improving readability and reducing error-prone comparisons.

Intuition

Use masks to isolate leading bits and compare against known UTF-8 patterns to determine character length, then verify continuation bytes similarly.

Algorithm

1. Define masks and expected patterns for 1 to 4 byte UTF-8 characters.
2. For each byte, apply masks to determine the character length.
3. Check if the following bytes are valid continuation bytes using bitmask.
4. Return false if any check fails; otherwise, return true.

💡 Using masks abstracts away bit shifts and makes the code easier to verify and extend.

</>

Code

def validUtf8(data):
    masks = [0b10000000, 0b11100000, 0b11110000, 0b11111000]
    patterns = [0b00000000, 0b11000000, 0b11100000, 0b11110000]
    n = len(data)
    i = 0
    while i < n:
        byte = data[i]
        num_bytes = 0
        for idx in range(4):
            if (byte & masks[idx]) == patterns[idx]:
                num_bytes = idx + 1
                break
        if num_bytes == 0 or num_bytes > 4:
            return False
        if i + num_bytes > n:
            return False
        for j in range(i + 1, i + num_bytes):
            if (data[j] & 0b11000000) != 0b10000000:
                return False
        i += num_bytes
    return True

# Driver code
if __name__ == '__main__':
    print(validUtf8([197, 130, 1]))  # True
    print(validUtf8([235, 140, 4]))  # False

Line Notes

masks = [0b10000000, 0b11100000, 0b11110000, 0b11111000]Masks isolate leading bits for 1-4 byte characters

patterns = [0b00000000, 0b11000000, 0b11100000, 0b11110000]Expected bit patterns for each character length

if (byte & masks[idx]) == patterns[idx]:Check if byte matches the pattern for this character length

if (data[j] & 0b11000000) != 0b10000000:Verify continuation bytes start with '10'

public class Solution {
    public boolean validUtf8(int[] data) {
        int[] masks = {0b10000000, 0b11100000, 0b11110000, 0b11111000};
        int[] patterns = {0b00000000, 0b11000000, 0b11100000, 0b11110000};
        int n = data.length;
        int i = 0;
        while (i < n) {
            int byteVal = data[i];
            int numBytes = 0;
            for (int idx = 0; idx < 4; idx++) {
                if ((byteVal & masks[idx]) == patterns[idx]) {
                    numBytes = idx + 1;
                    break;
                }
            }
            if (numBytes == 0 || numBytes > 4) return false;
            if (i + numBytes > n) return false;
            for (int j = i + 1; j < i + numBytes; j++) {
                if ((data[j] & 0b11000000) != 0b10000000) return false;
            }
            i += numBytes;
        }
        return true;
    }

    public static void main(String[] args) {
        Solution sol = new Solution();
        System.out.println(sol.validUtf8(new int[]{197, 130, 1})); // true
        System.out.println(sol.validUtf8(new int[]{235, 140, 4})); // false
    }
}

Line Notes

int[] masks = {0b10000000, 0b11100000, 0b11110000, 0b11111000};Masks for isolating leading bits

int[] patterns = {0b00000000, 0b11000000, 0b11100000, 0b11110000};Expected patterns for each UTF-8 character length

if ((byteVal & masks[idx]) == patterns[idx]) {Check if byte matches a valid UTF-8 leading pattern

if ((data[j] & 0b11000000) != 0b10000000) return false;Check continuation bytes start with '10'

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

class Solution {
public:
    bool validUtf8(vector<int>& data) {
        int masks[4] = {0b10000000, 0b11100000, 0b11110000, 0b11111000};
        int patterns[4] = {0b00000000, 0b11000000, 0b11100000, 0b11110000};
        int n = data.size();
        int i = 0;
        while (i < n) {
            int byteVal = data[i];
            int numBytes = 0;
            for (int idx = 0; idx < 4; idx++) {
                if ((byteVal & masks[idx]) == patterns[idx]) {
                    numBytes = idx + 1;
                    break;
                }
            }
            if (numBytes == 0 || numBytes > 4) return false;
            if (i + numBytes > n) return false;
            for (int j = i + 1; j < i + numBytes; j++) {
                if ((data[j] & 0b11000000) != 0b10000000) return false;
            }
            i += numBytes;
        }
        return true;
    }
};

int main() {
    Solution sol;
    vector<int> data1 = {197, 130, 1};
    vector<int> data2 = {235, 140, 4};
    cout << boolalpha << sol.validUtf8(data1) << endl; // true
    cout << boolalpha << sol.validUtf8(data2) << endl; // false
    return 0;
}

Line Notes

int masks[4] = {0b10000000, 0b11100000, 0b11110000, 0b11111000};Masks to isolate leading bits for UTF-8 bytes

int patterns[4] = {0b00000000, 0b11000000, 0b11100000, 0b11110000};Expected bit patterns for each character length

if ((byteVal & masks[idx]) == patterns[idx]) {Check if the byte matches a valid UTF-8 leading byte pattern

if ((data[j] & 0b11000000) != 0b10000000) return false;Verify continuation bytes start with '10'

var validUtf8 = function(data) {
    const masks = [0b10000000, 0b11100000, 0b11110000, 0b11111000];
    const patterns = [0b00000000, 0b11000000, 0b11100000, 0b11110000];
    let n = data.length;
    let i = 0;
    while (i < n) {
        let byte = data[i];
        let numBytes = 0;
        for (let idx = 0; idx < 4; idx++) {
            if ((byte & masks[idx]) === patterns[idx]) {
                numBytes = idx + 1;
                break;
            }
        }
        if (numBytes === 0 || numBytes > 4) return false;
        if (i + numBytes > n) return false;
        for (let j = i + 1; j < i + numBytes; j++) {
            if ((data[j] & 0b11000000) !== 0b10000000) return false;
        }
        i += numBytes;
    }
    return true;
};

// Test cases
console.log(validUtf8([197, 130, 1])); // true
console.log(validUtf8([235, 140, 4])); // false

Line Notes

const masks = [0b10000000, 0b11100000, 0b11110000, 0b11111000];Masks to isolate leading bits for UTF-8 bytes

const patterns = [0b00000000, 0b11000000, 0b11100000, 0b11110000];Expected bit patterns for each UTF-8 character length

if ((byte & masks[idx]) === patterns[idx]) {Check if the byte matches a valid UTF-8 leading byte pattern

if ((data[j] & 0b11000000) !== 0b10000000) return false;Verify continuation bytes start with '10'

⏱

Complexity

TimeO(n)

SpaceO(1)

We iterate through the array once, using bitmask operations which are constant time.

💡 For large inputs, this approach is just as efficient as brute force but easier to maintain.

Interview Verdict: Accepted

This approach is preferred for clarity and maintainability while maintaining optimal performance.

🧠

Single Pass with Continuation Byte Counter

💡 This approach optimizes by tracking how many continuation bytes are expected, reducing repeated pattern checks.

Intuition

Use a counter to track how many continuation bytes are needed after a leading byte, decrementing it as continuation bytes are validated.

Algorithm

1. Initialize a counter for expected continuation bytes to 0.
2. Iterate over each byte in the data array.
3. If counter is 0, determine the number of bytes in the character from leading bits and set counter accordingly.
4. If counter > 0, check if the byte is a valid continuation byte and decrement counter.
5. If any check fails or counter is not zero at the end, return false; else return true.

💡 This approach simplifies validation by linearizing the checks and avoiding nested loops.

</>

Code

def validUtf8(data):
    remaining = 0
    for byte in data:
        if remaining == 0:
            if (byte >> 5) == 0b110:
                remaining = 1
            elif (byte >> 4) == 0b1110:
                remaining = 2
            elif (byte >> 3) == 0b11110:
                remaining = 3
            elif (byte >> 7) == 0:
                remaining = 0
            else:
                return False
        else:
            if (byte >> 6) != 0b10:
                return False
            remaining -= 1
    return remaining == 0

# Driver code
if __name__ == '__main__':
    print(validUtf8([197, 130, 1]))  # True
    print(validUtf8([235, 140, 4]))  # False

Line Notes

remaining = 0Tracks how many continuation bytes are expected

if remaining == 0:If no continuation bytes expected, determine character length

elif (byte >> 4) == 0b1110:Check for 3-byte character leading bits

if (byte >> 6) != 0b10:Validate continuation byte pattern

public class Solution {
    public boolean validUtf8(int[] data) {
        int remaining = 0;
        for (int byteVal : data) {
            if (remaining == 0) {
                if ((byteVal >> 5) == 0b110) {
                    remaining = 1;
                } else if ((byteVal >> 4) == 0b1110) {
                    remaining = 2;
                } else if ((byteVal >> 3) == 0b11110) {
                    remaining = 3;
                } else if ((byteVal >> 7) == 0) {
                    remaining = 0;
                } else {
                    return false;
                }
            } else {
                if ((byteVal >> 6) != 0b10) return false;
                remaining--;
            }
        }
        return remaining == 0;
    }

    public static void main(String[] args) {
        Solution sol = new Solution();
        System.out.println(sol.validUtf8(new int[]{197, 130, 1})); // true
        System.out.println(sol.validUtf8(new int[]{235, 140, 4})); // false
    }
}

Line Notes

int remaining = 0;Counter for expected continuation bytes

if (remaining == 0) {Determine character length when no continuation bytes expected

else { if ((byteVal >> 6) != 0b10) return false;Validate continuation byte pattern

remaining--;Decrement continuation byte counter after validation

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

class Solution {
public:
    bool validUtf8(vector<int>& data) {
        int remaining = 0;
        for (int byteVal : data) {
            if (remaining == 0) {
                if ((byteVal >> 5) == 0b110) {
                    remaining = 1;
                } else if ((byteVal >> 4) == 0b1110) {
                    remaining = 2;
                } else if ((byteVal >> 3) == 0b11110) {
                    remaining = 3;
                } else if ((byteVal >> 7) == 0) {
                    remaining = 0;
                } else {
                    return false;
                }
            } else {
                if ((byteVal >> 6) != 0b10) return false;
                remaining--;
            }
        }
        return remaining == 0;
    }
};

int main() {
    Solution sol;
    vector<int> data1 = {197, 130, 1};
    vector<int> data2 = {235, 140, 4};
    cout << boolalpha << sol.validUtf8(data1) << endl; // true
    cout << boolalpha << sol.validUtf8(data2) << endl; // false
    return 0;
}

Line Notes

int remaining = 0;Tracks how many continuation bytes are expected

if (remaining == 0) {Determine the number of bytes for a new character

if ((byteVal >> 6) != 0b10) return false;Check if continuation byte starts with '10'

remaining--;Decrease the count of expected continuation bytes

var validUtf8 = function(data) {
    let remaining = 0;
    for (let byte of data) {
        if (remaining === 0) {
            if ((byte >> 5) === 0b110) {
                remaining = 1;
            } else if ((byte >> 4) === 0b1110) {
                remaining = 2;
            } else if ((byte >> 3) === 0b11110) {
                remaining = 3;
            } else if ((byte >> 7) === 0) {
                remaining = 0;
            } else {
                return false;
            }
        } else {
            if ((byte >> 6) !== 0b10) return false;
            remaining--;
        }
    }
    return remaining === 0;
};

// Test cases
console.log(validUtf8([197, 130, 1])); // true
console.log(validUtf8([235, 140, 4])); // false

Line Notes

let remaining = 0;Counter for how many continuation bytes are expected

if (remaining === 0) {When no continuation bytes expected, determine character length

if ((byte >> 6) !== 0b10) return false;Validate continuation byte pattern

remaining--;Decrement continuation byte counter after each valid continuation byte

⏱

Complexity

TimeO(n)

SpaceO(1)

Single pass through the data array with constant time checks per byte.

💡 This approach is the most efficient and clean for large inputs, minimizing overhead.

Interview Verdict: Accepted

This is the optimal approach to implement in interviews due to its simplicity and efficiency.

📊

All Approaches - One-Glance Tradeoffs

💡 The single pass approach with a continuation byte counter is recommended for interviews due to its clarity and efficiency.

Approach	Time	Space	Stack Risk	Reconstruct	Use In Interview
1. Brute Force	O(n)	O(1)	No	N/A	Good to mention for clarity, but avoid coding due to verbosity
2. Bitmasking with Predefined Masks	O(n)	O(1)	No	N/A	Good for clean code, acceptable to code if comfortable with bitmasks
3. Single Pass with Continuation Byte Counter	O(n)	O(1)	No	N/A	Best approach to code in interviews for simplicity and efficiency

💼

Interview Strategy

💡 Use this guide to understand the problem deeply, practice multiple approaches, and prepare to explain your reasoning clearly in interviews.

How to Present

Step 1: Clarify the problem and UTF-8 encoding rules with the interviewer.Step 2: Present the brute force approach to show understanding of the problem.Step 3: Discuss improvements using bitmasking for cleaner code.Step 4: Present the optimal single pass approach with a continuation byte counter.Step 5: Write clean, tested code and explain edge cases.

Time Allocation

Clarify: 3min → Approach: 5min → Code: 10min → Test: 5min. Total ~23min

What the Interviewer Tests

The interviewer tests your understanding of bit manipulation, attention to detail in pattern matching, and ability to optimize code while maintaining correctness.

Common Follow-ups

• How would you handle invalid input outside 0-255? → Validate input range before processing.
• Can you extend this to UTF-16 validation? → Discuss differences in encoding and validation rules.

💡 Follow-ups often test your ability to generalize or handle edge cases beyond the initial problem.

🔍

Pattern Recognition

When to Use

1) Input involves byte sequences or arrays of integers representing bytes; 2) Need to validate encoding or bit patterns; 3) Problem mentions leading bits or continuation bytes; 4) Requires bit manipulation to check patterns.

Signature Phrases

leading bitscontinuation bytesbyte pattern check

NOT This Pattern When

Problems involving string parsing without bit-level checks or problems focused on arithmetic bit operations without pattern validation.

Similar Problems

Valid Parentheses - pattern matching of sequencesBinary Number with Alternating Bits - checking bit patternsNumber of 1 Bits - counting bits in integers

Practice

1. Consider the following code snippet implementing the optimal approach for bitwise AND of numbers range. What is the output when calling rangeBitwiseAnd(5, 7)?

easy

A. 4

B. 5

C. 0

D. 7

Solution

1. Step 1: Trace initial values

   left = 5 (binary 0101), right = 7 (binary 0111)

2. Step 2: Execute loop iterations

   First iteration: right = 7 & 6 = 6 (0110) Second iteration: right = 6 & 5 = 4 (0100) Now right (4) <= left (5) is true, loop ends.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Final returned value is 4 after clearing differing bits [OK]

Hint: Bit-clearing reduces right until common prefix found [OK]

Common Mistakes:

• Confusing loop condition
• Off-by-one in bit clearing
• Returning right too early

2. You need to count how many bits are set to 1 in a 32-bit unsigned integer. Which approach guarantees the most efficient and reliable solution regardless of input size?

easy

A. Use a greedy algorithm that counts bits by repeatedly subtracting powers of two.

B. Use built-in bit count functions provided by the language or hardware instructions.

C. Use dynamic programming to store counts of smaller bit segments and combine them.

D. Use a brute force loop checking each bit one by one with bitwise operations.

Solution

1. Step 1: Understand problem constraints

   The problem requires counting set bits in a fixed-size integer, which can be done in constant time using hardware or built-in functions.

2. Step 2: Compare approaches

   Brute force and DP approaches work but are slower (O(1) but with more operations). Greedy subtraction is inefficient and incorrect for bit counting. Built-in functions leverage optimized CPU instructions for O(1) time.

3. Final Answer:

   Option B -> Option B

4. Quick Check:

   Built-in bit count functions are optimal and reliable [OK]

Hint: Built-in bit count functions use CPU instructions [OK]

Common Mistakes:

• Confusing greedy subtraction with bit counting
• Thinking DP is needed for fixed-size integers

3. Identify the bug in the following code that attempts to count the number of 1 bits in a signed 32-bit integer:

def hammingWeight(n: int) -> int:
    count = 0
    while n != 0:
        count += n & 1
        n >>= 1
    return count

medium

A. The loop condition 'while n != 0' causes infinite loop for negative numbers due to sign extension.

B. The bitwise AND operation 'n & 1' is incorrect; it should be logical AND 'n && 1'.

C. The right shift operator 'n >>= 1' should be replaced with left shift to count bits correctly.

D. The initial count variable should be initialized to 1 instead of 0.

Solution

1. Step 1: Understand signed right shift behavior

   In Python, right shifting a negative number propagates the sign bit, so n never becomes zero.

2. Step 2: Analyze loop condition

   Because n != 0 remains true for negative numbers, the loop never terminates, causing infinite loop.

3. Final Answer:

   Option A -> Option A

4. Quick Check:

   Signed right shift on negative numbers causes infinite loop [OK]

Hint: Signed right shift on negatives causes infinite loop [OK]

Common Mistakes:

• Confusing bitwise AND with logical AND
• Wrong shift direction
• Incorrect count initialization

4. Suppose you want to generate a Gray code sequence where each number can be repeated multiple times (i.e., reuse allowed). Which modification to the optimal bitwise formula approach correctly handles this variant?

hard

A. Use brute force backtracking to generate sequences with repeated numbers allowed.

B. Modify the formula to i ^ (i >> 1) but allow duplicates by not enforcing uniqueness in the output list.

C. Use the same XOR formula but generate only n numbers instead of 2^n.

D. No modification needed; the original formula already allows repeated numbers.

Solution

1. Step 1: Understand the reuse constraint

   Allowing repeated numbers breaks the Gray code uniqueness property, so the original formula which generates unique numbers won't work.

2. Step 2: Identify suitable approach

   Brute force backtracking can explore sequences with repeated numbers and enforce the one-bit difference condition without uniqueness constraints.

3. Step 3: Why other options fail

   Original formula always generates unique numbers; simply allowing duplicates in output breaks Gray code definition; generating fewer numbers misses full coverage.

4. Final Answer:

   Option A -> Option A

5. Quick Check:

   Backtracking is flexible enough to handle reuse constraints [OK]

Hint: Original formula assumes unique numbers; reuse needs backtracking [OK]

Common Mistakes:

• Assuming original formula can be tweaked trivially for reuse
• Thinking fewer numbers still form valid Gray code
• Ignoring uniqueness property of Gray code

5. If the problem changes so that elements can be reused unlimited times in subsets (i.e., multisets), which modification to the backtracking approach correctly generates all such subsets?

hard

A. Use bitmask enumeration as before without any changes

B. Change recursive call to backtrack(i + 1, path) to prevent reuse of elements

C. Sort the array and skip duplicates during recursion to avoid repeated subsets

D. Change recursive call to backtrack(i, path) to allow reuse of the current element

Solution

1. Step 1: Understand reuse means elements can appear multiple times

   Allowing reuse means after including nums[i], we can include it again in subsequent recursive calls.

2. Step 2: Modify recursion to call backtrack(i, path) instead of backtrack(i + 1, path)

   This keeps the index at i, allowing repeated inclusion of nums[i].

3. Step 3: Confirm other options break correctness

   Advancing index (i + 1) prevents reuse; skipping duplicates is unrelated to reuse; bitmask enumeration assumes unique inclusion per element.

4. Final Answer:

   Option D -> Option D

5. Quick Check:

   Recursing with same index enables unlimited reuse [OK]

Hint: Reuse requires recursive call with same index [OK]

Common Mistakes:

• Forgetting to keep index same to allow reuse
• Assuming bitmask works unchanged for reuse