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Minimum One Bit Operations to Make Integers Zero #8 Most Asked

Hard

Expedia Group

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Topics:

Bit ManipulationRecursionDynamic Programming

Given an integer n, you must transform it into 0 using the following operations any number of times:

Change the rightmost (0^th) bit in the binary representation of n.
Change the i^th bit in the binary representation of n if the (i-1)^th bit is set to 1 and the (i-2)^th through 0^th bits are set to 0.

Return the minimum number of operations to transform n into 0.

Example 1:

Input: n = 3
Output: 2
Explanation: The binary representation of 3 is "11".
"11" -> "01" with the 2^nd operation since the 0^th bit is 1.
"01" -> "00" with the 1^st operation.

Example 2:

Input: n = 6
Output: 4
Explanation: The binary representation of 6 is "110".
"110" -> "010" with the 2^nd operation since the 1^st bit is 1 and 0^th through 0^th bits are 0.
"010" -> "011" with the 1^st operation.
"011" -> "001" with the 2^nd operation since the 0^th bit is 1.
"001" -> "000" with the 1^st operation.

Constraints:

0 <= n <= 10⁹

Try coding it in LeetCode

Minimum One Bit Operations to Make Integers Zero #8 Most Asked

Hard

Expedia Group

2 views

Topics:

Bit ManipulationRecursionDynamic Programming

Given an integer n, you must transform it into 0 using the following operations any number of times:

Change the rightmost (0^th) bit in the binary representation of n.
Change the i^th bit in the binary representation of n if the (i-1)^th bit is set to 1 and the (i-2)^th through 0^th bits are set to 0.

Return the minimum number of operations to transform n into 0.

Example 1:

Input: n = 3
Output: 2
Explanation: The binary representation of 3 is "11".
"11" -> "01" with the 2^nd operation since the 0^th bit is 1.
"01" -> "00" with the 1^st operation.

Example 2:

Input: n = 6
Output: 4
Explanation: The binary representation of 6 is "110".
"110" -> "010" with the 2^nd operation since the 1^st bit is 1 and 0^th through 0^th bits are 0.
"010" -> "011" with the 1^st operation.
"011" -> "001" with the 2^nd operation since the 0^th bit is 1.
"001" -> "000" with the 1^st operation.

Constraints:

0 <= n <= 10⁹

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of possible values for the input integer n? Are there any constraints on its size?
Could you provide a small example to illustrate the operations more concretely?
If n is already 0, should I return 0?
Are there any specific constraints on the number of operations allowed, or am I only concerned with minimizing them?
Can you confirm that 'flipping the i-th bit' refers to standard 0-based indexing, where the rightmost bit is bit 0?

Brute Force Solution

Approach

The brute force approach tries all possible sequences of allowed operations to transform the given integer into zero. We explore every single path of operations, without trying to be smart or efficient, until we stumble upon the solution.

Here's how the algorithm would work step-by-step:

Start with the initial number.
Consider applying each of the two allowed operations on the number.
For each resulting number after applying those operations, again consider applying each of the two allowed operations.
Keep repeating this process, generating new numbers from the existing ones by applying the operations.
As you generate new numbers, check if any of them are zero.
If you find a zero, trace back the operations that were used to reach it from the starting number.
Keep doing this until you've explored all possible sequences of operations up to a certain length.
Out of all the sequences that lead to zero, choose the shortest one, which represents the minimum number of operations.

Code Implementation

def minimum_one_bit_operations_brute_force(number):
    queue = [(number, 0)]
    minimum_operations = float('inf')

    while queue:
        current_number, current_operations = queue.pop(0)

        if current_number == 0:
            minimum_operations = min(minimum_operations, current_operations)
            continue

        # Operation 1: Flip the rightmost bit
        operation_one_result = current_number ^ 1

        queue.append((operation_one_result, current_operations + 1))

        # Operation 2: Flip a bit if the bit to its right is one and all other bits to the right are zero
        for bit_index in range(current_number.bit_length()):
            if (current_number >> bit_index) & 1:
                if bit_index > 0 and ((current_number >> (bit_index - 1)) & 1) == 1:

                    is_all_zeros = True
                    for zero_index in range(bit_index - 1):
                        if ((current_number >> zero_index) & 1) == 1:
                            is_all_zeros = False
                            break

                    # Ensure only the right adjacent bit is 1 and the rest are zeros.
                    if is_all_zeros:
                        operation_two_result = current_number ^ (1 << bit_index)

                        # Add the new number to the exploration queue.
                        queue.append((operation_two_result, current_operations + 1))

    return minimum_operations

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach explores all possible sequences of operations. In the worst-case scenario, for a number n, each bit can potentially be flipped through a sequence of operations. Since at each step, we have two choices of operations to apply, and we explore all possible sequences of these operations until we reach zero or exhaust a maximum length, the number of possible sequences grows exponentially with the number of bits in the input integer n. This results in a time complexity of O(2^n) because we are essentially exploring a binary tree of depth proportional to n.

Space Complexity

O(2^N) – The brute force approach explores all possible sequences of operations. In the worst-case scenario, where N is the number of bits in the input integer, the algorithm could explore all 2^N possible operation sequences before finding a path to zero. This necessitates storing a queue or similar data structure to manage the generated numbers for exploration and potentially a visited set to avoid redundant computations. The space required to store these generated numbers and the visited set grows exponentially with N, leading to a space complexity of O(2^N).

Optimal Solution

Approach

The key to solving this problem efficiently is recognizing a pattern related to Gray codes. Instead of exploring all possible bit operations, we use the Gray code sequence to directly calculate the minimum number of operations needed to transform the given number to zero.

Here's how the algorithm would work step-by-step:

Understand the relationship between the given number and its corresponding Gray code representation.
Convert the given number to its Gray code equivalent. This can be done using a simple bitwise operation.
Because of the properties of Gray Code conversion, the resulting value now directly represents the minimum number of operations needed to convert the original number to zero.

Code Implementation

def minimum_one_bit_operations(number):
    # Convert the given number to its Gray code equivalent.
    gray_code = number ^ (number >> 1)

    # The Gray code value represents the minimum operations.
    return gray_code

Big(O) Analysis

Time Complexity

O(1) – The algorithm's core operation involves converting the given integer to its Gray code equivalent using a bitwise XOR operation (n ^ (n >> 1)). Bitwise operations take constant time. Since the number of operations doesn't depend on the size of the input number (n) but are fixed regardless of its value, the time complexity is O(1). The Gray code conversion happens in constant time, independent of input size.

Space Complexity

O(1) – The algorithm described converts the input number to its Gray code equivalent using a bitwise operation. This operation is performed in place using the same variable, so no additional memory scales with the input. Only a single variable to store the result of the Gray code conversion is used. Therefore, the auxiliary space complexity is constant, independent of the input number N.

Edge Cases

Case	How to Handle
Input n is 0	Return 0 immediately as no operations are needed.
Input n is 1	Return 1 immediately as only one flip is needed.
Input n is a large number (close to Integer.MAX_VALUE)	Ensure the solution avoids integer overflow when performing calculations involving n.
Input n is a power of 2	This represents a specific structure where the Gray code pattern simplifies.
Input n is a number with many consecutive 1s in its binary representation	Ensure the algorithm efficiently handles the cascading flips required for these cases.
Input n is negative (though problem states positive)	The problem definition should specify positive integers only; clarify the input range.
Maximum integer value Integer.MAX_VALUE	The solution should correctly calculate the operations without overflowing.
Numbers with alternating 0s and 1s in binary representation	The operations need to be correctly calculated for this alternating pattern.

Input n is 0

How to Handle:

Return 0 immediately as no operations are needed.

Input n is 1

How to Handle:

Return 1 immediately as only one flip is needed.

Input n is a large number (close to Integer.MAX_VALUE)

How to Handle:

Ensure the solution avoids integer overflow when performing calculations involving n.

Input n is a power of 2

How to Handle:

This represents a specific structure where the Gray code pattern simplifies.

Input n is a number with many consecutive 1s in its binary representation

How to Handle:

Ensure the algorithm efficiently handles the cascading flips required for these cases.

Input n is negative (though problem states positive)

How to Handle:

The problem definition should specify positive integers only; clarify the input range.

Maximum integer value Integer.MAX_VALUE

How to Handle:

The solution should correctly calculate the operations without overflowing.

Numbers with alternating 0s and 1s in binary representation

How to Handle:

The operations need to be correctly calculated for this alternating pattern.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of possible values for the input integer n? Are there any constraints on its size?
Could you provide a small example to illustrate the operations more concretely?
If n is already 0, should I return 0?
Are there any specific constraints on the number of operations allowed, or am I only concerned with minimizing them?
Can you confirm that 'flipping the i-th bit' refers to standard 0-based indexing, where the rightmost bit is bit 0?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Start with the initial number.
Consider applying each of the two allowed operations on the number.
For each resulting number after applying those operations, again consider applying each of the two allowed operations.
Keep repeating this process, generating new numbers from the existing ones by applying the operations.
As you generate new numbers, check if any of them are zero.
If you find a zero, trace back the operations that were used to reach it from the starting number.
Keep doing this until you've explored all possible sequences of operations up to a certain length.
Out of all the sequences that lead to zero, choose the shortest one, which represents the minimum number of operations.

Code Implementation

def minimum_one_bit_operations_brute_force(number):
    queue = [(number, 0)]
    minimum_operations = float('inf')

    while queue:
        current_number, current_operations = queue.pop(0)

        if current_number == 0:
            minimum_operations = min(minimum_operations, current_operations)
            continue

        # Operation 1: Flip the rightmost bit
        operation_one_result = current_number ^ 1

        queue.append((operation_one_result, current_operations + 1))

        # Operation 2: Flip a bit if the bit to its right is one and all other bits to the right are zero
        for bit_index in range(current_number.bit_length()):
            if (current_number >> bit_index) & 1:
                if bit_index > 0 and ((current_number >> (bit_index - 1)) & 1) == 1:

                    is_all_zeros = True
                    for zero_index in range(bit_index - 1):
                        if ((current_number >> zero_index) & 1) == 1:
                            is_all_zeros = False
                            break

                    # Ensure only the right adjacent bit is 1 and the rest are zeros.
                    if is_all_zeros:
                        operation_two_result = current_number ^ (1 << bit_index)

                        # Add the new number to the exploration queue.
                        queue.append((operation_two_result, current_operations + 1))

    return minimum_operations

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Understand the relationship between the given number and its corresponding Gray code representation.
Convert the given number to its Gray code equivalent. This can be done using a simple bitwise operation.
Because of the properties of Gray Code conversion, the resulting value now directly represents the minimum number of operations needed to convert the original number to zero.

Code Implementation

def minimum_one_bit_operations(number):
    # Convert the given number to its Gray code equivalent.
    gray_code = number ^ (number >> 1)

    # The Gray code value represents the minimum operations.
    return gray_code

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Input n is 0	Return 0 immediately as no operations are needed.
Input n is 1	Return 1 immediately as only one flip is needed.
Input n is a large number (close to Integer.MAX_VALUE)	Ensure the solution avoids integer overflow when performing calculations involving n.
Input n is a power of 2	This represents a specific structure where the Gray code pattern simplifies.
Input n is a number with many consecutive 1s in its binary representation	Ensure the algorithm efficiently handles the cascading flips required for these cases.
Input n is negative (though problem states positive)	The problem definition should specify positive integers only; clarify the input range.
Maximum integer value Integer.MAX_VALUE	The solution should correctly calculate the operations without overflowing.
Numbers with alternating 0s and 1s in binary representation	The operations need to be correctly calculated for this alternating pattern.

Input n is 0

How to Handle:

Return 0 immediately as no operations are needed.

Input n is 1

How to Handle:

Return 1 immediately as only one flip is needed.

Input n is a large number (close to Integer.MAX_VALUE)

How to Handle:

Ensure the solution avoids integer overflow when performing calculations involving n.

Input n is a power of 2

How to Handle:

This represents a specific structure where the Gray code pattern simplifies.

Input n is a number with many consecutive 1s in its binary representation

How to Handle:

Ensure the algorithm efficiently handles the cascading flips required for these cases.

Input n is negative (though problem states positive)

How to Handle:

The problem definition should specify positive integers only; clarify the input range.

Maximum integer value Integer.MAX_VALUE

How to Handle:

The solution should correctly calculate the operations without overflowing.

Numbers with alternating 0s and 1s in binary representation

How to Handle:

The operations need to be correctly calculated for this alternating pattern.

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