Minimum Flips To Make A OR B = C: LeetCode 1318 Guide

by Kenji Nakamura 54 views

Hey guys! Ever stumbled upon a tricky problem that just makes you scratch your head? Well, let's dive into a fascinating LeetCode problem: 1318. Minimum Flips to Make A OR B Equal to C. We'll break it down, explore the logic, and make sure you're equipped to tackle it like a pro. This article will not only explain the problem but also provide a detailed, SEO-optimized guide to ensure it's easy to find and understand.

Understanding the Problem

So, what's the deal with this "Minimum Flips" challenge? In essence, we're given three positive integers: a, b, and c. Our mission, should we choose to accept it, is to find the minimum number of bit flips required to make the bitwise OR of a and b equal to c. Think of it like this: we have two numbers, a and b, and we want their combined result (using the OR operation) to match a target number c. But, if they don't match, we need to flip some bits in a and b to make them match, and we want to do this with the fewest flips possible.

Let's break it down further with an example. Imagine a = 2, b = 6, and c = 5. First, we convert these numbers into their binary representations:

  • a = 2 (binary: 10)
  • b = 6 (binary: 110)
  • c = 5 (binary: 101)

Now, we perform the bitwise OR operation on a and b:

10 OR 110 = 110 (which is 6 in decimal)

We see that the result (110) is not equal to c (101). So, we need to figure out the minimum number of flips to make it equal. By observing the binary representations, we can deduce:

  • The second bit (from the right) in a OR b is 1, while in c it is 0. This means we need to flip this bit. To do that, we need to flip either the corresponding bit in a or b (or both if needed) from 1 to 0.
  • The first bit (from the right) in a OR b is 0, while in c it is 1. To achieve this, we need to ensure that after the flips, at least one of the corresponding bits in a or b is 1.

In this specific case, we could flip the second bit of b from 1 to 0 (one flip) and ensure the first bit of a is 1 by flipping its zero bit to 1. So, one optimal solution involves flipping one bit in b. Therefore, the minimum flips required is 2. This might sound a bit complex initially, but don't worry! We'll dissect the problem step-by-step and develop a robust approach.

The core challenge here is to efficiently compare the bits of a OR b with the bits of c and determine the optimal bit flips. We need a strategy that minimizes the flips while ensuring a OR b becomes equal to c. This involves carefully considering different scenarios and optimizing our approach.

To really nail this problem, we need to understand bitwise operations and how they influence the outcome. We will explore the binary representation of numbers and learn how bitwise OR works. We'll also cover the various scenarios that require bit flips and develop a logical strategy to minimize them. By the end of this article, you'll not only grasp the solution but also appreciate the underlying principles, making you a stronger problem-solver in the process.

Deconstructing the Solution: A Step-by-Step Approach

Now that we grasp the problem's essence, let's dive into the solution. The key to solving this problem lies in comparing the binary representations of a, b, and c bit by bit. We need to analyze each bit position and determine the necessary flips to make (a OR b) match c. This comparison is going to be the backbone of our solution.

The fundamental idea involves iterating through each bit position of the integers. For each bit position, we'll extract the corresponding bits from a, b, and c. Then, we'll apply some logical rules to determine the number of flips required at that position. The total number of flips will be the sum of flips needed at each bit position. Think of it as a meticulous, bit-by-bit audit to ensure everything aligns perfectly.

Here’s a more structured way to think about it:

  1. Iterate through bits: We loop through each bit position, typically from the least significant bit (LSB) to the most significant bit (MSB). For 32-bit integers, this means iterating 32 times. We will use bitwise operations to check each bit of the numbers.
  2. Extract bits: For each bit position i, we extract the i-th bit from a, b, and c. This can be done using the right shift operator (>>) and the bitwise AND operator (&). For instance, to get the i-th bit of a, we perform (a >> i) & 1. This shifts a to the right by i positions, effectively moving the i-th bit to the LSB, and then we use & 1 to isolate that bit.
  3. Compare and count flips: Now, we compare the extracted bits and apply the core logic:
    • If the i-th bit of c is 1: We need the i-th bit of (a OR b) to also be 1. This means at least one of the corresponding bits in a or b should be 1. If both bits in a and b are 0, we need to flip one of them, resulting in one flip.
    • If the i-th bit of c is 0: We need the i-th bit of (a OR b) to also be 0. This implies that both corresponding bits in a and b must be 0. If either or both bits in a and b are 1, we need to flip them to 0. If both bits are 1, we need two flips, and if only one bit is 1, we need one flip.
  4. Accumulate flips: We keep a running count of the flips required for each bit position. After iterating through all the bits, the total count gives us the minimum number of flips needed.

The beauty of this approach is its simplicity and efficiency. By breaking down the problem into individual bit comparisons, we can handle each case systematically. This approach is also very amenable to code, making the implementation straightforward. Understanding each step in this process is crucial for successfully applying it to the problem. So, let's solidify our understanding with an example and then transition to the code implementation.

Code Implementation: Turning Logic into Action

Alright, let's transform our step-by-step logic into actual code. We'll use Python for its readability, but the concepts translate easily to other languages like Java or C++. Remember, the goal is to iterate through the bits, compare them, and count the necessary flips. So, let's get coding!

Here's a Python function that implements the solution:

def minFlips(a: int, b: int, c: int) -> int:
    flips = 0
    for i in range(32): # Iterate through 32 bits
        bit_a = (a >> i) & 1 # Extract i-th bit of a
        bit_b = (b >> i) & 1 # Extract i-th bit of b
        bit_c = (c >> i) & 1 # Extract i-th bit of c

        if bit_c == 1:
            if bit_a | bit_b == 0: # If (a OR b) bit is 0, need 1 flip
                flips += 1
        else:
            flips += bit_a + bit_b # If c bit is 0, flips needed = number of 1s in a and b

    return flips

Let's break this code down line by line to understand what's happening:

  1. def minFlips(a: int, b: int, c: int) -> int:: This defines our function minFlips that takes three integers (a, b, c) as input and returns an integer representing the minimum flips.
  2. flips = 0: We initialize a variable flips to 0. This variable will accumulate the total number of flips required.
  3. for i in range(32):: This loop iterates 32 times, representing the 32 bits of an integer. We are going through each bit position.
  4. bit_a = (a >> i) & 1: This line extracts the i-th bit from integer a. The a >> i part right-shifts a by i positions, moving the i-th bit to the rightmost position. The & 1 part performs a bitwise AND with 1, effectively isolating the rightmost bit (which was originally the i-th bit).
  5. bit_b = (b >> i) & 1: Similarly, this line extracts the i-th bit from integer b.
  6. bit_c = (c >> i) & 1: And this line extracts the i-th bit from integer c.
  7. if bit_c == 1:: This starts a conditional block that handles the case when the i-th bit of c is 1.
  8. if bit_a | bit_b == 0:: Inside the if bit_c == 1 block, this condition checks if the i-th bit of (a OR b) is 0. The | operator is the bitwise OR. If both bit_a and bit_b are 0, then bit_a | bit_b will be 0. In this case, we need to flip one of the bits in a or b to make the i-th bit of (a OR b) become 1, so we increment flips by 1.
  9. flips += 1: This line increments the flips count if the condition bit_a | bit_b == 0 is true.
  10. else:: This starts a conditional block that handles the case when the i-th bit of c is 0.
  11. flips += bit_a + bit_b: If the i-th bit of c is 0, we need to ensure that the i-th bit of (a OR b) is also 0. This means both bit_a and bit_b must be 0. The expression bit_a + bit_b directly gives us the number of 1s among the i-th bits of a and b. If both are 1, it adds 2 to flips (since we need to flip both). If one of them is 1, it adds 1 to flips. If both are 0, it adds 0 to flips, which is exactly what we need.
  12. return flips: Finally, the function returns the total number of flips calculated.

This code efficiently implements our bit-by-bit comparison strategy. It is clear, concise, and directly reflects the logic we discussed earlier. To further illustrate, let's walk through the example we mentioned before (a = 2, b = 6, c = 5) to see how the code works in action.

Example Walkthrough: Seeing the Code in Action

Let's solidify our understanding by running our code with the example values we discussed earlier: a = 2, b = 6, and c = 5. We will trace the execution of the minFlips function step-by-step.

Remember the binary representations:

  • a = 2 (binary: 10)
  • b = 6 (binary: 110)
  • c = 5 (binary: 101)
  1. Initialization: flips = 0
  2. Loop (i = 0):
    • bit_a = (2 >> 0) & 1 = 0
    • bit_b = (6 >> 0) & 1 = 0
    • bit_c = (5 >> 0) & 1 = 1
    • bit_c == 1 is true
    • bit_a | bit_b == 0 (0 | 0 == 0) is true
    • flips += 1 (flips = 1)
  3. Loop (i = 1):
    • bit_a = (2 >> 1) & 1 = 1
    • bit_b = (6 >> 1) & 1 = 1
    • bit_c = (5 >> 1) & 1 = 0
    • bit_c == 1 is false
    • flips += bit_a + bit_b (flips += 1 + 1 = 2)
    • flips = 1 + 2 = 3
  4. Loop (i = 2):
    • bit_a = (2 >> 2) & 1 = 0
    • bit_b = (6 >> 2) & 1 = 1
    • bit_c = (5 >> 2) & 1 = 1
    • bit_c == 1 is true
    • bit_a | bit_b == 0 (0 | 1 == 1) is false
  5. Loop (i = 3 to 31):
    • For the remaining bits, bit_a, bit_b, and bit_c will all be 0.
    • bit_c == 1 will be false.
    • flips += bit_a + bit_b (flips += 0 + 0 = 0)
    • flips remains unchanged.
  6. Return: The function returns flips, which is 2.

As you can see, the code correctly determines that the minimum number of flips required to make (2 OR 6) equal to 5 is 2. This walkthrough should give you a clearer picture of how the bitwise operations and conditional logic work together to solve the problem. This example not only validates our code but also reinforces the underlying concepts.

Now that we have a solid understanding of the solution and its implementation, let's discuss the time and space complexity of our approach. Understanding the efficiency of our solution is crucial for real-world applications.

Complexity Analysis: Evaluating Efficiency

In the realm of algorithms, understanding the time and space complexity is crucial. It helps us evaluate how well our solution performs with varying input sizes. So, let's analyze the complexity of our minFlips function.

Time Complexity

The time complexity of our solution is O(1) (constant time). This might sound surprising, but let's break it down. Our code iterates through 32 bits in the for loop: for i in range(32):. The number 32 is a constant, representing the number of bits in an integer. Regardless of the input values of a, b, and c, the loop always runs 32 times. All the operations inside the loop, such as bitwise shifts (>>), bitwise AND (&), and comparisons, take constant time. Therefore, the entire function's execution time is bounded by a constant, making it O(1).

It's important to note that this constant time complexity is specific to the assumption that we're dealing with 32-bit integers. If we were working with integers of arbitrary size, the time complexity would be O(N), where N is the number of bits in the integer. However, for most practical purposes, integers are represented using a fixed number of bits, making O(1) a valid assessment.

Space Complexity

The space complexity of our solution is also O(1) (constant space). This is because we use a fixed number of variables (flips, bit_a, bit_b, bit_c, i) regardless of the input values. We are not using any auxiliary data structures that scale with the input size. The memory required by our function remains constant, regardless of how large the numbers a, b, and c are. This makes our solution very memory-efficient.

In summary, our minFlips function boasts both O(1) time complexity and O(1) space complexity. This means it's highly efficient, making it suitable for a wide range of applications where performance is critical. This analysis also highlights one of the advantages of working with bitwise operations – they often lead to efficient algorithms.

Optimizations and Alternatives: Exploring Further

While our solution is already quite efficient with O(1) time and space complexity, it's always a good practice to think about potential optimizations and alternative approaches. Could we make it even faster? Are there different ways to achieve the same result? Let's explore.

Potential Optimizations

  1. Early Exit: Although our loop runs a fixed number of times (32), we could potentially add an early exit condition. If a | b is already equal to c, we can return 0 immediately without iterating through any bits. This could save some computation time in certain cases, but the overall time complexity would still remain O(1).

  2. Bit Manipulation Tricks: There might be some clever bit manipulation tricks that could slightly reduce the number of operations within the loop. However, these optimizations would likely be micro-optimizations and might not significantly impact the overall performance, especially given the constant time complexity.

Alternative Approaches

  1. Built-in Functions: Some programming languages provide built-in functions for counting set bits (the number of 1s in a binary representation). We could potentially leverage these functions to simplify our code. However, this might not necessarily improve the performance, as these built-in functions also operate at the bit level.

  2. Lookup Tables: For a limited range of input values, we could precompute the minimum flips required for all possible combinations of a, b, and c and store them in a lookup table. This would allow us to solve the problem in O(1) time by simply looking up the result. However, this approach would require a significant amount of memory for larger input ranges, making it impractical in many scenarios.

When to Consider Alternatives

In general, our current solution is highly optimized for the problem constraints. The constant time and space complexity make it very efficient. Unless there are specific performance bottlenecks or highly specialized use cases, the current approach is likely the most practical and straightforward. Micro-optimizations can sometimes make the code harder to read and maintain without providing significant performance gains.

Thinking about optimizations and alternatives is still a valuable exercise. It helps us explore different problem-solving techniques and deepen our understanding of algorithms. However, it's important to balance optimization efforts with code readability and maintainability. Remember, the best solution is often the one that is clear, concise, and efficient, and our current implementation ticks all those boxes.

Conclusion: Mastering Bit Manipulation

Awesome! We've journeyed through the LeetCode problem 1318,