Problem Description

You are given three positive integers a, b, and c. Your task is to find the minimum number of bit flips needed in a and b to make the result of (a OR b) equal to c.

A bit flip operation means changing a single bit from 1 to 0 or from 0 to 1 in the binary representation of a number.

The bitwise OR operation works by comparing corresponding bits of two numbers. For each bit position, if at least one of the bits is 1, the result bit is 1. If both bits are 0, the result bit is 0.

To solve this problem, you need to examine each bit position in the binary representations of a, b, and c. For each bit position i:

• Extract the i-th bit from a, b, and c (let's call them x, y, and z)
• If z is 0, then both x and y must be 0 to satisfy (x OR y) = z. Count how many flips are needed (if x is 1, flip it; if y is 1, flip it)
• If z is 1, then at least one of x or y must be 1. If both are 0, you only need to flip one of them

The solution iterates through 32 bits (covering the range of typical integers) and accumulates the total number of flips required. When the target bit z is 0, the number of flips needed is x + y (since both must become 0). When z is 1, we need one flip only if both x and y are currently 0.

Intuition

The key insight is that we need to examine the problem bit by bit. Since the OR operation works independently on each bit position, we can solve the problem by checking each bit position separately and counting the minimum flips needed at each position.

For any bit position, there are only a few possible scenarios:

1. If the target bit c[i] is 1, we need at least one of a[i] or b[i] to be 1. If both are already 0, we only need to flip one of them (minimum cost = 1).

2. If the target bit c[i] is 0, we need both a[i] and b[i] to be 0 (since 0 OR 0 = 0). If either or both are 1, we must flip them all to 0.

This leads us to a simple counting strategy:

• When we need the result to be 0: Count how many 1s we have in positions a[i] and b[i]. Each 1 needs to be flipped to 0.
• When we need the result to be 1: Check if we already have at least one 1. If not (both are 0), we need exactly one flip.

The beauty of this approach is that each bit position is independent, so we can simply iterate through all 32 bits, apply our counting logic at each position, and sum up the total flips needed. This gives us an efficient O(32) = O(1) solution in terms of the number of bits we need to check.

Solution Approach

The implementation uses bit manipulation to extract and compare individual bits from the three numbers. Here's how the algorithm works:

1. Initialize a counter: Start with ans = 0 to track the total number of flips needed.

2. Iterate through each bit position: Loop through 32 bits (from position 0 to 31) to cover the entire range of a standard integer.

3. Extract individual bits: For each bit position i:

   • Extract the i-th bit from a: x = a >> i & 1
   • Extract the i-th bit from b: y = b >> i & 1
   • Extract the i-th bit from c: z = c >> i & 1

   The expression num >> i & 1 right-shifts the number by i positions and then masks with 1 to get only the least significant bit.

4. Count flips based on the target bit:

   • If z == 0: We need both x and y to be 0. The number of flips required is x + y because:
     • If x = 1, we need 1 flip to make it 0
     • If y = 1, we need 1 flip to make it 0
     • If both are 1, we need 2 flips total
   • If z == 1: We need at least one of x or y to be 1. We only need to flip if both are 0:
     • int(x == 0 and y == 0) returns 1 if both are 0 (need one flip), otherwise 0

5. Accumulate the result: Add the flips needed for the current bit position to ans.

6. Return the total: After checking all 32 bits, return the accumulated count.

The time complexity is O(1) since we always check exactly 32 bits, and the space complexity is O(1) as we only use a few variables regardless of input size.

Example Walkthrough

Let's walk through a concrete example with a = 2, b = 6, and c = 5.

First, let's look at the binary representations:

• a = 2 → 010 in binary
• b = 6 → 110 in binary
• c = 5 → 101 in binary (our target)

Currently, a OR b = 010 OR 110 = 110 (which is 6), but we need it to equal 101 (which is 5).

Now let's examine each bit position from right to left (positions 0, 1, 2):

Bit position 0 (rightmost bit):

• a[0] = 0, b[0] = 0, c[0] = 1
• Current: 0 OR 0 = 0, but we need 1
• Since both a[0] and b[0] are 0, we need to flip one of them to 1
• Flips needed: 1

Bit position 1 (middle bit):

• a[1] = 1, b[1] = 1, c[1] = 0
• Current: 1 OR 1 = 1, but we need 0
• To get 0, both bits must be 0, so we need to flip both a[1] and b[1]
• Flips needed: 2

Bit position 2 (leftmost bit):

• a[2] = 0, b[2] = 1, c[2] = 1
• Current: 0 OR 1 = 1, which matches what we need
• No flips needed: 0

Total flips = 1 + 2 + 0 = 3

After making these 3 flips:

• We could flip a[0] from 0 to 1, making a = 011 (3 in decimal)
• We flip a[1] from 1 to 0 and b[1] from 1 to 0, making b = 100 (4 in decimal)
• Now a OR b = 011 OR 100 = 111... wait, that's not right!

Let me recalculate. After the flips:

• If we flip b[0] from 0 to 1: b becomes 111
• If we flip a[1] from 1 to 0: a becomes 000
• If we flip b[1] from 1 to 0: b becomes 101
• Final: a = 000, b = 101, so a OR b = 000 OR 101 = 101 = 5 ✓

This matches our target c = 5 with exactly 3 bit flips.

Solution Implementation

Time and Space Complexity

The time complexity is O(log M), where M is the maximum value among the numbers a, b, and c. Since the loop iterates exactly 32 times (checking each bit position in a 32-bit integer), this can also be expressed as O(32) = O(1). However, in a more general sense where we consider arbitrary precision integers, the complexity would be O(log M) as we need to check each bit position up to the most significant bit.

The space complexity is O(1) as the algorithm only uses a constant amount of extra space. The variables ans, i, x, y, and z are the only additional storage used, regardless of the input size.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Handling When Target Bit is 0

A common mistake is thinking that when the target bit c is 0, you only need to flip one bit even if both a and b have 1s at that position. This is wrong because the OR operation requires BOTH bits to be 0 to produce a 0.

Incorrect approach:

if bit_c == 0:
    # WRONG: Only counting one flip maximum
    flip_count += 1 if (bit_a == 1 or bit_b == 1) else 0

Correct approach:

if bit_c == 0:
    # RIGHT: Both bits must be 0, so count each 1 that needs flipping
    flip_count += bit_a + bit_b

2. Not Considering the Full Bit Range

Another pitfall is iterating through fewer than 32 bits, which might miss significant bits in larger numbers. While Python handles arbitrary precision integers, the problem typically assumes 32-bit integers.

Incorrect approach:

# WRONG: May miss bits if numbers are large
while a > 0 or b > 0 or c > 0:
    bit_a = a & 1
    bit_b = b & 1
    bit_c = c & 1
    # ... process bits
    a >>= 1
    b >>= 1
    c >>= 1

Correct approach:

# RIGHT: Always check all 32 bits
for bit_position in range(32):
    bit_a = (a >> bit_position) & 1
    bit_b = (b >> bit_position) & 1
    bit_c = (c >> bit_position) & 1
    # ... process bits

3. Misunderstanding the OR Operation Logic

Some might incorrectly think that when the target bit is 1, you need to make both a and b equal to 1. This wastes flips since OR only requires at least one bit to be 1.

Incorrect approach:

if bit_c == 1:
    # WRONG: Making both bits 1 when only one is needed
    flip_count += (1 - bit_a) + (1 - bit_b)

Correct approach:

if bit_c == 1:
    # RIGHT: Only flip if both are 0 (need exactly one flip)
    flip_count += int(bit_a == 0 and bit_b == 0)

Complete Solution with Pitfall Avoidance:

class Solution:
    def minFlips(self, a: int, b: int, c: int) -> int:
        flip_count = 0
      
        # Always iterate through all 32 bits
        for i in range(32):
            bit_a = (a >> i) & 1
            bit_b = (b >> i) & 1
            bit_c = (c >> i) & 1
          
            if bit_c == 0:
                # Both must be 0 - count all 1s that need flipping
                flip_count += bit_a + bit_b
            else:
                # At least one must be 1 - only flip if both are 0
                flip_count += 1 if (bit_a == 0 and bit_b == 0) else 0
      
        return flip_count