The "Minimize XOR" interview question typically presents you with two integers, num1 and num2, and your task is to construct a third integer, let's call it X, such that X has a specific number of set bits (1s in its binary representation), and the bitwise XOR operation num1 XOR X is minimized. This problem delves deep into the properties of bitwise operations, especially XOR, and requires a strategic approach to constructing X bit by bit to achieve the minimum possible XOR result. The number of set bits in X is usually a given constraint, adding another layer of complexity to the optimization.
This "Minimize XOR" coding problem is a frequent visitor in interviews at leading tech companies like Microsoft, Amazon, Google, Bloomberg, and Adobe because it's an excellent way to gauge a candidate's understanding of bit manipulation and greedy algorithms. Bitwise operations are fundamental in many low-level optimizations and certain data structures. The problem assesses how effectively a candidate can reason about individual bits and their impact on the overall XOR sum. It requires a thoughtful, step-by-step approach to construct the optimal X, making it a strong indicator of careful algorithmic design and problem-solving skills.
The primary algorithmic pattern for solving the "Minimize XOR" coding problem is a Greedy Algorithm applied to Bit Manipulation. The core idea is to construct the integer X bit by bit, from the most significant bit (MSB) to the least significant bit (LSB). To minimize num1 XOR X, we want to make the resulting bits of num1 XOR X as many zeros as possible, starting from the MSB.
If the i-th bit of num1 is 0, we want the i-th bit of X to be 0 to make 0 XOR 0 = 0.
If the i-th bit of num1 is 1, we want the i-th bit of X to be 1 to make 1 XOR 1 = 0.
However, we are constrained by the total number of set bits required in X. The greedy strategy involves:
num1's set bits with X's set bits first, as this directly reduces the XOR sum. Do this from MSB to LSB.X (to meet the constraint), place them in positions where num1 has a 0, starting from the LSB. This will introduce 1s into the XOR sum, but doing it at LSBs minimizes their impact.X after matching, then you need to prioritize turning off X's bits from LSBs first, as long as it doesn't violate the constraint of matching num1's already considered MSBs.A more concrete greedy strategy usually starts by counting set bits in num2. Let this be set_bits_needed.
Iterate from MSB down to LSB (e.g., bit 30 down to 0):
num1 has a 1 at this bit position AND set_bits_needed > 0: Set this bit in X to 1, and decrement set_bits_needed. This makes the XOR result 0 at this position.
After this pass, if set_bits_needed > 0, it means num1 had fewer set bits than required for X. Now, iterate again from MSB down to LSB:num1 has a 0 at this bit position AND set_bits_needed > 0: Set this bit in X to 1, and decrement set_bits_needed. This introduces a 1 in the XOR result, but we do it at the lowest possible significant position remaining.
If set_bits_needed < 0 (meaning we set too many bits in X), iterate from LSB up to MSB:X has a 1 at this bit position (which we set from num1 having a 0) AND we have set_bits_needed to spare (i.e. we can remove this bit from X): Set this bit in X to 0, increment set_bits_needed.The exact construction depends on the specific interpretation of "number of set bits in X". If it refers to num2's set bits, then we build X to have the same number of set bits as num2.
A straightforward approach for X having k set bits (where k is the number of set bits in num2):
Initialize X = 0.
i-th bit of num1 is set AND we still need to set bits in X (k > 0), set the i-th bit of X and decrement k. This ensures num1 XOR X has a 0 at this bit position.k > 0, it means we still need to set k more bits in X. Iterate from LSB (0) up to MSB (30). If the i-th bit of X is NOT set AND k > 0, set the i-th bit of X and decrement k. This places the remaining required set bits in X at the lowest possible positions, minimizing their impact on the XOR sum.Let num1 = 5 (101 in binary) and num2 = 3 (011 in binary).
The problem statement implies we need to find X with k set bits, where k is the count of set bits in num2.
Number of set bits in num2 (3) is 2. So we need X to have 2 set bits.
Our goal is to minimize num1 XOR X.
Iterate from MSB to LSB (let's consider 3 bits for simplicity):
num1 = 101
X = 000 (initially)
k = 2 (set bits needed in X)
num1 has 1 here. We want X to have 1 here to make XOR 0.
Set bit 2 of X to 1. X = 100. k becomes 1.num1 has 0 here. We don't want X to have 1 here if we can avoid it. Since k > 0, we have a choice. We'll prioritize matching if num1 has a 1.num1 has 1 here. We want X to have 1 here to make XOR 0.
Set bit 0 of X to 1. X = 101. k becomes 0.After this pass, X = 101 (5) and k = 0.
num1 XOR X = 101 XOR 101 = 000 (0).
The minimum XOR is 0.
Let's try another example: num1 = 8 (1000), num2 = 2 (0010).
Number of set bits in num2 (2) is 1. So we need X to have 1 set bit. k = 1.
num1 = 1000
X = 0000
k = 1
Iterate from MSB (bit 3) down to 0:
num1 has 1 here. Set bit 3 of X to 1. X = 1000. k becomes 0.k is now 0, so we don't set any more bits in X based on num1's bits.After this pass, X = 1000 (8) and k = 0.
num1 XOR X = 1000 XOR 1000 = 0000 (0).
Minimum XOR is 0.
Consider num1 = 12 (1100), num2 = 4 (0100).
Number of set bits in num2 (4) is 1. So we need X to have 1 set bit. k = 1.
num1 = 1100
X = 0000
k = 1
Iterate from MSB (bit 3) down to 0:
num1 has 1 here. Set bit 3 of X to 1. X = 1000. k becomes 0.k is 0. Stop.After this pass, X = 1000 (8) and k = 0.
num1 XOR X = 1100 XOR 1000 = 0100 (4).
The minimum XOR is 4.
This greedy approach ensures that we try to make the most significant bits of the XOR result zero first, and if we still need to set bits in X (or unset them), we do so at less significant positions.
A common mistake in the "Minimize XOR" interview question is an unoptimized or incorrect greedy strategy. Candidates might try to set bits in X from LSB to MSB, or fail to prioritize matching num1's set bits first. Another pitfall is incorrectly managing the count of set bits required for X, leading to either too many or too few set bits in the final X. Misunderstanding how XOR behaves bit by bit (0^0=0, 1^1=0, 0^1=1, 1^0=1) can also lead to logical errors. Edge cases, such as when num1 or num2 are zero, or when num2 requires zero set bits for X, need careful consideration.
To prepare for the "Minimize XOR" coding problem, thoroughly understand bit manipulation operations. Practice problems involving counting set bits, setting/clearing specific bits, and understanding the effects of XOR. When approaching such a problem, always think about the binary representations of the numbers involved. Develop a clear, bit-by-bit greedy strategy, working from the most significant bit to the least significant bit, or vice versa, depending on the objective. Work through small examples by hand, tracing the binary changes and the k (set bits needed) count. This will solidify your understanding of how each bit decision impacts the overall XOR sum.