Minimize XOR - Video Solutions

Problem Overview: Given two integers num1 and num2, construct a number x such that x has the same number of set bits (1s) as num2 while minimizing x XOR num1. The challenge is choosing which bit positions to set so the XOR difference with num1 stays as small as possible.

Approach 1: Greedy Bit Setting Strategy (O(1) time, O(1) space)

This approach relies on a simple observation about XOR: the result is smallest when the most significant bits match. First compute the number of set bits required using popcount(num2). Then iterate through the bits of num1 from the most significant bit to the least significant bit. Whenever a bit in num1 is set, set the same bit in x while you still have remaining bits to place. This keeps the high-value bits aligned and minimizes the XOR contribution. If there are still bits left after this pass, fill them starting from the least significant positions where x currently has 0. This greedy ordering ensures minimal XOR while maintaining the required bit count. The algorithm processes a fixed number of bits (typically 32), so the time complexity is effectively constant.

The strategy works because higher bits contribute more to the XOR value. Matching them first avoids large penalties. Lower bits are used later to satisfy the remaining bit count with minimal impact. This is a classic greedy choice using properties of bit manipulation and bit significance.

Approach 2: Dynamic Bit Position Filling (O(log M) time, O(1) space)

This variation treats the problem as filling bit positions dynamically based on remaining capacity. First count the set bits in num2. Traverse the bits of num1 from the highest bit downward and copy matching set bits into x while decrementing the remaining count. After the first pass, iterate again from the least significant bit upward and place additional bits wherever x currently has 0 until the required count is reached.

Unlike the strict greedy framing, this version separates the matching phase and the filling phase more explicitly. It makes the implementation easier to reason about during interviews because each loop has a single responsibility: first minimize XOR by alignment, then satisfy the remaining bit constraint. The number of iterations depends on the bit length of the integers, which is O(log M) for maximum value M.

Both approaches rely on counting bits and manipulating individual positions, which are common patterns in greedy algorithms and bit manipulation problems.

Recommended for interviews: The greedy bit setting strategy is what most interviewers expect. It shows you understand how XOR magnitude depends on bit significance and how to apply a greedy choice to minimize it. Mentioning the two-phase process (match high bits first, then fill low bits) demonstrates strong control over bit-level reasoning.