This is a complex bit manipulation problem. You are given an integer n and two allowed operations:
i-th bit if the (i-1)-th bit is 1 and all bits from i-2 down to 0 are 0.
The goal is to find the minimum operations to change n to 0.Companies like Oracle and Bloomberg use this to test deep mathematical insight and pattern recognition. The rules for flipping bits are actually the rules for moving between consecutive numbers in a Gray Code sequence. The problem tests if a candidate can recognize this connection or derive a recursive solution through observation.
The pattern is Recursion with Memoization or Gray Code Math. The number of operations to turn the k-th bit on (with all lower bits being 0) is 2^(k+1) - 1. You can solve this by observing how the highest set bit dictates the number of steps and then recursively handling the remaining bits.
To change n = 3 (binary 11) to 0:
11 becomes 01. (1 step)01 becomes 00. (1 step)
Total = 2 steps.
For larger numbers, the sequence of rules creates a specific "climbing" and "descending" pattern that matches Gray Code properties.Trying to simulate this using BFS is a common trap—the state space is too large. Another mistake is failing to see the recursive sub-problem: to flip the i-th bit, you first have to transform the lower bits into a specific state (100...), which itself requires a known number of operations.
Gray Code is a niche but recurring topic in bit manipulation. If you see rules about flipping bits based on the state of lower bits, search for patterns. Often, the total operations f(n) can be found using the property f(n) = n ^ f(n >> 1), which is a very elegant O(log n) solution.
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