Minimum One Bit Operations to Make Integers Zero

How it works

This problem has a direct connection to Gray codes. A Gray code is a binary sequence where consecutive numbers differ by exactly one bit. The allowed operations in this problem exactly match how Gray codes transition between values. Converting a Gray code back to its binary index tells us how many steps away it is from zero. The conversion formula involves XORing n with all its right-shifted versions.

Steps

1. Initialize res = n.
2. While n is not zero:
   • Right shift n by 1.
   • XOR the result into res.
3. Return res, which represents the position of the original number in the Gray code sequence.

Complexity

• Time complexity: O(log n)
• Space complexity: O(1)

Things to watch for

Misunderstanding the Operations

The two operations allowed (flip the rightmost bit, flip bit i if bit i-1 is 1 and all bits to the right of i-1 are 0) are often misunderstood. Many solutions incorrectly assume you can flip any bit at any time, leading to incorrect operation counts. Carefully trace through small examples like n = 3 to understand the constraints.

Not Recognizing the Gray Code Pattern

This problem has a deep connection to Gray codes, where consecutive numbers differ by exactly one bit. Missing this insight leads to overly complex recursive solutions that are hard to derive and verify. The elegant solution converts the Gray code back to its binary index using iterative XOR.

Integer Overflow in Intermediate Calculations

When computing (k << 1) - 1 or similar expressions, intermediate values can overflow for large inputs near the 32-bit integer limit. In languages like C++ or Java, ensure you handle the bit shifting carefully, especially when k approaches 2^30.

Incorrect Sign Alternation

In iterative solutions that process bits from highest to lowest, the sign must alternate correctly between addition and subtraction. A common bug is initializing the sign incorrectly or flipping it at the wrong point in the loop, leading to off-by-one errors in the final count.

Off-by-One in Bit Position Calculation

Finding the highest set bit requires careful loop conditions. Using while (k < n) instead of while ((k << 1) <= n) or similar variations can cause the algorithm to select the wrong bit position, resulting in incorrect operation counts.

class Solution:
    def minimumOneBitOperations(self, n: int) -> int:
        res = n
        while n:
            n >>= 1
            res ^= n
        return res

class Solution {
public:
    int minimumOneBitOperations(int n) {
        int res = n;
        while (n != 0) {
            n >>= 1;
            res ^= n;
        }
        return res;
    }
};