LeetCode 1611 - Minimum One Bit Operations to Make Integers Zero

LeetCode Problem 1611

Difficulty: 🔴 Hard
Topics: Math, Dynamic Programming, Bit Manipulation, Recursion, Memoization

Solution

Problem Understanding

This problem asks us to transform a given integer n into 0 using a specialized set of bit-flipping operations. Each operation allows you to flip either the rightmost bit of the binary representation of n, or the i-th bit if the (i-1)-th bit is set to 1 and all lower bits (i-2 down to 0) are 0. The goal is to determine the minimum number of such operations required to reach 0.

The input n is a non-negative integer up to 10^9, which means its binary representation can be at most 30 bits. The output is an integer representing the fewest operations needed.

Important points are that not all bits can be flipped freely. There is a dependency pattern: a higher bit can only be flipped when a specific combination of lower bits satisfies a condition. This hints at a recursive or formulaic approach rather than brute-force iteration over all numbers.

Key edge cases include n = 0 (already zero, so no operations needed), powers of two (single bits set), and numbers with alternating bits, as these may require non-intuitive sequences of operations.

Approaches

The brute-force approach would attempt to flip bits recursively according to the rules, exploring every possible operation sequence and counting the minimum steps. While correct, this is extremely inefficient because the number of sequences grows exponentially with the number of bits. For n up to 10^9 (30 bits), this is impractical.

The optimal approach relies on a crucial observation: the sequence of minimum operations for n can be expressed recursively in terms of Gray codes. Specifically, the minimum operations to reduce n to 0 is the Gray code of n, which can be computed efficiently with a bitwise XOR formula: n ^ (n >> 1). This works because Gray codes encode the number of single-bit transitions needed to move from 0 to n in a specific order that respects the given bit-flipping constraints.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(2^B)	O(B)	Recursively explore all valid flips; B = number of bits
Optimal	O(1)	O(1)	Compute minimum operations directly using n ^ (n >> 1) formula

Algorithm Walkthrough

1. Observe that the given operation rules mimic the sequence of Gray code transitions. Each valid flip corresponds to moving along the Gray code sequence from 0 to n.
2. The minimum number of operations is equivalent to the Gray code of n, because Gray codes guarantee only one bit changes per step, which exactly matches the operation constraints.
3. Compute the Gray code directly using the formula gray = n ^ (n >> 1). This works because shifting n right by one and XORing with n generates the standard Gray code mapping.
4. Return the resulting integer as the minimum number of operations.

Why it works: Gray codes are constructed so that each consecutive value differs by exactly one bit. This aligns with the constraints of the problem. The XOR formula produces the count of transitions required to reach each bit pattern from 0, ensuring minimal operations while respecting the rules.

Python Solution

class Solution:
    def minimumOneBitOperations(self, n: int) -> int:
        # Compute the Gray code equivalent of n
        return n ^ (n >> 1)

In this Python implementation, we use a single line to compute the Gray code. The right shift (n >> 1) moves all bits one position to the right, and XORing with n generates the Gray code. No recursion, iteration, or extra memory is needed.

Go Solution

func minimumOneBitOperations(n int) int {
    // Compute the Gray code equivalent of n
    return n ^ (n >> 1)
}

The Go version is identical in logic to Python. Go handles integers efficiently, and the bitwise XOR and right shift operators work the same as in Python. No additional data structures or error handling are required since the input range is guaranteed.

Worked Examples

Example 1: n = 3

Binary representation: 11

Gray code: 3 ^ (3 >> 1) = 3 ^ 1 = 2

Minimum operations: 2

Step-by-step operations:

Step	n (binary)	Operation
0	11	Start
1	01	Flip 1st bit (rightmost 0th bit is 1)
2	00	Flip 0th bit

Example 2: n = 6

Binary representation: 110

Gray code: 6 ^ (6 >> 1) = 6 ^ 3 = 5

Step-by-step operations:

Step	n (binary)	Operation
0	110	Start
1	010	Flip 2nd bit
2	011	Flip 0th bit
3	001	Flip 1st bit
4	000	Flip 0th bit

Gray code calculation ensures minimal moves.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(1)	Single XOR and shift operation independent of n's value
Space	O(1)	No extra memory allocation, constant space usage

The algorithm is extremely efficient and scales to the largest possible input (n = 10^9) because it relies on bitwise operations, which are constant-time.

Test Cases

# test cases
assert Solution().minimumOneBitOperations(0) == 0  # already zero
assert Solution().minimumOneBitOperations(1) == 1  # single bit
assert Solution().minimumOneBitOperations(2) == 3  # 10 -> 00 in 3 ops
assert Solution().minimumOneBitOperations(3) == 2  # example 1
assert Solution().minimumOneBitOperations(6) == 5  # example 2
assert Solution().minimumOneBitOperations(10**9) == (10**9 ^ (10**9 >> 1))  # large number

Test	Why
0	Edge case, zero input
1	Single bit set
2	Simple 2-bit number
3	Provided example
6	Provided example
10^9	Large input to ensure no performance issues

Edge Cases

Edge Case 1: n = 0

This is the minimal input and requires no operations. The algorithm correctly returns 0 without any computation issues.

Edge Case 2: n is a power of two

When only one bit is set, e.g., n = 8 (1000 in binary), the Gray code formula still computes the minimal number of transitions efficiently. This case is tricky for naive recursion because it might explore unnecessary flips.

Edge Case 3: Alternating bits

Numbers like n = 10 (1010) require careful bit flipping. The Gray code method handles alternating bit patterns automatically, ensuring the operation sequence respects the flipping rules without explicit recursion or simulation.

This approach generalizes to all numbers within the input constraints, providing a reliable, optimal solution.