We use cookies (including Google cookies) to personalize ads and analyze traffic. By continuing to use our site, you accept our Privacy Policy.

SimplyLeet
TagsCompaniesDisclaimersPrivacyTermsContactAbout Us

Minimum One Bit Operations to Make Integers Zero

Number: 1732

Difficulty: Hard

Paid? No

Companies: Oracle, McKinsey, Expedia

Problem Description

Given an integer n, transform it into 0 using a sequence of bit-flip operations. You may:

  • Change the rightmost (0th) bit.
  • Change the ith bit if the (i-1)th bit is 1 and all bits from the (i-2)th down to 0th are 0. The goal is to determine the minimum number of operations required to transform n into 0.

Key Insights

  • The operations reveal a recursive structure similar to the generation of Gray Codes.
  • The highest set bit in n plays a central role; identifying it helps partition the problem.
  • A recurrence relation can be formulated by considering the contribution of the highest set bit and the transformation required for the remaining bits.
  • The recurrence is mainly based on bit manipulation and exploiting the binary structure of n.
  • The solution achieves an efficient O(log n) time complexity due to operations based on the number of bits.

Space and Time Complexity

Time Complexity: O(log n) – operations scale with the number of bits in n. Space Complexity: O(log n) – recursive calls (or equivalent state storage) depend on the bit-length of n.

Solution

We can observe that if n is a power of 2 (i.e. only one bit is set), the result is a specific number (derived from a pattern similar to full Gray Code sequence up to that bit). For non-power-of-2 numbers, we subtract the highest set bit and use a recurrence based on the flipped ordering.

Procedure:

  1. If n is 0, return 0 (base case).
  2. Find the highest set bit position k.
  3. If n equals 2^k (i.e. n is a pure power of 2), then operation count is (2^(k+1)-1).
  4. Otherwise, apply the recurrence: f(n) = (2^k) + f((2^(k+1)-1) - n)

This recurrence essentially “mirrors” the remaining number against the maximum number with k bits.

Key Data Structures and Techniques:

  • Recursive function to implement the recurrence.
  • Bitwise operators to find the highest set bit and compute power-of-2 values.

Code Solutions

Python:

JavaScript:

C++:

Java:

def minOneBitOps(n):
    # Base case: if n is zero, no operations are needed.
    if n == 0:
        return 0
    # Determine the highest set bit position.
    k = n.bit_length() - 1
    highest_bit = 1 << k
    # If n is exactly a power of 2, the answer follows the pattern: (2^(k+1) - 1)
    if n == highest_bit:
        return (highest_bit << 1) - 1
    # Otherwise, transform n by "mirroring" it relative to the maximum with k bits.
    return highest_bit + minOneBitOps((highest_bit << 1) - 1 - n)

# Example usage:
print(minOneBitOps(3))  # Expected output: 2
print(minOneBitOps(6))  # Expected output: 4
← Back to All Questions