How do you solve Minimum One Bit Operations to Make Integers Zero on LeetCode?

Minimum One Bit Operations to Make Integers Zero (LeetCode #1611) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(log n) time complexity and O(1) space complexity. Key insight: Understanding bit manipulation is crucial for optimizing operations on integers.

What is the time complexity of Minimum One Bit Operations to Make Integers Zero?

The optimal solution for Minimum One Bit Operations to Make Integers Zero runs in O(log n) time and uses O(1) space. The time complexity is O(log n) because we only need to process each bit of n, which is proportional to the number of bits in the binary representation of n.

What is the brute force solution for Minimum One Bit Operations to Make Integers Zero?

The brute force approach for Minimum One Bit Operations to Make Integers Zero is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves repeatedly applying the allowed operations to reduce the integer to zero. We will check each bit and perform operations until all bits are cleared. The optimal Optimal Solution approach improves this to O(log n).

What are the interview tips for Minimum One Bit Operations to Make Integers Zero?

Always explain your thought process clearly; it shows understanding. Practice bit manipulation problems to become comfortable with binary operations. Think about edge cases, such as when n is already zero or a power of two.

What are common mistakes when solving Minimum One Bit Operations to Make Integers Zero?

Not considering the implications of bit positions and how they affect operations. Overlooking the efficiency of operations based on the leftmost set bit.

#1611

Minimum One Bit Operations to Make Integers Zero

Hard

MathDynamic ProgrammingBit ManipulationRecursionMemoizationBit ManipulationDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(log n)
Space	O(1)	O(1)

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Intuition

Time O(log n)Space O(1)

The optimal approach leverages the observation that we can clear bits more efficiently by focusing on the leftmost set bit and counting the necessary operations to clear each bit in sequence.

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Algorithm

5 steps

1Step 1: Initialize a counter for operations.
2Step 2: While n is greater than 0, find the leftmost set bit.
3Step 3: Count how many bits are to the right of this bit (including the bit itself).
4Step 4: Increment the operations counter by the number of bits counted.
5Step 5: Clear the leftmost set bit and repeat until n is zero.

solution.py13 lines

1# Full working Python code
2
3def min_operations(n):
4    operations = 0
5    while n > 0:
6        leftmost_bit = n.bit_length() - 1
7        operations += leftmost_bit + 1
8        n &= ~(1 << leftmost_bit)
9    return operations
10
11# Example usage
12print(min_operations(3))  # Output: 2
13print(min_operations(6))  # Output: 4

ℹ

Complexity note: The time complexity is O(log n) because we only need to process each bit of n, which is proportional to the number of bits in the binary representation of n.

1Understanding bit manipulation is crucial for optimizing operations on integers.
2Recognizing patterns in binary representation can significantly reduce the number of operations needed.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.