How do you solve Minimum Bit Flips to Convert Number on LeetCode?

Minimum Bit Flips to Convert Number (LeetCode #2220) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Using XOR simplifies the problem of finding differing bits.

What is the brute force solution for Minimum Bit Flips to Convert Number?

The brute force approach for Minimum Bit Flips to Convert Number is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves comparing each bit of the two numbers and counting how many bits differ. This is straightforward but inefficient for larger numbers. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Bit Flips to Convert Number?

Minimum Bit Flips to Convert Number uses the following concepts: Bit Manipulation. The recommended patterns to study are: Bit Manipulation, XOR.

What are the interview tips for Minimum Bit Flips to Convert Number?

Always consider bit manipulation techniques for problems involving binary numbers. Be prepared to explain your thought process clearly, especially when using bitwise operations. Practice converting between binary and decimal to avoid mistakes during interviews.

What are common mistakes when solving Minimum Bit Flips to Convert Number?

Not using XOR to find differing bits, leading to unnecessary complexity. Miscounting the number of flips due to misunderstanding binary representation.

#2220

Minimum Bit Flips to Convert Number

Easy

Bit ManipulationBit ManipulationXOR

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

Using the XOR operation allows us to quickly identify which bits differ between the two numbers. Each differing bit corresponds to a needed flip, making this approach efficient.

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Algorithm

2 steps

1Step 1: Calculate the XOR of start and goal to identify differing bits.
2Step 2: Count the number of 1s in the result of the XOR operation, which indicates how many bits need to be flipped.

solution.py4 lines

1# Full working Python code
2
3def minBitFlips(start, goal):
4    return bin(start ^ goal).count('1')

ℹ

Complexity note: The time complexity is O(n) due to the need to count the bits in the result of the XOR operation, which is efficient compared to the brute-force method.

1Using XOR simplifies the problem of finding differing bits.
2Counting bits can be done efficiently with built-in functions.

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