How do you solve Number of Bit Changes to Make Two Integers Equal on LeetCode?

Number of Bit Changes to Make Two Integers Equal (LeetCode #3226) 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: Using XOR helps identify differing bits efficiently.

What is the time complexity of Number of Bit Changes to Make Two Integers Equal?

The optimal solution for Number of Bit Changes to Make Two Integers Equal runs in O(log n) time and uses O(1) space. The complexity is O(log n) because we are only processing the bits of the integers, which is proportional to the number of bits in the integers.

What is the brute force solution for Number of Bit Changes to Make Two Integers Equal?

The brute force approach for Number of Bit Changes to Make Two Integers Equal is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each bit of the binary representation of both integers and counts the number of changes needed. It simply compares the bits of n and k to determine how many bits need to be changed. The optimal Optimal Solution approach improves this to O(log n).

What algorithm or data structure is used to solve Number of Bit Changes to Make Two Integers Equal?

Number of Bit Changes to Make Two Integers Equal uses the following concepts: Bit Manipulation. The recommended patterns to study are: Bit Manipulation, XOR, Counting Bits.

What are the interview tips for Number of Bit Changes to Make Two Integers Equal?

Always check for edge cases, like when n equals k. Understand bit manipulation basics, as they are often used in optimization. Practice counting bits and using bitwise operations.

What are common mistakes when solving Number of Bit Changes to Make Two Integers Equal?

Not checking if n and k are already equal. Overlooking the condition where bits in k cannot be matched by bits in n.

#3226

Number of Bit Changes to Make Two Integers Equal

Easy

Bit ManipulationBit ManipulationXORCounting Bits

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 uses bit manipulation to directly calculate the number of changes needed without iterating through each bit. By using the XOR operation, we can quickly identify which bits differ between n and k.

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Algorithm

5 steps

1Step 1: Calculate the XOR of n and k to find differing bits.
2Step 2: Count the number of 1s in the result of the XOR operation.
3Step 3: Check if the bits in k that are 1 can be matched by bits in n that are 1.
4Step 4: If any bit in k is 1 and the corresponding bit in n is 0, return -1.
5Step 5: Return the count of differing bits.

solution.py11 lines

1# Full working Python code
2
3def count_bit_changes(n, k):
4    if n == k:
5        return 0
6    xor = n ^ k
7    changes = bin(xor).count('1')
8    if (n & k) != k:
9        return -1
10    return changes
11

ℹ

Complexity note: The complexity is O(log n) because we are only processing the bits of the integers, which is proportional to the number of bits in the integers.

1Using XOR helps identify differing bits efficiently.
2Counting bits can be done using built-in functions, simplifying the process.

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