What is the brute force solution for Minimum Number of Operations to Make Array XOR Equal to K?

The brute force approach for Minimum Number of Operations to Make Array XOR Equal to K is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking all possible combinations of flipping bits in the array to see if we can achieve the desired XOR value of k. This method is straightforward but inefficient as it explores too many possibilities. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Number of Operations to Make Array XOR Equal to K?

Minimum Number of Operations to Make Array XOR Equal to K uses the following concepts: Array, Bit Manipulation. The recommended patterns to study are: Bit Manipulation, XOR properties.

What are the interview tips for Minimum Number of Operations to Make Array XOR Equal to K?

Always start by calculating the initial state (like the XOR in this case) before making changes. Think about how bit manipulation can simplify the problem — often it can reduce complexity significantly. Practice explaining your thought process clearly; it shows understanding and can help you catch mistakes.

What are common mistakes when solving Minimum Number of Operations to Make Array XOR Equal to K?

Students often try to brute force without recognizing the efficiency of counting differing bits. Not considering the initial XOR comparison with k can lead to unnecessary calculations.

#2997

Minimum Number of Operations to Make Array XOR Equal to K

Medium

ArrayBit ManipulationBit ManipulationXOR properties

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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)

The optimal solution focuses on calculating the difference between the initial XOR and k. Each differing bit indicates a necessary flip. This approach is efficient because it directly counts the required flips without exploring all combinations.

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Algorithm

3 steps

1Step 1: Calculate the initial XOR of all elements in the array.
2Step 2: Calculate the XOR difference between the initial XOR and k.
3Step 3: Count the number of differing bits in the XOR difference, which gives the minimum number of flips required.

solution.py9 lines

1# Full working Python code
2
3def min_operations_optimal(nums, k):
4    initial_xor = 0
5    for num in nums:
6        initial_xor ^= num
7    xor_diff = initial_xor ^ k
8    return bin(xor_diff).count('1')
9

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Complexity note: The complexity is O(n) because we only make a single pass through the array to calculate the initial XOR and then a constant time operation to count the differing bits.

1The XOR operation is reversible and can be used to determine the bits that need to be flipped.
2Each differing bit between the initial XOR and k directly corresponds to a necessary flip.

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