How do you solve Minimum Operations to Equalize Array on LeetCode?

Minimum Operations to Equalize Array (LeetCode #3674) 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: All elements equal means 0 operations.

What is the time complexity of Minimum Operations to Equalize Array?

The optimal solution for Minimum Operations to Equalize Array runs in O(n) time and uses O(1) space. We only need to check each element once, leading to linear time complexity.

What is the brute force solution for Minimum Operations to Equalize Array?

The brute force approach for Minimum Operations to Equalize Array is Brute Force, with O(n²) time complexity and O(1) space complexity. We can try every possible subarray and see how many operations it takes to make all elements equal. This approach is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Operations to Equalize Array?

Minimum Operations to Equalize Array uses the following concepts: Array, Bit Manipulation, Brainteaser. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Operations to Equalize Array?

Clarify the problem statement. Discuss edge cases upfront. Optimize your solution after understanding the brute force.

What are common mistakes when solving Minimum Operations to Equalize Array?

Assuming multiple operations are needed when one suffices. Overlooking the case where all elements are already equal.

#3674

Minimum Operations to Equalize Array

Easy

ArrayBit ManipulationBrainteaserHash MapArray

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)

If all elements are already equal, no operations are needed. Otherwise, one operation can make them equal by replacing the entire array with the AND of all elements.

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Algorithm

3 steps

1Step 1: Check if all elements in nums are equal.
2Step 2: If they are equal, return 0.
3Step 3: If not, return 1, as one operation can equalize the array.

solution.py4 lines

1def min_operations(nums):
2    if all(x == nums[0] for x in nums):
3        return 0
4    return 1

ℹ

Complexity note: We only need to check each element once, leading to linear time complexity.

1All elements equal means 0 operations.
2One operation can make the array equal if not already.

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