How do you solve Minimum Operations to Make Array Equal on LeetCode?

Minimum Operations to Make Array Equal (LeetCode #1551) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(1) time complexity and O(1) space complexity. Key insight: The sum of the array must be evenly divisible by n for all elements to be equal.

What is the time complexity of Minimum Operations to Make Array Equal?

The optimal solution for Minimum Operations to Make Array Equal runs in O(1) time and uses O(1) space. The time complexity is O(1) because we are performing a constant number of arithmetic operations regardless of the size of n.

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

The brute force approach for Minimum Operations to Make Array Equal is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves repeatedly adjusting the elements of the array until they all become equal. This method is straightforward but inefficient, as it requires multiple operations to balance the array. The optimal Optimal Solution approach improves this to O(1).

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

Minimum Operations to Make Array Equal uses the following concepts: Math. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Operations to Make Array Equal?

Always consider if there's a mathematical shortcut to avoid brute-force solutions. Clarify the problem constraints and guarantees before diving into coding. Practice explaining your thought process clearly, as this can help you catch mistakes.

What are common mistakes when solving Minimum Operations to Make Array Equal?

Students often try to simulate each operation without realizing the mathematical properties of the problem. Forgetting to account for the fact that each operation affects two elements.

#1551

Minimum Operations to Make Array Equal

Medium

MathHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(1)Space O(1)

The optimal solution leverages the mathematical properties of the array to directly calculate the number of operations needed without simulating each operation. This drastically reduces the time complexity.

⚙️

Algorithm

3 steps

1Step 1: Calculate the total sum of the array, which is n * n (since arr[i] = (2 * i) + 1).
2Step 2: Determine the target value, which is the average of the array: target = sum(arr) / n.
3Step 3: Calculate the number of operations needed by summing the differences of elements greater than the target divided by 2 (since each operation balances two elements).

solution.py2 lines

1def minOperations(n):
2    return (n // 2) * (n // 2) if n % 2 == 0 else (n // 2) * (n // 2 + 1)

ℹ

Complexity note: The time complexity is O(1) because we are performing a constant number of arithmetic operations regardless of the size of n.

1The sum of the array must be evenly divisible by n for all elements to be equal.
2Each operation adjusts two elements, so the total number of operations is half the total difference.

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