How do you solve Minimum Moves to Equal Array Elements on LeetCode?

Minimum Moves to Equal Array Elements (LeetCode #453) 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: The problem can be viewed as reducing all elements to the minimum instead of incrementing others.

What is the time complexity of Minimum Moves to Equal Array Elements?

The optimal solution for Minimum Moves to Equal Array Elements runs in O(n) time and uses O(1) space. The time complexity is O(n) because we only need to traverse the array a couple of times: once to find the minimum and once to calculate the moves.

What is the brute force solution for Minimum Moves to Equal Array Elements?

The brute force approach for Minimum Moves to Equal Array Elements is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves repeatedly incrementing all but one element until all elements are equal. This method is straightforward but inefficient, as it requires multiple iterations over the array. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Moves to Equal Array Elements?

Minimum Moves to Equal Array Elements uses the following concepts: Array, Math. The recommended patterns to study are: Array, Mathematical Operations.

What are the interview tips for Minimum Moves to Equal Array Elements?

Always consider both incrementing and decrementing perspectives. Clarify the problem statement to ensure you understand the operations allowed. Practice similar problems to recognize patterns in array manipulation.

What are common mistakes when solving Minimum Moves to Equal Array Elements?

Not recognizing that incrementing n-1 elements is equivalent to reducing one element. Overcomplicating the problem by trying to simulate each move.

#453

Minimum Moves to Equal Array Elements

Medium

ArrayMathArrayMathematical Operations

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)

Instead of incrementing elements, we can think of the problem as reducing all elements to the minimum element in the array. The number of moves required will be the difference between each element and the minimum element, summed up.

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Algorithm

3 steps

1Step 1: Find the minimum element in the array.
2Step 2: Calculate the total moves needed by summing the differences between each element and the minimum element.
3Step 3: Return the total moves.

solution.py5 lines

1# Full working Python code
2
3def minMoves(nums):
4    min_num = min(nums)
5    return sum(num - min_num for num in nums)

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Complexity note: The time complexity is O(n) because we only need to traverse the array a couple of times: once to find the minimum and once to calculate the moves.

1The problem can be viewed as reducing all elements to the minimum instead of incrementing others.
2Understanding the operation's effect is crucial to finding the optimal solution.

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