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

Minimum Moves to Equal Array Elements III (LeetCode #3736) 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 should reach the maximum value to minimize moves.

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

The optimal solution for Minimum Moves to Equal Array Elements III runs in O(n) time and uses O(1) space. We only traverse the array twice: once to find the max and once to calculate moves.

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

The brute force approach for Minimum Moves to Equal Array Elements III is Brute Force, with O(n²) time complexity and O(1) space complexity. We can calculate the number of moves needed for each possible target value (from the minimum to maximum of the array) and choose the one with the least moves. 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 III?

Minimum Moves to Equal Array Elements III uses the following concepts: Array, Math. The recommended patterns to study are: Hash Map, Array.

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

Explain your thought process clearly. Consider edge cases like all elements being equal. Practice similar problems to build intuition.

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

Forgetting to consider the maximum value. Overcomplicating the calculation of moves.

#3736

Minimum Moves to Equal Array Elements III

Easy

ArrayMathHash 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)

To minimize moves, all elements should be raised to the maximum value in the array, as this requires the least total increase.

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Algorithm

3 steps

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

solution.py3 lines

1def minMoves(nums):
2    max_val = max(nums)
3    return sum(max_val - num for num in nums)

ℹ

Complexity note: We only traverse the array twice: once to find the max and once to calculate moves.

1All elements should reach the maximum value to minimize moves.
2The number of moves is simply the sum of differences from the maximum.

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