How do you solve Minimum Cost to Move Chips to The Same Position on LeetCode?

Minimum Cost to Move Chips to The Same Position (LeetCode #1217) 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: Chips can move between even positions at no cost, and between odd positions at no cost.

What is the brute force solution for Minimum Cost to Move Chips to The Same Position?

The brute force approach for Minimum Cost to Move Chips to The Same Position is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute-force approach, we consider moving each chip to every possible position and calculate the total cost for each option. This is straightforward but inefficient, as it requires checking every position for every chip. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Cost to Move Chips to The Same Position?

Minimum Cost to Move Chips to The Same Position uses the following concepts: Array, Math, Greedy. The recommended patterns to study are: Greedy, Counting.

What are the interview tips for Minimum Cost to Move Chips to The Same Position?

Always consider edge cases, such as all chips being in the same position. Explain your thought process clearly; it helps interviewers understand your reasoning. Practice counting and categorizing elements in arrays, as this is a common pattern.

What are common mistakes when solving Minimum Cost to Move Chips to The Same Position?

Forgetting to consider the parity of positions when calculating costs. Not recognizing that moving between even or odd positions incurs no cost.

#1217

Minimum Cost to Move Chips to The Same Position

Easy

ArrayMathGreedyGreedyCounting

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)

The optimal solution leverages the fact that chips can move between even positions at no cost and odd positions at no cost. Thus, we only need to count how many chips are on even and odd positions and move the smaller group to minimize the cost.

⚙️

Algorithm

4 steps

1Step 1: Initialize two counters for even and odd positioned chips.
2Step 2: Iterate through the position array and increment the respective counter based on whether the position is even or odd.
3Step 3: The minimum cost will be the smaller of the two counters (either move all odd chips to even or all even chips to odd).
4Step 4: Return the minimum cost.

solution.py4 lines

1def minCostToMoveChips(position):
2    even_count = sum(1 for pos in position if pos % 2 == 0)
3    odd_count = len(position) - even_count
4    return min(even_count, odd_count)

ℹ

Complexity note: The time complexity is O(n) as we only traverse the list once to count even and odd chips. The space complexity is O(1) since we only use a fixed number of variables.

1Chips can move between even positions at no cost, and between odd positions at no cost.
2The minimum cost to align all chips is determined by the smaller group (even or odd).

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