LeetCode 1217: Minimum Cost to Move Chips to The Same Position Solution

Master LeetCode problem 1217 (Minimum Cost to Move Chips to The Same Position), a easy challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1217. Minimum Cost to Move Chips to The Same Position

Easy

Array Math Greedy

Problem Explanation

Explanation:

To minimize the cost of moving all chips to the same position, we need to consider the parity of the chips' positions. Since moving a chip by 2 units incurs no cost, we can group the chips based on whether they are at even or odd positions. Calculating the number of chips at even and odd positions, we can determine the minimum cost by finding the smaller group size.

Algorithm:

Count the number of chips at even and odd positions.
Return the minimum of the two counts as the minimum cost.

Time Complexity:

The time complexity of this algorithm is O(n), where n is the number of chips.

Space Complexity:

The space complexity of this algorithm is O(1) as we are using constant extra space.

Solution Code

class Solution {
    public int minCostToMoveChips(int[] position) {
        int evenCount = 0, oddCount = 0;
        
        for (int chip : position) {
            if (chip % 2 == 0) {
                evenCount++;
            } else {
                oddCount++;
            }
        }
        
        return Math.min(evenCount, oddCount);
    }
}

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