LeetCode in Kotlin

1217. Minimum Cost to Move Chips to The Same Position

Easy

We have n chips, where the position of the ith chip is position[i].

We need to move all the chips to the same position. In one step, we can change the position of the ith chip from position[i] to:

• position[i] + 2 or position[i] - 2 with cost = 0.
• position[i] + 1 or position[i] - 1 with cost = 1.

Return the minimum cost needed to move all the chips to the same position.

Example 1:

Input: position = [1,2,3]

Output: 1

Explanation: First step: Move the chip at position 3 to position 1 with cost = 0. Second step: Move the chip at position 2 to position 1 with cost = 1. Total cost is 1.

Example 2:

Input: position = [2,2,2,3,3]

Output: 2

Explanation: We can move the two chips at position 3 to position 2. Each move has cost = 1. The total cost = 2.

Example 3:

Input: position = [1,1000000000]

Output: 1

Constraints:

• 1 <= position.length <= 100
• 1 <= position[i] <= 10^9

Solution

class Solution {
    fun minCostToMoveChips(position: IntArray): Int {
        var chipsAtOddPosition = 0
        var chipsAtEvenPosition = 0
        for (j in position) {
            if (j % 2 == 0) {
                chipsAtEvenPosition++
            } else {
                chipsAtOddPosition++
            }
        }
        return Math.min(chipsAtEvenPosition, chipsAtOddPosition)
    }
}

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