1217. Minimum Cost to Move Chips to The Same Position
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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] + 2orposition[i] - 2withcost = 0.position[i] + 1orposition[i] - 1withcost = 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 <= 1001 <= position[i] <= 10^9
Hints
Hint 1
The first move keeps the parity of the element as it is.
Hint 2
The second move changes the parity of the element.
Hint 3
Since the first move is free, if all the numbers have the same parity, the answer would be zero.
Hint 4
Find the minimum cost to make all the numbers have the same parity.