#Problem 1217. Minimum Cost to Move Chips to The Same Position


'''
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
'''
def minCostToMoveChips(position):
    #count the number of the odd and even chips
    even_cnt = 0
    odd_cnt = 0
    for i in position:
        #if the position is odd or even, we will add them to the counter 
        if i % 2 == 0:
            even_cnt += 1

        else: 
            odd_cnt += 1

    return min(even_cnt, odd_cnt)

#Time complexity: O(N), where N is the number of i in position, because we just need to loop through them
#Space complexity: O(1), we only need to store even_cnt and odd_cnt


#main function to run the test case
def main():
    print("TESTING MINIMUM COST TO MOVE CHIP INTO THE SAME POSITION")
    p_01 = [1,2,3]
    p_02 = [2,2,2,3,3]
    p_03 = [1,1000000000]

    print(minCostToMoveChips(p_01))
    print(minCostToMoveChips(p_02))
    print(minCostToMoveChips(p_03))
    
    print("END OF TESTING...")
main()