LeetCode //C - 1217. Minimum Cost to Move Chips to The Same Position

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

From: LeetCode Link: 1217. Minimum Cost to Move Chips to The Same Position

Solution:

Ideas:

Moving by 2 is free, so chips on odd positions can move to any odd position for free, and even to any even for free. Only moving odd ↔ even costs 1.

Code:

int minCostToMoveChips(int* position, int positionSize) {
int odd = 0, even = 0;

for (int i = 0; i < positionSize; i++) {
if (position[i] % 2 == 0)
even++;
else
odd++;
}

return odd < even ? odd : even;
}