LeetCode-in-Java
3587. Minimum Adjacent Swaps to Alternate Parity
Medium
You are given an array nums of distinct integers.
In one operation, you can swap any two adjacent elements in the array.
An arrangement of the array is considered valid if the parity of adjacent elements alternates, meaning every pair of neighboring elements consists of one even and one odd number.
Return the minimum number of adjacent swaps required to transform nums into any valid arrangement.
If it is impossible to rearrange nums such that no two adjacent elements have the same parity, return -1.
Example 1:
Input: nums = [2,4,6,5,7]
Output: 3
Explanation:
Swapping 5 and 6, the array becomes [2,4,5,6,7]
Swapping 5 and 4, the array becomes [2,5,4,6,7]
Swapping 6 and 7, the array becomes [2,5,4,7,6]. The array is now a valid arrangement. Thus, the answer is 3.
Example 2:
Input: nums = [2,4,5,7]
Output: 1
Explanation:
By swapping 4 and 5, the array becomes [2,5,4,7], which is a valid arrangement. Thus, the answer is 1.
Example 3:
Input: nums = [1,2,3]
Output: 0
Explanation:
The array is already a valid arrangement. Thus, no operations are needed.
Example 4:
Input: nums = [4,5,6,8]
Output: -1
Explanation:
No valid arrangement is possible. Thus, the answer is -1.
Constraints:
1 <= nums.length <= 1051 <= nums[i] <= 109- All elements in
numsare distinct.
Solution
import java.util.ArrayList;
import java.util.List;
public class Solution {
private static int helper(List<Integer> indices) {
int swaps = 0;
for (int i = 0; i < indices.size(); i++) {
swaps += Math.abs(indices.get(i) - 2 * i);
}
return swaps;
}
public int minSwaps(int[] nums) {
List<Integer> evenIndices = new ArrayList<>();
List<Integer> oddIndices = new ArrayList<>();
for (int i = 0; i < nums.length; i++) {
if (nums[i] % 2 == 0) {
evenIndices.add(i);
} else {
oddIndices.add(i);
}
}
int evenCount = evenIndices.size();
int oddCount = oddIndices.size();
if (Math.abs(evenCount - oddCount) > 1) {
return -1;
}
int ans = Integer.MAX_VALUE;
if (evenCount >= oddCount) {
ans = Math.min(ans, helper(evenIndices));
}
if (oddCount >= evenCount) {
ans = Math.min(ans, helper(oddIndices));
}
return ans;
}
}