数组的最小偏移量

/**
 * Problem: Minimize the deviation in an array using Java Streams
 * Approach:
 * 1. Transform odd numbers to even by multiplying by 2 and find the minimum element.
 * 2. Use a TreeSet to keep sorted elements and find the current deviation.
 * 3. Continuously update the max element in the set by dividing it by 2 until all elements are odd.
 */
import java.util.*;
import java.util.stream.Collectors;

public class Solution {
    public int minimumDeviation(int[] nums) {
        TreeSet<Integer> set = new TreeSet<>();

        // Preprocess the array to handle all numbers and create a sorted set
        Arrays.stream(nums)
              .map(num -> (num % 2 == 1) ? num * 2 : num) // Make odd numbers even
              .forEach(set::add);

        int minDeviation = Integer.MAX_VALUE;

        while (true) {
            int max = set.last();
            int min = set.first();
            minDeviation = Math.min(minDeviation, max - min);

            if (max % 2 == 1) break; // Stop when max is odd

            set.remove(max);
            set.add(max / 2);
        }

        return minDeviation;
    }
}