This approach utilizes a max-heap to efficiently retrieve the largest element while adjusting it within the array to reduce deviation.
1. First, we convert all numbers to even (since an odd number can only become larger when even via multiplication). This is done by multiplying odd numbers by 2.
2. We utilize a max-heap where each iteration involves reducing the maximum number (if even) to reduce the deviation.
3. Throughout this process, track the minimum seen number to calculate and update the minimum deviation at every stage.
Time Complexity: O(n log n) due to sorting operations on each iteration.
Space Complexity: O(n) for the heap structure representation.
1import java.util.*;
2
3public class MinimumDeviation {
4 public int minimumDeviation(int[] nums) {
5 PriorityQueue<Integer> maxHeap = new PriorityQueue<>(Collections.reverseOrder());
6 int minVal = Integer.MAX_VALUE;
7
8 for (int num : nums) {
9 if (num % 2 == 1) num *= 2;
10 minVal = Math.min(minVal, num);
11 maxHeap.add(num);
12 }
13
14 int minDevi = Integer.MAX_VALUE;
15
16 while (true) {
17 int maxVal = maxHeap.poll();
18 minDevi = Math.min(minDevi, maxVal - minVal);
19 if (maxVal % 2 == 1) break;
20 maxVal /= 2;
21 minVal = Math.min(minVal, maxVal);
22 maxHeap.add(maxVal);
23 }
24
25 return minDevi;
26 }
27}This Java implementation also relies on a reverse-priority queue to manage and mutate the largest current element, adjusting it to drive the deviation towards a minimum value.
We loop until the largest element to process is an odd number, making it an exit condition.
In this alternative approach, the strategy focuses on an incremental adjustment instead of heap reliance. It involves directly mutating and managing the adjustment in a sorted sequence approach to target deviation reduction without standard heap operations.
1. First, transform any odd number by doubling it. Track the preliminary minimum number.
2. Sort the array and iteratively adjust the maximum through division by halving until reaching an odd number.
The minimum deviation is kept through direct comparison evaluations across the deviations achieved after each max adjustment.
Time Complexity: O(n^2 log n) due to multiple sorts.
Space Complexity: O(1) or O(n) considering in-place sorts might not require extra space.
1#include <iostream>
2#include <vector>
3#include <algorithm>
4
5using namespace std;
6
7int minimumDeviation(vector<int>& nums) {
8 for (auto &num : nums)
9 if (num % 2 == 1) num *= 2;
10
11 sort(nums.begin(), nums.end());
12 int minDeviation = nums.back() - nums.front();
13
14 while (nums.back() % 2 == 0) {
15 nums.back() /= 2;
16 sort(nums.begin(), nums.end());
17 minDeviation = min(minDeviation, nums.back() - nums.front());
18 }
19
20 return minDeviation;
21}This C++ method establishes deviation reduction by direct manipulation within a sorting framework, enhancing deviation tracking without reliance on a priority queue.
It leverages basic direct list sorting after direct maximal element alterations.