Given an array nums, you are allowed to choose one element of nums and change it by any value in one move.
Return the minimum difference between the largest and smallest value of nums after perfoming at most 3 moves.
Example 1:
**Input:** nums = [5,3,2,4]
**Output:** 0
**Explanation:** Change the array [5,3,2,4] to [**2**,**2**,2,**2**].
The difference between the maximum and minimum is 2-2 = 0.
Example 2:
**Input:** nums = [1,5,0,10,14]
**Output:** 1
**Explanation:** Change the array [1,5,0,10,14] to [1,**1**,0,**1**,**1**].
The difference between the maximum and minimum is 1-0 = 1.
Example 3:
**Input:** nums = [6,6,0,1,1,4,6]
**Output:** 2
Example 4:
**Input:** nums = [1,5,6,14,15]
**Output:** 1
Constraints:
1 <= nums.length <= 10^5
-10^9 <= nums[i] <= 10^9
ac1: PriorityQueue
The idea is we have 4 options: in a sorted array, remove 1) 0 min, 3 max; 2) 1 min, 2 max; 3) 2 min, 1 max; 4) 3 min, 0 max. Get 4 min + 4 max via 2 PriorityQueue.
class Solution {
public int minDifference(int[] nums) {
if (nums == null || nums.length <= 4) {
return 0;
}
int[] arr = getMinMaxArray(nums);
int min = Integer.MAX_VALUE;
for (int i = 0; i < 4; i++) {
min = Math.min(min, arr[4+i] - arr[i]);
}
return min;
}
private int[] getMinMaxArray(int[] nums) {
PriorityQueue<Integer> min = new PriorityQueue<>();
PriorityQueue<Integer> max = new PriorityQueue<>(Collections.reverseOrder());
for (int n : nums) {
min.offer(n);
if (min.size() > 4) min.poll();
max.offer(n);
if (max.size() > 4) max.poll();
}
return queueToArray(min, max);
}
private int[] queueToArray(PriorityQueue<Integer> min, PriorityQueue<Integer> max) {
int[] arr = new int[8];
int i = 3;
while (!max.isEmpty()) {
arr[i--] = max.poll();
}
i = 4;
while (!min.isEmpty()) {
arr[i++] = min.poll();
}
return arr;
}
}
// O(N) time, O(1) space.
ac2: Sorting
class Solution {
public int minDifference(int[] nums) {
if (nums == null || nums.length <= 4) {
return 0;
}
Arrays.sort(nums);
int min = Integer.MAX_VALUE;
for (int i = 0; i < 4; i++) {
min = Math.min(min, nums[nums.length-4+i] - nums[i]);
}
return min;
}
}
// O(NlogN) time, O(1) space.