2020-06-06

## 1471. The k Strongest Values in an Array

### Question:

Given an array of integers `arr` and an integer `k`.

A value `arr[i]` is said to be stronger than a value `arr[j]` if `|arr[i] - m| > |arr[j] - m|` where `m` is the median of the array.
If `|arr[i] - m| == |arr[j] - m|`, then `arr[i]` is said to be stronger than `arr[j]` if `arr[i] > arr[j]`.

Return a list of the strongest `k` values in the array. return the answer in any arbitrary order.

Median is the middle value in an ordered integer list. More formally, if the length of the list is n, the median is the element in position `((n - 1) / 2)` in the sorted list (0-indexed).

• For `arr = [6, -3, 7, 2, 11]``n = 5` and the median is obtained by sorting the array `arr = [-3, 2, 6, 7, 11]` and the median is `arr[m]` where `m = ((5 - 1) / 2) = 2`. The median is `6`.
• For `arr = [-7, 22, 17, 3]``n = 4` and the median is obtained by sorting the array `arr = [-7, 3, 17, 22]` and the median is `arr[m]` where `m = ((4 - 1) / 2) = 1`. The median is `3`.

#### Example 1:

``````Input: arr = [1,2,3,4,5], k = 2
Output: [5,1]
Explanation: Median is 3, the elements of the array sorted by the strongest are [5,1,4,2,3]. The strongest 2 elements are [5, 1]. [1, 5] is also accepted answer.
Please note that although |5 - 3| == |1 - 3| but 5 is stronger than 1 because 5 > 1.
``````

#### Example 2:

``````Input: arr = [1,1,3,5,5], k = 2
Output: [5,5]
Explanation: Median is 3, the elements of the array sorted by the strongest are [5,5,1,1,3]. The strongest 2 elements are [5, 5].
``````

#### Example 3:

``````Input: arr = [6,7,11,7,6,8], k = 5
Output: [11,8,6,6,7]
Explanation: Median is 7, the elements of the array sorted by the strongest are [11,8,6,6,7,7].
Any permutation of [11,8,6,6,7] is accepted.
``````

#### Example 4:

``````Input: arr = [6,-3,7,2,11], k = 3
Output: [-3,11,2]
``````

#### Example 5:

``````Input: arr = [-7,22,17,3], k = 2
Output: [22,17]
``````

Constraints:

• `1 <= arr.length <= 10^5`
• `-10^5 <= arr[i] <= 10^5`
• `1 <= k <= arr.length`

### Solution:

Sort the array in the ascending order, and use two pointers to add element to the result

``````class Solution {
public int[] getStrongest(int[] arr, int k) {
int[] result = new int[k];
Arrays.sort(arr);
int median = arr[(arr.length - 1) / 2];
int left = 0, right = arr.length - 1;
for (int i = 0; i < k; i++) {
if (Math.abs(arr[left] - median) > Math.abs(arr[right] - median)) {
result[i] = arr[left];
left++;
} else {
result[i] = arr[right];
right--;
}
}
return result;
}
}
``````

Using PrioirtyQueue to solve the question with a comparator.

``````class Solution {
public int[] getStrongest(int[] arr, int k) {
int[] result = new int[k];
Arrays.sort(arr);
int median = arr[(arr.length - 1) / 2];
PriorityQueue<Integer> pq = new PriorityQueue<Integer>(arr.length, new Comparator<Integer>(){
@Override
public int compare(Integer a, Integer b) {
if (Math.abs(a - median) == Math.abs(b - median)) {
return b - a;
} else {
return Math.abs(b - median) - Math.abs(a - median);
}
}
});
for (int i : arr) pq.offer(i);
for (int i = 0; i < k; i++) {
result[i] = pq.poll();
}
return result;
}
}
``````

Similarly, change the array to arraylist, and use a comparator to figure out the result.

``````class Solution {
public int[] getStrongest(int[] arr, int k) {
int[] result = new int[k];
Arrays.sort(arr);
int median = arr[(arr.length - 1) / 2];
List<Integer> list = new ArrayList<>();
for (int i : arr)
{
}
Collections.sort(list, new Comparator<Integer>(){
@Override
public int compare(Integer a, Integer b) {
if (Math.abs(a - median) == Math.abs(b - median)) {
return b - a;
} else {
return Math.abs(b - median) - Math.abs(a - median);
}
}
});
for (int i = 0; i < k; i++) {
result[i] = list.get(i);
}
return result;
}
}
``````