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Medium Array

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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)
        {
            list.add(i);
        }
        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;
    }
}