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1471. The k Strongest Values in an Array

Description

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 centre 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.

The centre is the middle value in an ordered integer list. More formally, if the length of the list is n, the centre 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 centre is obtained by sorting the array arr = [-3, 2, 6, 7, 11] and the centre is arr[m] where m = ((5 - 1) / 2) = 2. The centre is 6.
  • For arr = [-7, 22, 17, 3], n = 4 and the centre is obtained by sorting the array arr = [-7, 3, 17, 22] and the centre is arr[m] where m = ((4 - 1) / 2) = 1. The centre is 3.
 
 

 

 

 

Example 1:

Input: arr = [1,2,3,4,5], k = 2
Output: [5,1]
Explanation: Centre 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: Centre 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: Centre 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.

 

Constraints:

  • 1 <= arr.length <= 105
  • -105 <= arr[i] <= 105
  • 1 <= k <= arr.length

Solutions

Solution 1: Sorting

We first sort the array \(\textit{arr}\) and then find the median \(m\) of the array.

Next, we sort the array according to the rules described in the problem, and finally return the first \(k\) elements of the array.

The time complexity is \(O(n \times \log n)\), and the space complexity is \(O(n)\). Here, \(n\) is the length of the array \(\textit{arr}\).

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class Solution:
    def getStrongest(self, arr: List[int], k: int) -> List[int]:
        arr.sort()
        m = arr[(len(arr) - 1) >> 1]
        arr.sort(key=lambda x: (-abs(x - m), -x))
        return arr[:k]

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class Solution {
    public int[] getStrongest(int[] arr, int k) {
        Arrays.sort(arr);
        int m = arr[(arr.length - 1) >> 1];
        List<Integer> nums = new ArrayList<>();
        for (int v : arr) {
            nums.add(v);
        }
        nums.sort((a, b) -> {
            int x = Math.abs(a - m);
            int y = Math.abs(b - m);
            return x == y ? b - a : y - x;
        });
        int[] ans = new int[k];
        for (int i = 0; i < k; ++i) {
            ans[i] = nums.get(i);
        }
        return ans;
    }
}

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class Solution {
public:
    vector<int> getStrongest(vector<int>& arr, int k) {
        sort(arr.begin(), arr.end());
        int m = arr[(arr.size() - 1) >> 1];
        sort(arr.begin(), arr.end(), [&](int a, int b) {
            int x = abs(a - m), y = abs(b - m);
            return x == y ? a > b : x > y;
        });
        vector<int> ans(arr.begin(), arr.begin() + k);
        return ans;
    }
};

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func getStrongest(arr []int, k int) []int {
    sort.Ints(arr)
    m := arr[(len(arr)-1)>>1]
    sort.Slice(arr, func(i, j int) bool {
        x, y := abs(arr[i]-m), abs(arr[j]-m)
        if x == y {
            return arr[i] > arr[j]
        }
        return x > y
    })
    return arr[:k]
}

func abs(x int) int {
    if x < 0 {
        return -x
    }
    return x
}

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function getStrongest(arr: number[], k: number): number[] {
    arr.sort((a, b) => a - b);
    const m = arr[(arr.length - 1) >> 1];
    return arr.sort((a, b) => Math.abs(b - m) - Math.abs(a - m) || b - a).slice(0, k);
}

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