LeetCode #1471 — MEDIUM

The k Strongest Values in an Array

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

Given an array of integers arr and an integer k.

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

Step 01

Brute Force Baseline

Problem summary: 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

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Two Pointers

Example 1

[1,2,3,4,5]
2

Example 2

[1,1,3,5,5]
2

Example 3

[6,7,11,7,6,8]
5

Step 02

Core Insight

What unlocks the optimal approach

Calculate the centre of the array as defined in the statement.
Use custom sort function to sort values (Strongest first), then slice the first k.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1471: The k Strongest Values in an Array
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;
    }
}

// Accepted solution for LeetCode #1471: The k Strongest Values in an Array
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
}

# Accepted solution for LeetCode #1471: The k Strongest Values in an Array
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]

// Accepted solution for LeetCode #1471: The k Strongest Values in an Array
struct Solution;

impl Solution {
    fn get_strongest(mut arr: Vec<i32>, mut k: i32) -> Vec<i32> {
        arr.sort_unstable();
        let n = arr.len();
        let median = arr[(n - 1) / 2];
        let mut l = 0;
        let mut r = n - 1;
        let mut res = vec![];
        while k > 0 {
            if (arr[l] - median).abs() <= (arr[r] - median).abs() {
                res.push(arr[r]);
                r -= 1;
            } else {
                res.push(arr[l]);
                l += 1;
            }
            k -= 1;
        }
        res
    }
}

#[test]
fn test() {
    let arr = vec![1, 2, 3, 4, 5];
    let k = 2;
    let mut res = vec![5, 1];
    let mut ans = Solution::get_strongest(arr, k);
    res.sort_unstable();
    ans.sort_unstable();
    assert_eq!(ans, res);
    let arr = vec![1, 1, 3, 5, 5];
    let k = 2;
    let mut res = vec![5, 5];
    let mut ans = Solution::get_strongest(arr, k);
    res.sort_unstable();
    ans.sort_unstable();
    assert_eq!(ans, res);
    let arr = vec![6, 7, 11, 7, 6, 8];
    let k = 5;
    let mut res = vec![11, 8, 6, 6, 7];
    let mut ans = Solution::get_strongest(arr, k);
    res.sort_unstable();
    ans.sort_unstable();
    assert_eq!(ans, res);
    let arr = vec![6, -3, 7, 2, 11];
    let k = 3;
    let mut res = vec![-3, 11, 2];
    let mut ans = Solution::get_strongest(arr, k);
    res.sort_unstable();
    ans.sort_unstable();
    assert_eq!(ans, res);
    let arr = vec![-7, 22, 17, 3];
    let k = 2;
    let mut res = vec![22, 17];
    let mut ans = Solution::get_strongest(arr, k);
    res.sort_unstable();
    ans.sort_unstable();
    assert_eq!(ans, res);
}

// Accepted solution for LeetCode #1471: The k Strongest Values in an Array
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);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.