The k Strongest Values in an Array - Practice Coding Problems

The k Strongest Values in an Array - Problem

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 k strongest values in the array. Return the answer in any arbitrary order.

The 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).

Input & Output

Example 1 — Basic Case

$ Input: arr = [1,2,3,4,5], k = 2

› Output: [5,1]

💡 Note: Array sorted: [1,2,3,4,5], median = 3 at position (5-1)/2 = 2. Strengths: |1-3|=2, |2-3|=1, |3-3|=0, |4-3|=1, |5-3|=2. Elements 1 and 5 have highest strength 2. Since both have equal strength, we can return either order.

Example 2 — Equal Strengths with Tie-Breaking

$ Input: arr = [1,1,3,5,5], k = 2

› Output: [5,5]

💡 Note: Sorted: [1,1,3,5,5], median = 3. Strengths: |1-3|=2, |1-3|=2, |3-3|=0, |5-3|=2, |5-3|=2. Multiple elements with strength 2, but 5 > 1, so we pick the 5s first.

Example 3 — Small Array

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

› Output: [-3,11,2]

💡 Note: Sorted: [-3,2,6,7,11], median = 6. Strengths: |-3-6|=9, |2-6|=4, |6-6|=0, |7-6|=1, |11-6|=5. Top 3 strongest are -3 (strength 9), 11 (strength 5), and 2 (strength 4).

Constraints

1 ≤ arr.length ≤ 10⁴
1 ≤ k ≤ arr.length
-10⁵ ≤ arr[i] ≤ 10⁵

Visualization

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Understanding the Visualization

Input

Array [6,-3,7,2,11] and k=2

Find Median

Sort to get [-3,2,6,7,11], median = 6

Calculate Strengths

Distance from median: 9,4,0,1,5

Select Top k

Choose 2 strongest: -3(strength 9) and 11(strength 5)

Key Takeaway

🎯 Key Insight: The strongest elements are those farthest from the median, making sorted array extremes the prime candidates

Asked in

F Facebook 15 a Amazon 12 G Google 8

The key insight is that strength is determined by distance from median, and stronger elements tend to be at array extremes after sorting. Best approach uses two pointers after sorting to efficiently select from both ends. Time: O(n log n), Space: O(1).

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force - Calculate All Strengths	O(n log n)	O(n)	Calculate strength for each element, sort by strength, return top k
Two Pointers - Optimal Selection	O(n log n)	O(1)	Sort once, then use two pointers from ends to select k strongest values

Brute Force - Calculate All Strengths — Algorithm Steps

Step 1: Sort array to find median at position (n-1)/2
Step 2: Calculate strength |arr[i] - median| for each element
Step 3: Sort elements by strength (distance first, then value)
Step 4: Return first k elements

Visualization

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Step-by-Step Walkthrough

Find Median

Sort array and find median at position (n-1)/2

Calculate Strengths

For each element, calculate |element - median|

Sort & Select

Sort by strength and pick top k

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>

typedef struct {
    int value;
    int strength;
} Pair;

int compare(const void* a, const void* b) {
    Pair* pa = (Pair*)a;
    Pair* pb = (Pair*)b;
    if (pa->strength != pb->strength) {
        return pb->strength - pa->strength;
    }
    return pb->value - pa->value;
}

int* solution(int* arr, int arrSize, int k, int* returnSize) {
    *returnSize = k;
    
    // Find median
    int* sortedArr = (int*)malloc(arrSize * sizeof(int));
    memcpy(sortedArr, arr, arrSize * sizeof(int));
    
    // Simple bubble sort for median
    for (int i = 0; i < arrSize - 1; i++) {
        for (int j = 0; j < arrSize - i - 1; j++) {
            if (sortedArr[j] > sortedArr[j + 1]) {
                int temp = sortedArr[j];
                sortedArr[j] = sortedArr[j + 1];
                sortedArr[j + 1] = temp;
            }
        }
    }
    
    int median = sortedArr[(arrSize - 1) / 2];
    free(sortedArr);
    
    // Calculate strengths
    Pair* strengths = (Pair*)malloc(arrSize * sizeof(Pair));
    for (int i = 0; i < arrSize; i++) {
        strengths[i].value = arr[i];
        strengths[i].strength = abs(arr[i] - median);
    }
    
    // Sort by strength
    qsort(strengths, arrSize, sizeof(Pair), compare);
    
    // Extract first k elements
    int* result = (int*)malloc(k * sizeof(int));
    for (int i = 0; i < k; i++) {
        result[i] = strengths[i].value;
    }
    
    free(strengths);
    return result;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array manually
    int arr[100];
    int arrSize = 0;
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        arr[arrSize++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int k;
    scanf("%d", &k);
    
    int returnSize;
    int* result = solution(arr, arrSize, k, &returnSize);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        printf("%d", result[i]);
        if (i < returnSize - 1) printf(",");
    }
    printf("]\n");
    
    free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n log n)

Sorting array for median O(n log n) plus sorting by strength O(n log n)

⚡ Linearithmic

Space Complexity

O(n)

Extra space for sorted array and strength pairs

⚡ Linearithmic Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Calculate All Strengths — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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