The k Strongest Values in an Array

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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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

✓ Brute Force - Calculate All Strengths

⏱️ Time: O(n log n) Space: O(n)

Find the median, then calculate the strength of each element as |element - median|. Create pairs of (element, strength) and sort by strength rules. Finally, extract the first k elements.

Two Pointers - Optimal Selection

⏱️ Time: O(n log n) Space: O(1)

After finding the median, sort the array and use two pointers from both ends. Since stronger elements have larger distances from median, they tend to be at the extremes. Compare elements at both ends and pick the stronger one.

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 (pb->strength != pa->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));
    for (int i = 0; i < arrSize; i++) {
        sortedArr[i] = arr[i];
    }
    
    // Simple bubble sort
    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 and create pairs
    Pair* strengths = (Pair*)malloc(arrSize * sizeof(Pair));
    for (int i = 0; i < arrSize; i++) {
        int strength = abs(arr[i] - median);
        strengths[i].value = arr[i];
        strengths[i].strength = strength;
    }
    
    // Sort by strength (distance first, then by value descending)
    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;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    int arr[100];
    int arrSize;
    parseArray(line, arr, &arrSize);
    
    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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Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Calculate All Strengths — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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