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Divide Array Into Arrays With Max Difference - Problem

Imagine you're a team captain dividing players into groups for a tournament! You have an array of n integers (where n is divisible by 3) and need to create exactly n/3 teams, each containing exactly 3 players.

Here's the challenge: within each team of 3, the difference between any two players' scores must not exceed k. This ensures balanced, competitive teams!

Your mission: Return a 2D array where each sub-array represents a valid team of 3 players. If it's impossible to form such balanced teams, return an empty array.

Example: With scores [1,3,3,2,7,3] and k=3, you could form teams like [[1,2,3],[3,3,7]] since within each team, no two scores differ by more than 3.

Input & Output

example_1.py — Basic Valid Case

$ Input: nums = [1,3,3,2,7,3], k = 3

› Output: [[1,2,3],[3,3,7]]

💡 Note: After sorting: [1,2,3,3,3,7]. First group [1,2,3] has max difference 3-1=2 ≤ 3. Second group [3,3,7] has max difference 7-3=4 > 3, so this would actually return [].

example_2.py — Valid Grouping

$ Input: nums = [1,3,4,8,7,9,3,5,1], k = 2

› Output: [[1,1,3],[3,4,5],[7,8,9]]

💡 Note: After sorting: [1,1,3,3,4,5,7,8,9]. Groups: [1,1,3] (diff=2), [3,4,5] (diff=2), [7,8,9] (diff=2). All differences ≤ k=2.

example_3.py — Impossible Case

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

› Output: []

💡 Note: Not enough elements (need multiple of 3). But if we had [1,2,3,10,11,12], the group [3,10,11] would have difference 11-3=8 > k=1, making it impossible.

Constraints

n == nums.length
1 ≤ n ≤ 10⁵
n is a multiple of 3
1 ≤ nums[i] ≤ 10⁵
1 ≤ k ≤ 10⁵

Visualization

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Understanding the Visualization

Line Up Players

Sort all players by skill rating from lowest to highest

Form Teams Greedily

Group every 3 consecutive players into a team

Check Team Balance

Ensure skill gap within each team doesn't exceed k

Validate Solution

If any team fails the balance check, tournament is impossible

Key Takeaway

🎯 Key Insight: Sorting enables greedy grouping - if consecutive elements can't form a valid team, no other arrangement will work either!

Asked in

G Google 45 a Amazon 32 f Meta 28 ⊞ Microsoft 21

The optimal solution sorts the array and greedily groups consecutive elements into teams of 3. This works because if consecutive elements can't form a valid team (difference > k), then no other arrangement with those elements will work. Time complexity: O(n log n) for sorting.

Common Approaches

Approach	Time	Space	Notes
✓ Greedy Sorting + Sequential Grouping (Optimal)	O(n log n)	O(1)	Sort array and group consecutive elements, checking if each group satisfies the constraint
Brute Force (Try All Combinations)	O((n!)/(3!)^(n/3))	O(n)	Generate all possible ways to divide array into groups of 3 and check validity

Greedy Sorting + Sequential Grouping (Optimal) — Algorithm Steps

Sort the entire array in ascending order
Group consecutive elements into teams of 3: [0,1,2], [3,4,5], [6,7,8], etc.
For each group of 3, check if (max - min) ≤ k
If any group fails, return empty array; otherwise return all groups

Visualization

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Step-by-Step Walkthrough

Sort Array

Sort all elements in ascending order

Group Consecutively

Take groups of 3 consecutive elements

Validate Groups

Check if largest - smallest ≤ k for each group

Success or Failure

All groups valid = success, any invalid = impossible

Code -

solution.c — C

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

int compare(const void *a, const void *b) {
    return (*(int*)a - *(int*)b);
}

int** divideArray(int* nums, int numsSize, int k, int* returnSize, int** returnColumnSizes) {
    *returnSize = 0;
    qsort(nums, numsSize, sizeof(int), compare);
    
    int groups = numsSize / 3;
    int** result = (int**)malloc(groups * sizeof(int*));
    *returnColumnSizes = (int*)malloc(groups * sizeof(int));
    
    // Group consecutive elements into teams of 3
    for (int i = 0; i < numsSize; i += 3) {
        // Check if max difference in group is <= k
        if (nums[i+2] - nums[i] > k) {
            free(result);
            free(*returnColumnSizes);
            *returnSize = 0;
            return NULL;
        }
        
        result[*returnSize] = (int*)malloc(3 * sizeof(int));
        result[*returnSize][0] = nums[i];
        result[*returnSize][1] = nums[i+1];
        result[*returnSize][2] = nums[i+2];
        (*returnColumnSizes)[*returnSize] = 3;
        (*returnSize)++;
    }
    
    return result;
}

int main() {
    int nums[] = {1,3,3,2,7,3};
    int numsSize = 6;
    int k = 3;
    int returnSize;
    int* returnColumnSizes;
    
    int** result = divideArray(nums, numsSize, k, &returnSize, &returnColumnSizes);
    
    for (int i = 0; i < returnSize; i++) {
        printf("[");
        for (int j = 0; j < returnColumnSizes[i]; j++) {
            printf("%d", result[i][j]);
            if (j < returnColumnSizes[i] - 1) printf(",");
        }
        printf("] ");
    }
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n log n)

Sorting takes O(n log n), grouping and validation takes O(n)

⚡ Linearithmic

Space Complexity

O(1)

Only constant extra space if we exclude the output array

✓ Linear Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Greedy Sorting + Sequential Grouping (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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