702995. Stuffs Division

Easy

Here is the generated Markdown blog post:

Stuffs Division

Slug: stuffs-division5735

Difficulty: Easy

Id: 702995

Topic Tags: Arrays, Mathematical, Data Structures, Algorithms

Company Tags: None

Summary

The problem "Stuffs Division" involves dividing an array of integers into groups based on certain conditions. The key concepts involved are arrays and mathematical operations.

Detailed Explanation

To solve this problem, we need to iterate through the input array and group the elements based on their remainder when divided by a certain number (let's call it k). We can use a simple loop to achieve this. Here's a step-by-step breakdown of the solution:

  1. Initialize an empty array result to store the grouped elements.
  2. Iterate through the input array arr.
  3. For each element x in arr, calculate its remainder when divided by k using the modulo operator (%).
  4. Use this remainder as an index into result and append x to the corresponding group.
  5. Return the result array.

Here's a sample diagram to illustrate the process:

  arr = [3, 6, 9, 12, 15]
  k = 3

  result = [[3], [6, 9], [12, 15]]

  remainder 0: [3]
  remainder 1: [6, 9]
  remainder 2: [12, 15]

The time complexity of this solution is O(n), where n is the length of the input array. The space complexity is also O(n) since we need to store the grouped elements in the result array.

Optimized Solutions

Java

View on GeeksforGeeks
public int[][] stuffsDivision(int[] arr, int k) {
    int n = arr.length;
    int[][] result = new int[(n + k - 1) / k][k];
    for (int i = 0; i < n; i++) {
        result[i % k][i / k] = arr[i];
    }
    return result;
}

DevExCode

Master coding interviews with Leetcode, system design, TechBit, QuickLearn, DevTips, Tech Battles, and career services.

© 2026 DevExCode. All rights reserved.