Java Program to Group numbers by remainder with Explanation

Java Medium Collections 28 views

2 min read 229 words

This problem helps you practice core Java fundamentals in a practical way. It builds intuition around remainder, list, group. Let’s break it down step by step so you can implement it confidently.

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Problem Statement

Task: group numbers by (value % m) and return a map remainder -> list of numbers.

Input Format

An integer n (size of array)
Then n integers (array elements)
An integer m

Output Format

A map where:

• key → value % m (remainder)

• value → list of numbers having that remainder

Constraints

• n ≥ 0

• m > 0

• Maintain the original order of numbers inside each list

Concept Explanation

Each number is grouped based on its remainder when divided by m.
All numbers with the same remainder go into the same list.

Step-by-Step Logic

1.Read integer n and the array elements.

2.Read integer m.

3.Create an empty Map.

4.Loop through each number in the array:

• Compute rem = number % m.

5.If rem is not already a key in the map:

• Create a new empty list and put it in the map.

6.Add the current number to the list for key rem.

7.Continue for all numbers.

8.Return or print the map.

Code Solution

This explanation is written for learning purposes and to help beginners understand the concept clearly.

static java.util.Map<Integer,java.util.List<Integer>> groupByRemainder(int[] a,int m){java.util.HashMap<Integer,java.util.List<Integer>> map=new java.util.HashMap<>();for(int x:a){int r=x%m; if(r<0) r+=m; map.computeIfAbsent(r,k->new java.util.ArrayList<>()).add(x);}return map;}

Output Example

Input:

6 1 2 3 4 5 6 3

Output:

0 -> [3, 6] 1 -> [1, 4] 2 -> [2, 5]

Common Mistakes

- Misreading input/output format.
- Not handling constraints and edge cases.
- Off-by-one errors in loops.
- Forgetting to reset variables between test cases (if any).

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