Group numbers by remainder
Medium
Java
Collections
27 views
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
Then n integers (array elements)
An integer m
Output Format
A map where:
• key → value % m (remainder)
• value → list of numbers having that remainder
• key → value % m (remainder)
• value → list of numbers having that remainder
Example
6 1 2 3 4 5 6 3
0 -> [3, 6] 1 -> [1, 4] 2 -> [2, 5]
Constraints
• n ≥ 0
• m > 0
• Maintain the original order of numbers inside each list
• 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.
All numbers with the same remainder go into the same list.
Step-by-Step Explanation
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.
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.
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.
All numbers with the same remainder go into the same list.
Step-by-Step Explanation
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.
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.
Input / Output Format
Input Format
An integer n (size of array)
Then n integers (array elements)
An integer m
Then n integers (array elements)
An integer m
Output Format
A map where:
• key → value % m (remainder)
• value → list of numbers having that remainder
• 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
• m > 0
• Maintain the original order of numbers inside each list
Examples
Input:
6
1 2 3 4 5 6
3
Output:
0 -> [3, 6]
1 -> [1, 4]
2 -> [2, 5]
Example Solution (Public)
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;}
Official Solution Code
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;}
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