Sort elements by modulo with K

Given an array, arr[] of integers and an integer K. The task is to sort the elements of the given array in the increasing order of their modulo with K. If two numbers have the same remainder then the smaller number should come first.

Examples: 

Input: arr[] = {10, 3, 2, 6, 12}, K = 4 
Output: 12 2 6 10 3 
{12, 2, 6, 10, 3} is the required sorted order as the modulo 
of these elements with K = 4 is {0, 2, 2, 2, 3}.

Input: arr[] = {3, 4, 5, 10, 11, 1}, K = 3 
Output: 3 1 4 10 5 11 

Approach:

• Create K empty vectors.
• Traverse the array from left to right and update the vectors such that the i^th vector contains the elements that give i as the remainder when divided by K.
• Sort all the vectors separately as all the elements that give the same modulo value with K have to be sorted in ascending.
• Now, starting from the first vector to the last vector and going from left to right in the vectors will give the elements in the required sorted order.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Utility function to print the
// contents of an array
void printArr(int arr[], int n)
{
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
}

// Function to sort the array elements
// based on their modulo with K
void sortWithRemainder(int arr[], int n, int k)
{

    // Create K empty vectors
    vector<int> v[k];

    // Update the vectors such that v[i]
    // will contain all the elements
    // that give remainder as i
    // when divided by k
    for (int i = 0; i < n; i++) {
        v[arr[i] % k].push_back(arr[i]);
    }

    // Sorting all the vectors separately
    for (int i = 0; i < k; i++)
        sort(v[i].begin(), v[i].end());

    // Replacing the elements in arr[] with
    // the required modulo sorted elements
    int j = 0;
    for (int i = 0; i < k; i++) {

        // Add all the elements of the
        // current vector to the array
        for (vector<int>::iterator it = v[i].begin();
            it != v[i].end(); it++) {

            arr[j] = *it;
            j++;
        }
    }

    // Print the sorted array
    printArr(arr, n);
}

// Driver code
int main()
{
    int arr[] = { 10, 7, 2, 6, 12, 3, 33, 46 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int k = 4;

    sortWithRemainder(arr, n, k);

    return 0;
}

Java

// Java implementation of the approach
import java.util.*;

class GFG{

// Utility function to print the
// contents of an array
static void printArr(int[] arr, int n)
{
    for(int i = 0; i < n; i++)
        System.out.print(arr[i] + " ");
}

// Function to sort the array elements
// based on their modulo with K
static void sortWithRemainder(int[] arr,
                              int n, int k)
{

    // Create K empty vectors
    ArrayList<
    ArrayList<Integer>> v = new ArrayList<
                                ArrayList<Integer>>(k);

    for(int i = 0; i < k; i++)
        v.add(new ArrayList<Integer>());

    // Update the vectors such that v[i]
    // will contain all the elements
    // that give remainder as i
    // when divided by k
    for(int i = 0; i < n; i++)
    {
        int t = arr[i] % k;
        v.get(t).add(arr[i]);
    }

    // Sorting all the vectors separately
    for(int i = 0; i < k; i++)
    {
        Collections.sort(v.get(i));
    }

    // Replacing the elements in
    // arr[] with the required
    // modulo sorted elements
    int j = 0;

    for(int i = 0; i < k; i++)
    {

        // Add all the elements of the
        // current vector to the array
        for(int x : v.get(i))
        {
            arr[j] = x;
            j++;
        }
    }

    // Print the sorted array
    printArr(arr, n);
}

// Driver Code
public static void main(String[] args)
{
    int[] arr = { 10, 7, 2, 6,
                  12, 3, 33, 46 };
    int n = arr.length;
    int k = 4;

    sortWithRemainder(arr, n, k);
}
}

// This code is contributed by grand_master

Python3

# Python3 implementation of the approach

# Utility function to print
# contents of an array
def printArr(arr, n):

    for i in range(n):
        print(arr[i], end = ' ')

# Function to sort the array elements
# based on their modulo with K
def sortWithRemainder(arr, n, k):

    # Create K empty vectors
    v = [[] for i in range(k)]

    # Update the vectors such that v[i]
    # will contain all the elements
    # that give remainder as i
    # when divided by k
    for i in range(n):
        v[arr[i] % k].append(arr[i])

    # Sorting all the vectors separately
    for i in range(k):
        v[i].sort()

    # Replacing the elements in arr[] with
    # the required modulo sorted elements
    j = 0

    for i in range(k):

        # Add all the elements of the
        # current vector to the array
        for it in v[i]:
            arr[j] = it
            j += 1

    # Print the sorted array
    printArr(arr, n)

# Driver code
if __name__=='__main__':

    arr = [ 10, 7, 2, 6, 12, 3, 33, 46 ]
    n = len(arr)
    k = 4

    sortWithRemainder(arr, n, k)

# This code is contributed by pratham76

C#

// C# implementation of the
// above approach
using System;
using System.Collections;
class GFG{

// Utility function to print the
// contents of an array
static void printArr(int []arr,
                     int n)
{
  for (int i = 0; i < n; i++)
    Console.Write(arr[i] + " ");
}

// Function to sort the array elements
// based on their modulo with K
static void sortWithRemainder(int []arr,
                              int n, int k)
{
  // Create K empty vectors
  ArrayList []v = new ArrayList[k];

  for(int i = 0; i < k; i++)
  {
    v[i] = new ArrayList();
  }

  // Update the vectors such that v[i]
  // will contain all the elements
  // that give remainder as i
  // when divided by k
  for (int i = 0; i < n; i++)
  {
    v[arr[i] % k].Add(arr[i]);
  }

  // Sorting all the vectors separately
  for (int i = 0; i < k; i++)
  {
    v[i].Sort();
  }

  // Replacing the elements in
  // arr[] with the required
  // modulo sorted elements
  int j = 0;
  for (int i = 0; i < k; i++)
  {
    // Add all the elements of the
    // current vector to the array
    foreach(int x in v[i])
    {
      arr[j] = x;
      j++;
    }
  }

  // Print the sorted array
  printArr(arr, n);
}

// Driver Code
public static void Main(string[] args)
{
  int []arr = {10, 7, 2, 6,
               12, 3, 33, 46};
  int n = arr.Length;
  int k = 4;
  sortWithRemainder(arr, n, k);
}
}

// This code is contributed by rutvik_56

Javascript

<script>
// Javascript implementation of the approach

// Utility function to print the
// contents of an array
function printArr(arr, n)
{
    for (let i = 0; i < n; i++)
        document.write(arr[i] + " ");
}

// Function to sort the array elements
// based on their modulo with K
function sortWithRemainder(arr, n, k)
{

    // Create K empty vectors
    let v = new Array();

    for (let i = 0; i < k; i++)
    {
        v.push([])
    }

    // Update the vectors such that v[i]
    // will contain all the elements
    // that give remainder as i
    // when divided by k
    for (let i = 0; i < n; i++) {
        v[arr[i] % k].push(arr[i]);
    }

    // Sorting all the vectors separately
    for (let i = 0; i < k; i++)
        v[i].sort((a, b) => a - b);

        console.log(v)

    // Replacing the elements in arr[] with
    // the required modulo sorted elements
    let j = 0;
    for (let i = 0; i < k; i++) {

        // Add all the elements of the
        // current vector to the array
        for (let it of v[i]) {

            arr[j] = it;
            j++;
        }
    }

    // Print the sorted array
    printArr(arr, n);
}

// Driver code

let arr = [10, 7, 2, 6, 12, 3, 33, 46];
let n = arr.length;
let k = 4;

sortWithRemainder(arr, n, k);

// This code is contributed by _saurabh_jaiswal
</script>

Output

12 33 2 6 10 46 3 7 

Time Complexity: O(nlogn)
Auxiliary Space: O(k), where k is a given integer.