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 ith 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 arrayvoid 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 Kvoid 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 codeint 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 approachimport java.util.*;class GFG{// Utility function to print the// contents of an arraystatic 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 Kstatic 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 Codepublic 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 arraydef printArr(arr, n): for i in range(n): print(arr[i], end = ' ')# Function to sort the array elements# based on their modulo with Kdef 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 codeif __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 arraystatic 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 Kstatic 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 Codepublic 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 arrayfunction 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 Kfunction 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 codelet 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.
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