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 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++

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// 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;
}

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Output:

12 33 2 6 10 46 3 7

Time Complexity: O(nlogn)



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An Btech Computer Engineering undergraduate from Aligarh Muslim University

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