Sort elements of array whose modulo with K yields P
Last Updated :
08 Sep, 2022
Given an array of integers and a number K. The task is to sort only those elements of the array which yields remainder P upon division by K . Sorting must be done at their relative positions only without affecting any other elements.
Examples:
Input : arr[] = {10, 3, 2, 6, 12}, K = 4, P = 2
Output : 2 3 6 10 12
Input : arr[] = {3, 4, 5, 10, 11, 1}, K = 3, P = 1
Output : 3 1 5 4 11 10
Approach:
- Initialise two empty vectors.
- Traverse the array, from left to right and check modulo of each element with K.
- In first vector, insert the index of all elements which yields remainder P.
- In second vector, insert the elements which yields remainder P.
- Sort the second vector.
- Now, we have the index of all required elements and also all of the required elements in sorted order.
- So, insert the elements of the second vector into the array at the indices present in first vector one by one.
Below is the implementation of the above approach:
C++
// C++ program for sorting array elements
// whose modulo with K yields P
#include <bits/stdc++.h>
using namespace std;
// Function to sort elements
// whose modulo with K yields P
void sortWithRemainderP(int arr[], int n, int k, int p)
{
// initialise two vectors
vector<int> v1, v2;
for (int i = 0; i < n; i++) {
if (arr[i] % k == p) {
// first vector contains indices of
// required element
v1.push_back(i);
// second vector contains
// required elements
v2.push_back(arr[i]);
}
}
// sorting the elements in second vector
sort(v2.begin(), v2.end());
// replacing the elements whose modulo with K yields P
// with the sorted elements
for (int i = 0; i < v1.size(); i++)
arr[v1[i]] = v2[i];
// printing the new sorted array elements
for (int i = 0; i < n; i++)
cout << arr[i] << " ";
}
// Driver code
int main()
{
int arr[] = { 8, 255, 16, 2, 4, 0 };
int n = sizeof(arr) / sizeof(arr[0]);
int k = 2;
int p = 0;
sortWithRemainderP(arr, n, k, p);
return 0;
}
// Java program for sorting array elements
// whose modulo with K yields P
import java.util.*;
class GFG
{
// Function to sort elements
// whose modulo with K yields P
static void sortWithRemainderP(int arr[], int n, int k, int p)
{
// initialise two vectors
Vector<Integer> v1 = new Vector<Integer>();
Vector<Integer> v2 = new Vector<Integer>();
for (int i = 0; i < n; i++)
{
if (arr[i] % k == p)
{
// first vector contains indices of
// required element
v1.add(i);
// second vector contains
// required elements
v2.add(arr[i]);
}
}
// sorting the elements in second vector
Collections.sort(v2);
// replacing the elements whose modulo with K yields P
// with the sorted elements
for (int i = 0; i < v1.size(); i++)
arr[v1.get(i)] = v2.get(i);
// printing the new sorted array elements
for (int i = 0; i < n; i++)
System.out.print(arr[i]+" ");
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 8, 255, 16, 2, 4, 0 };
int n = arr.length;
int k = 2;
int p = 0;
sortWithRemainderP(arr, n, k, p);
}
}
// This code is contributed by 29AjayKumar
# Python 3 program for sorting array
# elements whose modulo with K yields P
# Function to sort elements whose modulo
# with K yields P
def sortWithRemainderP(arr, n, k, p):
# initialise two vectors
v1 = []
v2 = []
for i in range(0, n, 1):
if (arr[i] % k == p):
# first vector contains indices
# of required element
v1.append(i)
# second vector contains
# required elements
v2.append(arr[i])
# sorting the elements in second vector
v2.sort(reverse = False)
# replacing the elements whose modulo
# with K yields P with the sorted elements
for i in range(0, len(v1), 1):
arr[v1[i]] = v2[i]
# printing the new sorted array elements
for i in range(0, n, 1):
print(arr[i], end = " ")
# Driver code
if __name__ == '__main__':
arr = [8, 255, 16, 2, 4, 0]
n = len(arr)
k = 2
p = 0
sortWithRemainderP(arr, n, k, p)
# This code is contributed by
# Sahil_Shelangia
// C# program for sorting array elements
// whose modulo with K yields P
using System;
using System.Collections.Generic;
class GFG
{
// Function to sort elements
// whose modulo with K yields P
static void sortWithRemainderP(int []arr, int n,
int k, int p)
{
// initialise two vectors
List<int> v1 = new List<int>();
List<int> v2 = new List<int>();
for (int i = 0; i < n; i++)
{
if (arr[i] % k == p)
{
// first vector contains indices of
// required element
v1.Add(i);
// second vector contains
// required elements
v2.Add(arr[i]);
}
}
// sorting the elements in second vector
v2.Sort();
// replacing the elements whose modulo with
// K yields P with the sorted elements
for (int i = 0; i < v1.Count; i++)
arr[v1[i]] = v2[i];
// printing the new sorted array elements
for (int i = 0; i < n; i++)
Console.Write(arr[i] + " ");
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 8, 255, 16, 2, 4, 0 };
int n = arr.Length;
int k = 2;
int p = 0;
sortWithRemainderP(arr, n, k, p);
}
}
// This code is contributed by PrinciRaj1992
<?php
// PHP program for sorting array elements
// whose modulo with K yields P
// Function to sort elements
// whose modulo with K yields P
function sortWithRemainderP($arr, $n, $k, $p)
{
// initialise two vectors
$v1 = array();
$v2 = array();
for ($i = 0; $i < $n; $i++)
{
if ($arr[$i] % $k == $p)
{
// first vector contains indices of
// required element
array_push($v1, $i);
// second vector contains
// required elements
array_push($v2, $arr[$i]);
}
}
// sorting the elements in second vector
sort($v2);
// replacing the elements whose modulo with K
// yields P with the sorted elements
for ($i = 0; $i < count($v1); $i++)
$arr[$v1[$i]] = $v2[$i];
// printing the new sorted array elements
for ($i = 0; $i < $n; $i++)
echo $arr[$i] . " ";
}
// Driver code
$arr = array( 8, 255, 16, 2, 4, 0 );
$n = count($arr);
$k = 2;
$p = 0;
sortWithRemainderP($arr, $n, $k, $p);
// This code is contributed by mits
?>
<script>
// Javascript program for sorting array elements
// whose modulo with K yields P
// Function to sort elements
// whose modulo with K yields P
function sortWithRemainderP(arr, n, k, p)
{
// initialise two vectors
var v1 = [], v2 = [];
for (var i = 0; i < n; i++) {
if (arr[i] % k == p) {
// first vector contains indices of
// required element
v1.push(i);
// second vector contains
// required elements
v2.push(arr[i]);
}
}
// sorting the elements in second vector
v2.sort((a,b)=> a-b)
// replacing the elements whose modulo with K yields P
// with the sorted elements
for (var i = 0; i < v1.length; i++)
arr[v1[i]] = v2[i];
// printing the new sorted array elements
for (var i = 0; i < n; i++)
document.write( arr[i] + " ");
}
// Driver code
var arr = [8, 255, 16, 2, 4, 0 ];
var n = arr.length;
var k = 2;
var p = 0;
sortWithRemainderP(arr, n, k, p);
</script>
Time Complexity: O(nlogn)
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