Sort M elements of given circular array starting from index K
Last Updated :
23 Jul, 2025
Given a circular array arr[] of size N and two integers K and M, the task is to sort M array elements starting from the index K.
Examples:
Input: arr[] = {4, 1, 6, 5, 3}, K = 2, M = 3
Output: 4 1 3 5 6
Explanation: After sorting 3 array elements starting from index 2 modifies arr[] to {4, 1, 3, 5, 6}.
Input: arr[] = {67, 2, 9, 7, 1}, K = 4, M = 3
Output: 2 67 9 7 1
Explanation: After sorting 3 array elements starting from index 4 modifies arr[] to {2, 67, 9, 7, 1}.
Naive Approach: The idea is to swap the adjacent elements in the circular array if the elements of them are not in the correct order. Follow the steps below to solve the given problem:
Below is the implementation of the above approach:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to print the circular array
void printCircularArray(int arr[], int n)
{
// Print the array
for (int i = 0; i < n; i++) {
cout << arr[i] << " ";
}
}
// Function to sort m elements of diven
// circular array starting from index k
void sortCircularArray(int arr[], int n,
int k, int m)
{
// Traverse M elements
for (int i = 0; i < m; i++) {
// Iterate from index k to k + m - 1
for (int j = k; j < k + m - 1; j++) {
// Check if the next element
// in the circular array is
// less than the current element
if (arr[j % n]
> arr[(j + 1) % n]) {
// Swap current element
// with the next element
swap(arr[j % n], arr[(j + 1) % n]);
}
}
}
// Print the circular array
printCircularArray(arr, n);
}
// Driver Code
int main()
{
int arr[] = { 4, 1, 6, 5, 3 };
int K = 2, M = 3;
int N = sizeof(arr) / sizeof(arr[0]);
// Function Call
sortCircularArray(arr, N, K, M);
return 0;
}
// Java program for the above approach
class GFG{
// Function to print the circular array
static void printCircularArray(int arr[], int n)
{
// Print the array
for (int i = 0; i < n; i++) {
System.out.print(arr[i] + " ");
}
}
// Function to sort m elements of diven
// circular array starting from index k
static void sortCircularArray(int arr[], int n,
int k, int m)
{
// Traverse M elements
for (int i = 0; i < m; i++) {
// Iterate from index k to k + m - 1
for (int j = k; j < k + m - 1; j++) {
// Check if the next element
// in the circular array is
// less than the current element
if (arr[j % n]
> arr[(j + 1) % n]) {
// Swap current element
// with the next element
int t = arr[j % n];
arr[j % n] = arr[(j + 1) % n];
arr[(j + 1) % n] = t;
}
}
}
// Print the circular array
printCircularArray(arr, n);
}
// Driver Code
public static void main (String[] args)
{
int[] arr = { 4, 1, 6, 5, 3 };
int K = 2, M = 3;
int N = arr.length;
// Function Call
sortCircularArray(arr, N, K, M);
}
}
// This code is contributed by susmitakundugoaldanga
# Python3 program for the above approach
# Function to print the circular array
def printCircularArray(arr, n):
# Print the array
for i in range(n):
print(arr[i], end = " ")
# Function to sort m elements of diven
# circular array starting from index k
def sortCircularArray(arr, n, k, m):
# Traverse M elements
for i in range(m):
# Iterate from index k to k + m - 1
for j in range(k, k + m - 1):
# Check if the next element
# in the circular array is
# less than the current element
if (arr[j % n] > arr[(j + 1) % n]):
# Swap current element
# with the next element
arr[j % n], arr[(j + 1) % n] = (arr[(j + 1) % n],
arr[j % n])
# Print the circular array
printCircularArray(arr, n)
# Driver Code
if __name__ == "__main__" :
arr = [ 4, 1, 6, 5, 3 ]
K = 2
M = 3
N = len(arr)
# Function Call
sortCircularArray(arr, N, K, M)
# This code is contributed by AnkThon
// C# program for the above approach
using System;
class GFG
{
// Function to print the circular array
static void printCircularArray(int []arr, int n)
{
// Print the array
for (int i = 0; i < n; i++)
{
Console.Write(arr[i] + " ");
}
}
// Function to sort m elements of diven
// circular array starting from index k
static void sortCircularArray(int []arr, int n,
int k, int m)
{
// Traverse M elements
for (int i = 0; i < m; i++)
{
// Iterate from index k to k + m - 1
for (int j = k; j < k + m - 1; j++)
{
// Check if the next element
// in the circular array is
// less than the current element
if (arr[j % n]
> arr[(j + 1) % n]) {
// Swap current element
// with the next element
int t = arr[j % n];
arr[j % n] = arr[(j + 1) % n];
arr[(j + 1) % n] = t;
}
}
}
// Print the circular array
printCircularArray(arr, n);
}
// Driver Code
public static void Main (string[] args)
{
int[] arr = { 4, 1, 6, 5, 3 };
int K = 2, M = 3;
int N = arr.Length;
// Function Call
sortCircularArray(arr, N, K, M);
}
}
// This code is contributed by AnkThon
<script>
// JavaScript program for the above approach
// Function to print the circular array
function printCircularArray(arr, n)
{
// Print the array
for (let i = 0; i < n; i++) {
document.write(arr[i] + " ");
}
}
// Function to sort m elements of diven
// circular array starting from index k
function sortCircularArray(arr, n, k, m)
{
// Traverse M elements
for (let i = 0; i < m; i++) {
// Iterate from index k to k + m - 1
for (let j = k; j < k + m - 1; j++) {
// Check if the next element
// in the circular array is
// less than the current element
if (arr[j % n]
> arr[(j + 1) % n]) {
// Swap current element
// with the next element
let t = arr[j % n];
arr[j % n] = arr[(j + 1) % n];
arr[(j + 1) % n] = t;
}
}
}
// Print the circular array
printCircularArray(arr, n);
}
// Driver Code
let arr = [ 4, 1, 6, 5, 3 ];
let K = 2, M = 3;
let N = arr.length;
// Function Call
sortCircularArray(arr, N, K, M);
// This code is contributed by Surbhi Tyagi.
</script>
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach:
Step-by-step approach:
- Create a new array of size M and copy the M elements starting from index K of the circular array arr[] to the new array.
- Sort the new array.
- Traverse the circular array arr[] from index K to index (K + M - 1) mod N and replace each element with the corresponding element in the sorted new array.
- If K + M is greater than N, traverse the remaining elements of the new array and replace the corresponding elements in the circular array arr[].
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void sortCircularArray(int arr[], int n, int k, int m) {
// Create a new array of size M and copy the
// M elements starting from index K
int* tempArr = new int[m];
for (int i = 0; i < m; i++) {
tempArr[i] = arr[(k + i) % n];
}
// Sort the new array
sort(tempArr, tempArr + m);
// Traverse the circular array and replace the
// corresponding elements with sorted elements
for (int i = 0; i < m; i++) {
arr[(k + i) % n] = tempArr[i];
}
// If K + M is greater than N, traverse the remaining elements of the new array
// and replace the corresponding elements in the circular array arr[]
if (k + m > n) {
for (int i = m; i < n - k; i++) {
arr[(k + i) % n] = tempArr[i - m];
}
}
}
int main() {
int arr[] = {4, 1, 6, 5, 3};
int K = 2, M = 3;
int N = sizeof(arr) / sizeof(arr[0]);
sortCircularArray(arr, N, K, M);
// Print the modified circular array
for (int i = 0; i < N; i++) {
cout << arr[i] << " ";
}
cout << endl;
return 0;
}
import java.util.Arrays;
public class Main {
public static void sortCircularArray(int[] arr, int n,
int k, int m)
{
// Create a new array of size M and copy the
// M elements starting from index K
int[] tempArr = new int[m];
for (int i = 0; i < m; i++) {
tempArr[i] = arr[(k + i) % n];
}
// Sort the new array
Arrays.sort(tempArr);
// Traverse the circular array and replace the
// corresponding elements with sorted elements
for (int i = 0; i < m; i++) {
arr[(k + i) % n] = tempArr[i];
}
// If K + M is greater than N, traverse the
// remaining elements of the new array and replace
// the corresponding elements in the circular array
// arr[]
if (k + m > n) {
for (int i = m; i < n - k; i++) {
arr[(k + i) % n] = tempArr[i - m];
}
}
}
public static void main(String[] args)
{
int[] arr = { 4, 1, 6, 5, 3 };
int K = 2, M = 3;
int N = arr.length;
sortCircularArray(arr, N, K, M);
// Print the modified circular array
for (int i = 0; i < N; i++) {
System.out.print(arr[i] + " ");
}
}
}
def sortCircularArray(arr, n, k, m):
# Create a new array of size M and copy the
# M elements starting from index K
tempArr = [0] * m
for i in range(m):
tempArr[i] = arr[(k + i) % n]
# Sort the new array
tempArr.sort()
# Traverse the circular array and replace the
# corresponding elements with sorted elements
for i in range(m):
arr[(k + i) % n] = tempArr[i]
# If K + M is greater than N, traverse the remaining elements of the new array
# and replace the corresponding elements in the circular array arr[]
if k + m > n:
for i in range(m, n - k):
arr[(k + i) % n] = tempArr[i - m]
if __name__ == '__main__':
arr = [4, 1, 6, 5, 3]
K = 2
M = 3
N = len(arr)
sortCircularArray(arr, N, K, M)
# Print the modified circular array
for i in range(N):
print(arr[i], end=" ")
using System;
public class GFG {
public static void SortCircularArray(int[] arr, int n,
int k, int m)
{
// Create a new array of size M and copy the
// M elements starting from index K
int[] tempArr = new int[m];
for (int i = 0; i < m; i++) {
tempArr[i] = arr[(k + i) % n];
}
// Sort the new array
Array.Sort(tempArr);
// Traverse the circular array and replace the
// corresponding elements with sorted elements
for (int i = 0; i < m; i++) {
arr[(k + i) % n] = tempArr[i];
}
// If K + M is greater than N, traverse the
// remaining elements of the new array and replace
// the corresponding elements in the circular array
// arr[]
if (k + m > n) {
for (int i = m; i < n - k; i++) {
arr[(k + i) % n] = tempArr[i - m];
}
}
}
public static void Main()
{
int[] arr = { 4, 1, 6, 5, 3 };
int K = 2, M = 3;
int N = arr.Length;
SortCircularArray(arr, N, K, M);
// Print the modified circular array
for (int i = 0; i < N; i++) {
Console.Write(arr[i] + " ");
}
Console.WriteLine();
}
}
function sortCircularArray(arr, n, k, m) {
// Create a new array of size M and copy the
// M elements starting from index K
let tempArr = [];
for (let i = 0; i < m; i++) {
tempArr[i] = arr[(k + i) % n];
}
// Sort the new array
tempArr.sort((a, b) => a - b);
// Traverse the circular array and replace the
// corresponding elements with sorted elements
for (let i = 0; i < m; i++) {
arr[(k + i) % n] = tempArr[i];
}
// If K + M is greater than N, traverse the remaining elements of the new array
// and replace the corresponding elements in the circular array arr[]
if (k + m > n) {
for (let i = m; i < n - k; i++) {
arr[(k + i) % n] = tempArr[i - m];
}
}
}
// Main function
let arr = [4, 1, 6, 5, 3];
let K = 2, M = 3;
let N = arr.length;
sortCircularArray(arr, N, K, M);
// Print the modified circular array
for (let i = 0; i < N; i++) {
console.log(arr[i] + " ");
}
Time Complexity: O(M log M), where M is the number of elements to be sorted.
Auxiliary Space: O(M), since we are creating a new array of size M to store the M elements to be sorted.
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