Check if an array can be sorted by rearranging odd and even-indexed elements or not
Last Updated :
23 Jul, 2025
Given an array arr[] of size N, the task is to check if it is possible to sort the array using the following operations:
- Swap(arr[i], arr[j]), if i & 1 = 1 and j & 1 = 1.
- Swap(arr[i], arr[j]), if i & 1 = 0 and j & 1 = 0.
Examples:
Input: arr[] = {3, 5, 1, 2, 6}
Output: Yes
Explanation:
Swap(3, 1) --> {1, 5, 3, 2, 6}
Swap(5, 2) --> {1, 2, 3, 5, 6}
Input: arr[] = {3, 1, 5, 2, 6}
Output: No
Naive Approach: The idea is to find the minimum element for the even indexes or odd indexes and swap it from the current element if the index of the current element is even or odd respectively.
- Traverse the array arr[] and perform the following operations:
- If the current index is even, traverse the remaining even indices.
- Find the minimum element present in the even-indexed elements.
- Swap the minimum with the current array element.
- Repeat the above steps for all odd-indexed elements also.
- After completing the above operations, if the array is sorted, then it is possible to sort the array.
- Otherwise, it is not possible to sort the array.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
// Function to check if array
// can be sorted or not
bool isSorted(int arr[], int n)
{
for(int i = 0; i < n - 1; i++)
{
if (arr[i] > arr[i + 1])
return false;
}
return true;
}
// Function to check if given
// array can be sorted or not
bool sortPoss(int arr[], int n)
{
// Traverse the array
for(int i = 0; i < n; i++)
{
int idx = -1;
int minVar = arr[i];
// Traverse remaining elements
// at indices separated by 2
int j = i;
while (j < n)
{
// If current element
// is the minimum
if (arr[j] < minVar)
{
minVar = arr[j];
idx = j;
}
j = j + 2;
}
// If any smaller minimum exists
if (idx != -1)
{
// Swap with current element
swap(arr[i], arr[idx]);
}
}
// If array is sorted
if (isSorted(arr, n))
return true;
// Otherwise
else
return false;
}
// Driver Code
int main()
{
// Given array
int arr[] = { 3, 5, 1, 2, 6 };
int n = sizeof(arr) / sizeof(arr[0]);
if (sortPoss(arr, n))
cout << "True";
else
cout << "False";
return 0;
}
// This code is contributed by ukasp
class GFG{
// Function to check if array
// can be sorted or not
public static boolean isSorted(int arr[], int n)
{
for(int i = 0; i < n - 1; i++)
{
if (arr[i] > arr[i + 1])
return false;
}
return true;
}
// Function to check if given
// array can be sorted or not
public static boolean sortPoss(int arr[], int n)
{
// Traverse the array
for(int i = 0; i < n; i++)
{
int idx = -1;
int minVar = arr[i];
// Traverse remaining elements
// at indices separated by 2
int j = i;
while (j < n)
{
// If current element
// is the minimum
if (arr[j] < minVar)
{
minVar = arr[j];
idx = j;
}
j = j + 2;
}
// If any smaller minimum exists
if (idx != -1)
{
// Swap with current element
int t;
t = arr[i];
arr[i] = arr[idx];
arr[idx] = t;
}
}
// If array is sorted
if (isSorted(arr, n))
return true;
// Otherwise
else
return false;
}
// Driver Code
public static void main(String args[])
{
// Given array
int arr[] = { 3, 5, 1, 2, 6 };
int n = arr.length;
if (sortPoss(arr, n))
System.out.println("True");
else
System.out.println("False");
}
}
// This code is contributed by SoumikMondal
# Function to check if array
# can be sorted or not
def isSorted(arr):
for i in range(len(arr)-1):
if arr[i]>arr[i + 1]:
return False
return True
# Function to check if given
# array can be sorted or not
def sortPoss(arr):
# Traverse the array
for i in range(len(arr)):
idx = -1
minVar = arr[i]
# Traverse remaining elements
# at indices separated by 2
for j in range(i, len(arr), 2):
# If current element
# is the minimum
if arr[j]<minVar:
minVar = arr[j]
idx = j
# If any smaller minimum exists
if idx != -1:
# Swap with current element
arr[i], arr[idx] = arr[idx], arr[i]
# If array is sorted
if isSorted(arr):
return True
# Otherwise
else:
return False
# Driver Code
# Given array
arr = [ 3, 5, 1, 2, 6 ]
print(sortPoss(arr))
using System;
class GFG{
// Function to check if array
// can be sorted or not
public static bool isSorted(int[] arr, int n)
{
for(int i = 0; i < n - 1; i++)
{
if (arr[i] > arr[i + 1])
return false;
}
return true;
}
// Function to check if given
// array can be sorted or not
public static bool sortPoss(int[] arr, int n)
{
// Traverse the array
for(int i = 0; i < n; i++)
{
int idx = -1;
int minVar = arr[i];
// Traverse remaining elements
// at indices separated by 2
int j = i;
while (j < n)
{
// If current element
// is the minimum
if (arr[j] < minVar)
{
minVar = arr[j];
idx = j;
}
j = j + 2;
}
// If any smaller minimum exists
if (idx != -1)
{
// Swap with current element
int t;
t = arr[i];
arr[i] = arr[idx];
arr[idx] = t;
}
}
// If array is sorted
if (isSorted(arr, n))
return true;
// Otherwise
else
return false;
}
// Driver code
static public void Main()
{
// Given array
int[] arr = { 3, 5, 1, 2, 6 };
int n = arr.Length;
if (sortPoss(arr, n))
Console.WriteLine("True");
else
Console.WriteLine("False");
}
}
// This code is contributed by offbeat
<script>
// Function to check if array
// can be sorted or not
function isSorted(arr , n) {
for (i = 0; i < n - 1; i++) {
if (arr[i] > arr[i + 1])
return false;
}
return true;
}
// Function to check if given
// array can be sorted or not
function sortPoss(arr , n) {
// Traverse the array
for (i = 0; i < n; i++) {
var idx = -1;
var minVar = arr[i];
// Traverse remaining elements
// at indices separated by 2
var j = i;
while (j < n) {
// If current element
// is the minimum
if (arr[j] < minVar) {
minVar = arr[j];
idx = j;
}
j = j + 2;
}
// If any smaller minimum exists
if (idx != -1) {
// Swap with current element
var t;
t = arr[i];
arr[i] = arr[idx];
arr[idx] = t;
}
}
// If array is sorted
if (isSorted(arr, n))
return true;
// Otherwise
else
return false;
}
// Driver Code
// Given array
var arr = [ 3, 5, 1, 2, 6 ];
var n = arr.length;
if (sortPoss(arr, n))
document.write("True");
else
document.write("False");
// This code contributed by umadevi9616
</script>
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: The idea is to check if utilize the fact that we can arrange all the even indexed and odd indexed elements the way we want to use the swap operations.
- Initialize an array, say dupArr[], to store the contents of the given array.
- Sort the array dupArr[].
- Check if all even-indexed elements in the original array are the same as the even-indexed elements in dupArr[].
- If found to be true, then sorting is possible. Otherwise, sorting is not possible.
Below is the implementation of the above approach:
C++
// C++ implementation of the
// above approach
#include<bits/stdc++.h>
using namespace std;
// Function to check if array can
// be sorted by given operations
bool isEqual(vector<int>&A,vector<int>&B){
if(A.size() != B.size())return false;
for(int i = 0; i < A.size(); i++){
if(A[i] != B[i])return false;
}
return true;
}
bool sortPoss(vector<int>arr){
// Copy contents
// of the array
vector<int>dupArr(arr.begin(),arr.end());
// Sort the duplicate array
sort(dupArr.begin(),dupArr.end());
vector<int>evenOrg;
vector<int>evenSort;
// Traverse the array
for(int i=0;i<arr.size();i+=2){
// Append even-indexed elements
// of the original array
evenOrg.push_back(arr[i]);
// Append even-indexed elements
// of the duplicate array
evenSort.push_back(dupArr[i]);
}
// Sort the even-indexed elements
sort(evenOrg.begin(),evenOrg.end());
sort(evenSort.begin(),evenSort.end());
// Return true if even-indexed
// elements are identical
return isEqual(evenOrg,evenSort);
}
// Driver Code
int main(){
// Given array
vector<int>arr = {3, 5, 1, 2, 6};
cout << sortPoss(arr) << endl;
}
// This code is contributed by shinjanpatra.
// Java implementation of the
// above approach
import java.io.*;
import java.util.*;
import java.util.ArrayList;
class GFG {
// Function to check if array can
// be sorted by given operations
public static boolean isEqual(ArrayList<Integer> A,ArrayList<Integer> B){
if(A.size() != B.size())return false;
for(int i = 0; i < A.size(); i++){
if(A.get(i) != B.get(i))return false;
}
return true;
}
public static boolean sortPoss(int[] arr){
// Copy contents
// of the array
ArrayList<Integer> dupArr = new ArrayList<Integer>();
for(int i = 0; i < arr.length; i++)
{
dupArr.add(arr[i]);
}
// Sort the duplicate array
Collections.sort(dupArr);
ArrayList<Integer> evenOrg = new ArrayList<Integer>();
ArrayList<Integer> evenSort = new ArrayList<Integer>();
// Traverse the array
for(int i = 0; i < arr.length; i += 2){
// Append even-indexed elements
// of the original array
evenOrg.add(arr[i]);
// Append even-indexed elements
// of the duplicate array
evenSort.add(dupArr.get(i));
}
// Sort the even-indexed elements
Collections.sort(evenOrg);
Collections.sort(evenSort);
// Return true if even-indexed
// elements are identical
return isEqual(evenOrg,evenSort);
}
// Driver Code
public static void main (String[] args)
{
// Given array
int[] arr = {3, 5, 1, 2, 6};
System.out.println(sortPoss(arr));
}
}
// This code is contributed by Aman Kumar.
# Python Program to implement
# the above approach
# Function to check if array can
# be sorted by given operations
def sortPoss(arr):
# Copy contents
# of the array
dupArr = list(arr)
# Sort the duplicate array
dupArr.sort()
evenOrg = []
evenSort = []
# Traverse the array
for i in range(0, len(arr), 2):
# Append even-indexed elements
# of the original array
evenOrg.append(arr[i])
# Append even-indexed elements
# of the duplicate array
evenSort.append(dupArr[i])
# Sort the even-indexed elements
evenOrg.sort()
evenSort.sort()
# Return true if even-indexed
# elements are identical
return evenOrg == evenSort
# Driver Code
# Given array
arr = [3, 5, 1, 2, 6]
print(sortPoss(arr))
// C# code to implement the approach
using System;
using System.Linq;
using System.Collections.Generic;
class GFG {
// Function to check if array can
// be sorted by given operations
public static bool isEqual(List<int> A,List<int> B){
if(A.Count != B.Count) return false;
for(int i = 0; i < A.Count; i++){
if(A[i] != B[i]) return false;
}
return true;
}
public static bool sortPoss(int[] arr){
// Copy contents of the array
List<int> dupArr = arr.ToList();
// Sort the duplicate array
dupArr.Sort();
List<int> evenOrg = new List<int>();
List<int> evenSort = new List<int>();
// Traverse the array
for(int i = 0; i < arr.Length; i += 2){
// Append even-indexed elements
// of the original array
evenOrg.Add(arr[i]);
// Append even-indexed elements
// of the duplicate array
evenSort.Add(dupArr[i]);
}
// Sort the even-indexed elements
evenOrg.Sort();
evenSort.Sort();
// Return true if even-indexed
// elements are identical
return isEqual(evenOrg,evenSort);
}
// Driver Code
public static void Main(string[] args)
{
// Given array
int[] arr = {3, 5, 1, 2, 6};
Console.WriteLine(sortPoss(arr));
}
}
// This code is contributed by phasing17
<script>
// JavaScript Program to implement
// the above approach
// Function to check if array can
// be sorted by given operations
function isEqual(A,B){
if(A.length != B.length)return false;
for(let i = 0; i < A.length; i++){
if(A[i] != B[i])return false;
}
return true;
}
function sortPoss(arr){
// Copy contents
// of the array
let dupArr = arr.slice();
// Sort the duplicate array
dupArr.sort()
let evenOrg = []
let evenSort = []
// Traverse the array
for(let i=0;i<arr.length;i+=2){
// Append even-indexed elements
// of the original array
evenOrg.push(arr[i])
// Append even-indexed elements
// of the duplicate array
evenSort.push(dupArr[i])
}
// Sort the even-indexed elements
evenOrg.sort()
evenSort.sort()
// Return true if even-indexed
// elements are identical
return isEqual(evenOrg,evenSort);
}
// Driver Code
// Given array
let arr = [3, 5, 1, 2, 6]
document.write(sortPoss(arr),"</br>")
// This code is contributed by shinjanpatra.
</script>
Time Complexity: O(N*log(N))
Auxiliary Space: O(N)
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