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Check if an array can be sorted by rearranging odd and even-indexed elements or not
  • Last Updated : 22 Apr, 2021

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


Java




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


Python3




# 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))


C#




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


Output: 

True

 

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:

Python3




# 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))


Output: 

True

 

Time Complexity: O(N*log(N))
Auxiliary Space: O(N)

 

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