# Check if an array can be sorted by rearranging odd and even-indexed elements or not

• Last Updated : 25 Apr, 2022

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 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]

## 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

## Javascript



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:

## C++

 // C++ implementation of the // above approach #include using namespace std;   // Function to check if array can // be sorted by given operations bool isEqual(vector&A,vector&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(vectorarr){       // Copy contents     // of the array     vectordupArr(arr.begin(),arr.end());       // Sort the duplicate array     sort(dupArr.begin(),dupArr.end());       vectorevenOrg;     vectorevenSort;       // Traverse the array     for(int i=0;iarr = {3, 5, 1, 2, 6};       cout << sortPoss(arr) << endl;   }   // This code is contributed by shinjanpatra.

## 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))

## Javascript



Output:

True

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

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