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Check if it is possible to sort an array with conditional swapping of adjacent allowed

We are given an unsorted array of integers in the range from 0 to n-1. We are allowed to swap adjacent elements in array many number of times but only if the the absolute difference between these element is 1. Check if it is possible to sort the array.If yes then print “yes” else “no”.

Examples:

Input : arr[] = {1, 0, 3, 2}
Output : yes
Explanation:- We can swap arr[0] and arr[1].
Again we swap arr[2] and arr[3]. 
Final arr[] = {0, 1, 2, 3}.

Input : arr[] = {2, 1, 0}
Output : no


Although the problems looks complex at first look, there is a simple solution to it. If we traverse array from left to right and we make sure elements before an index i are sorted before we reach i, we must have maximum of arr[0..i-1] just before i. And this maximum must be either smaller than arr[i] or just one greater than arr[i]. In first case, we simply move ahead. In second case, we swap and move ahead.

Compare the current element with the next element in array.If current element is greater than next element then do following:-
…a) Check if difference between two numbers is 1 then swap it.
…b) else Return false.
If we reach end of array, we return true.

C++

// C++ program to check if we can sort
// an array with adjacent swaps allowed
#include<bits/stdc++.h>
using namespace std;
  
// Returns true if it is possible to sort
// else false/
bool checkForSorting(int arr[], int n)
{
    for (int i=0; i<n-1; i++)
    {
        // We need to do something only if
        // previousl element is greater
        if (arr[i] > arr[i+1])
        {
            if (arr[i] - arr[i+1] == 1)
                swap(arr[i], arr[i+1]);
  
            // If difference is more than
            // one, then not possible
            else
                return false;
        }
    }
    return true;
}
  
// Driver code
int main()
{
    int arr[] = {1,0,3,2};
    int n = sizeof(arr)/sizeof(arr[0]);
    if (checkForSorting(arr, n))
       cout << "Yes";
    else
       cout << "No";
}

Java

class Main
{
    // Returns true if it is possible to sort
    // else false/
    static boolean checkForSorting(int arr[], int n)
    {
        for (int i=0; i<n-1; i++)
        {
            // We need to do something only if
            // previousl element is greater
            if (arr[i] > arr[i+1])
            {
                if (arr[i] - arr[i+1] == 1)
                    {
                        // swapping
                        int temp = arr[i];
                        arr[i] = arr[i+1];
                        arr[i+1] = temp;
                    }
       
                // If difference is more than
                // one, then not possible
                else
                    return false;
            }
        }
        return true;
    }
      
    // Driver function
    public static void main(String args[])
    {
        int arr[] = {1,0,3,2};
        int n = arr.length;
        if (checkForSorting(arr, n))
           System.out.println("Yes");
        else
           System.out.println("No");
    }
}

Python3

# Python 3 program to
# check if we can sort
# an array with adjacent
# swaps allowed
  
# Returns true if it
# is possible to sort
# else false/
def checkForSorting(arr, n):
  
    for i in range(0,n-1):
      
        # We need to do something only if
        # previousl element is greater
        if (arr[i] > arr[i+1]):
          
            if (arr[i] - arr[i+1] == 1):
                arr[i], arr[i+1] = arr[i+1], arr[i]
  
            # If difference is more than
            # one, then not possible
            else:
                return False
  
    return True
  
# Driver code
arr = [1,0,3,2]
n = len(arr)
if (checkForSorting(arr, n)):
    print("Yes")
else:
    print("No")
  
# This code is contributed by
# Smitha Dinesh Semwal

/div>

C#

// C# program to check if we can sort
// an array with adjacent swaps allowed
using System;
  
class GFG
{
    // Returns true if it is 
    // possible to sort else false
    static bool checkForSorting(int []arr, int n)
    {
        for (int i=0; i<n-1; i++)
        {
            // We need to do something only if
            // previousl element is greater
            if (arr[i] > arr[i+1])
            {
                if (arr[i] - arr[i+1] == 1)
                    {
                        // swapping
                        int temp = arr[i];
                        arr[i] = arr[i+1];
                        arr[i+1] = temp;
                    }
      
                // If difference is more than
                // one, then not possible
                else
                    return false;
            }
        }
        return true;
    }
      
    // Driver function
    public static void Main()
    {
        int []arr = {1, 0, 3, 2};
        int n = arr.Length;
        if (checkForSorting(arr, n))
        Console.Write("Yes");
        else
        Console.Write("No");
    }
}
  
// This code is contributed by nitin mittal.

PHP

<?php
// PHP program to check if we can sort
// an array with adjacent swaps allowed
// Returns true if it is possible to sort
// else false
  
function checkForSorting($arr, $n)
{
    $temp = 0;
    for ($i = 0; $i < $n - 1; $i++)
    {
          
        // We need to do something only if
        // previousl element is greater
        if ($arr[$i] > $arr[$i + 1])
        {
            if ($arr[$i] - $arr[$i + 1] == 1)
                {
                        // swapping
                        $temp = $arr[$i];
                        $arr[$i] = $arr[$i + 1];
                        $arr[$i + 1] = $temp;
                }
  
            // If difference is more than
            // one, then not possible
            else
                return false;
        }
    }
    return true;
}
  
    // Driver Code
    $arr = array(1,0,3,2);
    $n = sizeof($arr);
    if (checkForSorting($arr, $n))
        echo "Yes";
    else
        echo "No";
  
// This code is contributed
// by nitin mittal.
?>


Output:

Yes

Time Complexity=O(n)

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



This article is attributed to GeeksforGeeks.org

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