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

17 Nov, 2021
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 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++
Java
Python3
C#
PHP
Javascript
// 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 codeint main(){    int arr[] = {1,0,3,2};    int n = sizeof(arr)/sizeof(arr[0]);    if (checkForSorting(arr, n))       cout << "Yes";    else       cout << "No";}
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");    }}
# 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 codearr = [1,0,3,2]n = len(arr)if (checkForSorting(arr, n)):    print("Yes")else:    print("No") # This code is contributed by# Smitha Dinesh Semwal
// C# program to check if we can sort// an array with adjacent swaps allowedusing 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 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.?>
<script> // JavaScript 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){    let temp = 0;    for (let 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    let arr = new Array(1,0,3,2);    let n = arr.length;    if (checkForSorting(arr, n))        document.write("Yes");    else        document.write("No"); // This code is contributed// by nitin gfgking </script>
Output:  
Yes
Time Complexity=O(n)
Auxiliary Space=O(1)This article is contributed by Roshni Agarwal. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
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