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

Last Updated : 11 Jul, 2022

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:-

1.  Check if difference between two numbers is 1 then swap it.
2. 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` `using` `namespace` `std;`   `// Returns true if it is possible to sort` `// else false/` `bool` `checkForSorting(``int` `arr[], ``int` `n)` `{` `    ``for` `(``int` `i=0; 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 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`

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 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

 ` ``\$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.` `?>`

Javascript

 ``

Output

`Yes`

Time Complexity=O(n)
Auxiliary Space=O(1)

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