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# Minimum number of adjacent swaps required to convert a permutation to another permutation by given condition

• Difficulty Level : Easy
• Last Updated : 09 Aug, 2021

Given a permutation P of size N, having values from 1 to N. the task is to find the minimum number of adjacent swaps required such that for all i in the range [1, N], P[i] does not equal i.
Examples:

Input: P = [1, 4, 3, 5, 2]
Output:
Explanation:
Here P = [1, 4, 3, 5, 2] at index 1, 2, 3, 4, 5. As we can see, P = 1 and P = 3. Hence, we need to get rid of this invariant.
Swap 1: Swap index 1 and index 2 => [4, 1, 3, 5, 2]
Swap 2: Swap index 2 and index 3 => [4, 3, 1, 5, 2]
The final array has no i where P[i] = i. Hence, a minimum of 2 swaps is required.
Input: P = [1, 2, 4, 9, 5, 8, 7, 3, 6]
Output:
Explanation:
Swap 1: Swap index 1 and index 2 => [2, 1, 4, 9, 5, 8, 7, 3, 6]
Swap 2: Swap index 5 and index 6 => [2, 1, 4, 9, 8, 5, 7, 3, 6]
Swap 2: Swap index 7 and index 8 => [2, 1, 4, 9, 8, 5, 3, 7, 6]
Hence, a minimum of 3 swaps is required.

Approach: Let us consider the positions where P[i] = i be denoted by X and the other positions by O. Below are three basic observation for the question:

• If values at any two adjacent index of the permutation is of the form XO, we can simply swap the 2 indexes to get ‘OO’.
• If values at any two adjacent index of the permutation is of the form XX, we can simply swap the 2 indexes to get ‘OO’.
• If values at any two adjacent index of the permutation is of the form OX, it is simply ‘XO’ or ‘XX’ once the pointer reaches index at X.

Below are the steps:

1. Iterate from 1 to N – 1 and check if P[i] = i then we simply swap(P[i], P[i + 1]), otherwise continue the process for the next adjacent pairs.
2. The Corner Case for the given question is when i = N, if P[i] = i, then we swap(P[i], P[i – 1]).

Below is the implementation of above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to find the minimum``// number of swaps``void` `solve(vector<``int``>& P, ``int` `n)``{` `    ``// New array to convert``    ``// to 1-based indexing``    ``vector<``int``> arr;` `    ``arr.push_back(0);` `    ``for` `(``auto` `x : P)``        ``arr.push_back(x);` `    ``// Keeps count of swaps``    ``int` `cnt = 0;` `    ``for` `(``int` `i = 1; i < n; i++) {` `        ``// Check if it is an 'X' position``        ``if` `(arr[i] == i) {``            ``swap(arr[i], arr[i + 1]);``            ``cnt++;``        ``}``    ``}` `    ``// Corner Case``    ``if` `(arr[n] == n) {` `        ``swap(arr[n - 1], arr[n]);``        ``cnt++;``    ``}` `    ``// Print the minimum swaps``    ``cout << cnt << endl;``}` `// Driver Code``signed` `main()``{``    ``// Given Number N``    ``int` `N = 9;` `    ``// Given Permutation of N numbers``    ``vector<``int``> P = { 1, 2, 4, 9, 5,``                      ``8, 7, 3, 6 };` `    ``// Function Call``    ``solve(P, N);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.io.*;` `class` `GFG{` `// Function to find the minimum``// number of swaps``static` `void` `solve(``int` `P[], ``int` `n)``{` `    ``// New array to convert``    ``// to 1-based indexing``    ``int` `arr[] = ``new` `int``[n + ``1``];` `    ``arr[``0``] = ``0``;` `    ``for``(``int` `i = ``0``; i < n; i++)``       ``arr[i + ``1``] = P[i];` `    ``// Keeps count of swaps``    ``int` `cnt = ``0``;` `    ``for``(``int` `i = ``1``; i < n; i++)``    ``{``       ` `       ``// Check if it is an 'X' position``       ``if` `(arr[i] == i)``       ``{``           ``int` `t = arr[i + ``1``];``           ``arr[i + ``1``] = arr[i];``           ``arr[i] = t;``           ``cnt++;``       ``}``    ``}` `    ``// Corner Case``    ``if` `(arr[n] == n)``    ``{``        ` `        ``// Swap``        ``int` `t = arr[n - ``1``];``        ``arr[n - ``1``] = arr[n];``        ``arr[n] = t;``        ``cnt++;``    ``}` `    ``// Print the minimum swaps``    ``System.out.println(cnt);``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ` `    ``// Given Number N``    ``int` `N = ``9``;` `    ``// Given Permutation of N numbers``    ``int` `P[] = ``new` `int``[]{ ``1``, ``2``, ``4``, ``9``, ``5``,``                         ``8``, ``7``, ``3``, ``6` `};` `    ``// Function Call``    ``solve(P, N);``}``}` `// This code is contributed by Pratima Pandey`

## Python3

 `# Python3 program for the above approach` `# Function to find the minimum``# number of swaps``def` `solve(P, n):` `    ``# New array to convert``    ``# to 1-based indexing``    ``arr ``=` `[]` `    ``arr.append(``0``)` `    ``for` `x ``in` `P:``        ``arr.append(x)` `    ``# Keeps count of swaps``    ``cnt ``=` `0` `    ``for` `i ``in` `range``(``1``, n):` `        ``# Check if it is an 'X' position``        ``if` `(arr[i] ``=``=` `i):``            ``arr[i], arr[i ``+` `1``] ``=` `arr[i ``+` `1``], arr[i]``            ``cnt ``+``=` `1` `    ``# Corner Case``    ``if` `(arr[n] ``=``=` `n):``        ``arr[n ``-` `1``], arr[n] ``=` `arr[n] , arr[n ``-` `1``]``        ``cnt ``+``=` `1` `    ``# Print the minimum swaps``    ``print``(cnt)` `# Driver Code` `# Given number N``N ``=` `9` `# Given permutation of N numbers``P ``=` `[ ``1``, ``2``, ``4``, ``9``, ``5``,``      ``8``, ``7``, ``3``, ``6` `]` `# Function call``solve(P, N)` `# This code is contributed by chitranayal`

## C#

 `// C# program for the above approach``using` `System;` `class` `GFG{` `// Function to find the minimum``// number of swaps``static` `void` `solve(``int` `[]P, ``int` `n)``{``    ` `    ``// New array to convert``    ``// to 1-based indexing``    ``int` `[]arr = ``new` `int``[n + 1];` `    ``arr = 0;` `    ``for``(``int` `i = 0; i < n; i++)``        ``arr[i + 1] = P[i];` `    ``// Keeps count of swaps``    ``int` `cnt = 0;` `    ``for``(``int` `i = 1; i < n; i++)``    ``{``        ` `        ``// Check if it is an 'X' position``        ``if` `(arr[i] == i)``        ``{``            ``int` `t = arr[i + 1];``            ``arr[i + 1] = arr[i];``            ``arr[i] = t;``            ``cnt++;``        ``}``    ``}` `    ``// Corner Case``    ``if` `(arr[n] == n)``    ``{``        ` `        ``// Swap``        ``int` `t = arr[n - 1];``        ``arr[n - 1] = arr[n];``        ``arr[n] = t;``        ``cnt++;``    ``}` `    ``// Print the minimum swaps``    ``Console.WriteLine(cnt);``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ` `    ``// Given Number N``    ``int` `N = 9;` `    ``// Given Permutation of N numbers``    ``int` `[]P = { 1, 2, 4, 9, 5,``                ``8, 7, 3, 6 };` `    ``// Function Call``    ``solve(P, N);``}``}` `// This code is contributed by Princi Singh`

## Javascript

 ``
Output:
`3`

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

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