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Maximum number of fixed points using atmost 1 swap
• Difficulty Level : Medium
• Last Updated : 02 Jun, 2021

Given a permutation of N elements (Elements are in range 0 to N-1). A fixed point is an index at which the value is same as the index (That is, a[i]=i). You are allowed to make atmost 1 swap. Find the maximum number of fixed points that you can get.
Examples

```Input : N = 5
arr[] = {0, 1, 3, 4, 2}
Output : 3
2 and 3 can be swapped to get:
0 1 2 4 3
which has 3 fixed points.

Input : N = 5
a[] = {0, 1, 2, 4, 3}
Output : 5 ```

Since we are allowed to make only 1 swap, the number of fixed points can be increased by atmost 2.
Let’s have an array pos which keeps the position of each element in the input array. Now, we traverse the array and have the following cases:

• If, a[i] = i. We can simply increment the count and move on.
• If, pos[i] = a[i] which means that swapping the 2 terms would make i and a[i] fixed points, hence increasing the count by 2. Keep in mind that swap can be done atmost once.

At the end of the traversal, if we haven’t made any swap, it means that our swap was not able to increase count by 2, so now if there are at least 2 elements which are not fixed points, we can make a swap to increase count by 1, i.e make one of those points a fixed point.
Below is the implementation of the above approach:

## C++

 `// CPP program to find maximum number of``// fixed points using atmost 1 swap` `#include ``using` `namespace` `std;` `// Function to find maximum number of``// fixed points using atmost 1 swap``int` `maximumFixedPoints(``int` `a[], ``int` `n)``{``    ``int` `i, pos[n], count = 0, swapped = 0;` `    ``// Store position of each element in``    ``// input array``    ``for` `(i = 0; i < n; i++)``        ``pos[a[i]] = i;` `    ``for` `(i = 0; i < n; i++) {` `        ``// If fixed point, incremenmt count``        ``if` `(a[i] == i)``            ``count++;` `        ``// Else check if swapping increments``        ``// count by 2``        ``else` `if` `(swapped == 0 && pos[i] == a[i]) {``            ``count += 2;``            ``swapped = 1;``        ``}``    ``}` `    ``// If not swapped yet and elements remaining``    ``if` `(swapped == 0 && count < n - 1)``        ``count++;` `    ``return` `count;``}` `// Driver Code``int` `main()``{``    ``int` `a[] = { 0, 1, 3, 4, 2 };` `    ``int` `n = ``sizeof``(a) / ``sizeof``(a[0]);` `    ``cout << maximumFixedPoints(a, n);` `    ``return` `0;``}`

## Java

 `// Java program to find maximum number of``// fixed points using atmost 1 swap``import` `java.io.*;` `class` `GFG {`` `  `// Function to find maximum number of``// fixed points using atmost 1 swap``static` `int` `maximumFixedPoints(``int` `a[], ``int` `n)``{``    ``int` `i, count = ``0``, swapped = ``0``;``    ``int` `pos[] = ``new` `int``[n];` `    ``// Store position of each element in``    ``// input array``    ``for` `(i = ``0``; i < n; i++)``        ``pos[a[i]] = i;` `    ``for` `(i = ``0``; i < n; i++) {` `        ``// If fixed point, incremenmt count``        ``if` `(a[i] == i)``            ``count++;` `        ``// Else check if swapping increments``        ``// count by 2``        ``else` `if` `(swapped == ``0` `&& pos[i] == a[i]) {``            ``count += ``2``;``            ``swapped = ``1``;``        ``}``    ``}` `    ``// If not swapped yet and elements remaining``    ``if` `(swapped == ``0` `&& count < n - ``1``)``        ``count++;` `    ``return` `count;``}` `// Driver Code` `    ``public` `static` `void` `main (String[] args) {``            ``int` `[]a= { ``0``, ``1``, ``3``, ``4``, ``2` `};` `    ``int` `n = a.length;` `    ``System.out.println(maximumFixedPoints(a, n));``    ``}``}`  `// This code is contributed``// by shs`

## Python3

 `# Python3 program to find the maximum number``# of fixed points using at most 1 swap` `# Function to find maximum number of``# fixed points using atmost 1 swap``def` `maximumFixedPoints(a, n):` `    ``pos ``=` `[``None``] ``*` `n``    ``count, swapped ``=` `0``, ``0` `    ``# Store position of each element``    ``# in input array``    ``for` `i ``in` `range``(``0``, n):``        ``pos[a[i]] ``=` `i` `    ``for` `i ``in` `range``(``0``, n):``    ` `        ``# If fixed point, incremenmt count``        ``if` `a[i] ``=``=` `i:``            ``count ``+``=` `1` `        ``# Else check if swapping increments``        ``# count by 2``        ``elif` `swapped ``=``=` `0` `and` `pos[i] ``=``=` `a[i]:``            ``count ``+``=` `2``            ``swapped ``=` `1` `    ``# If not swapped yet and elements remaining``    ``if` `swapped ``=``=` `0` `and` `count < n ``-` `1``:``        ``count ``+``=` `1` `    ``return` `count` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:` `    ``a ``=` `[``0``, ``1``, ``3``, ``4``, ``2``]``    ``n ``=` `len``(a)` `    ``print``(maximumFixedPoints(a, n))` `# This code is contributed by Rituraj Jain`

## C#

 `// C# program to find maximum number of``// fixed points using atmost 1 swap``using` `System;` `class` `Program {` `// Function to find maximum number of``// fixed points using atmost 1 swap``static` `int` `maximumFixedPoints(``int` `[]a, ``int` `n)``{``    ``int` `i, count = 0, swapped = 0;``    ``int` `[]pos = ``new` `int``[n];` `    ``// Store position of each element in``    ``// input array``    ``for` `(i = 0; i < n; i++)``        ``pos[a[i]] = i;` `    ``for` `(i = 0; i < n; i++) {` `        ``// If fixed point, incremenmt count``        ``if` `(a[i] == i)``            ``count++;` `        ``// Else check if swapping increments``        ``// count by 2``        ``else` `if` `(swapped == 0 && pos[i] == a[i]) {``            ``count += 2;``            ``swapped = 1;``        ``}``    ``}` `    ``// If not swapped yet and elements remaining``    ``if` `(swapped == 0 && count < n - 1)``        ``count++;` `    ``return` `count;``}` `    ``// Driver Code``    ``static` `void` `Main()``    ``{``        ``int` `[]a= { 0, 1, 3, 4, 2 };` `        ``int` `n = a.Length;` `        ``Console.WriteLine(maximumFixedPoints(a, n));``    ``}``}` `// This code is contributed``// by ANKITRAI1`

## PHP

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## Javascript

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Output:

`3`

Time Complexity: O(N)

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