Maximum number of fixed points using atmost 1 swap

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++

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// CPP program to find maximum number of
// fixed points using atmost 1 swap
  
#include <bits/stdc++.h>
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;
}

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Java

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// 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

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Python3

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# 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] *
    count, swapped = 0, 0
  
    # Store position of each element 
    # in input array 
    for i in range(0, n): 
        pos[a[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

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C#

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// 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

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PHP

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<?php
// PHP program to find maximum number of 
// fixed points using atmost 1 swap 
  
// Function to find maximum number of 
// fixed points using atmost 1 swap 
function  maximumFixedPoints($a, $n
    $i; $pos[$n]=array();
    $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 
     $a = array (0, 1, 3, 4, 2 ); 
        $n = sizeof($a) / sizeof($a[0]); 
  
    echo maximumFixedPoints($a, $n); 
  
  
// This code is contributed by Sachin
?>

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Output:

3

Time Complexity: O(N)



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