Minimum adjacent swaps to move maximum and minimum to corners

Given N number of elements, find the minimum number of swaps required so that the maximum element is at the beginning and the minimum element is at last with the condition that only swapping of adjacent elements is allowed.

Examples:

Input: a[] = {3, 1, 5, 3, 5, 5, 2}
Output: 6
Step 1: Swap 5 with 1 to make the array as {3, 5, 1, 3, 5, 5, 2}
Step 2: Swap 5 with 3 to make the array as {5, 3, 1, 3, 5, 5, 2}
Step 3: Swap 1 with 3 at its right to make the array as {5, 3, 3, 1, 5, 5, 2}
Step 4: Swap 1 with 5 at its right to make the array as {5, 3, 3, 5, 1, 5, 2}
Step 5: Swap 1 with 5 at its right to make the array as {5, 3, 3, 5, 5, 1, 2}
Step 6: Swap 1 with 2 at its right to make the array as {5, 3, 3, 5, 5, 2, 1}
After performing 6 swapping operations 5 is at the beginning and 1 at the end

Input: a[] = {5, 6, 1, 3}
Output: 2



The approach will be to find the index of the largest element(let l). Find the index of the leftmost largest element if largest element appears in the array more than once. Similarly, find the index of the rightmost smallest element(let r). There exists two cases to solve this problem.

  1. Case 1: If l < r: Number of swaps = l + (n-r-1)
  2. Case 2: If l > r: Number of swaps = l + (n-r-2), as one swap has already been performed while swapping the larger element to the front

C++

// CPP program to count Minimum number
// of adjacent /swaps so that the largest element
// is at beginning and the smallest element 
// is at last
#include <bits/stdc++.h>
using namespace std;

// Function that returns the minimum swaps
void solve(int a[], int n)
{
    int maxx = -1, minn = a[0], l = 0, r = 0;
    for (int i = 0; i < n; i++) {

        // Index of leftmost largest element
        if (a[i] > maxx) {
            maxx = a[i];
            l = i;
        }

        // Index of rightmost smallest element
        if (a[i] <= minn) {
            minn = a[i];
            r = i;
        }
    }
    if (r < l)
        cout << l + (n - r - 2);
    else
        cout << l + (n - r - 1);
}

// Driver Code
int main()
{
    int a[] = { 5, 6, 1, 3 };
    int n = sizeof(a)/sizeof(a[0]);
    solve(a, n);
    return 0;
}

Java

// Java program to count Minimum number
// of swaps so that the largest element
// is at beginning and the
// smallest element is at last
import java.io.*;
class GFG {
    // Function performing calculations
    public static void minimumSwaps(int a[], int n)
    {
        int maxx = -1, minn = a[0], l = 0, r = 0;
        for (int i = 0; i < n; i++) {

            // Index of leftmost largest element
            if (a[i] > maxx) {
                maxx = a[i];
                l = i;
            }

            // Index of rightmost smallest element
            if (a[i] <= minn) {
                minn = a[i];
                r = i;
            }
        }
        if (r < l)
            System.out.println(l + (n - r - 2));
        else
            System.out.println(l + (n - r - 1));
    }

    // Driver Code
    public static void main(String args[]) throws IOException
    {
        int a[] = { 5, 6, 1, 3 };
        int n = a.length;
        minimumSwaps(a, n);
    }
}

Python3

# Python3 program to count 
# Minimum number of adjacent 
# swaps so that the largest 
# element is at beginning and 
# the smallest element is at last.
def minSwaps(arr):
    '''Function that returns 
       the minimum swaps'''
    
    n = len(arr)
    maxx, minn, l, r = -1, arr[0], 0, 0

    for i in range(n):
        
        # Index of leftmost
        # largest element
        if arr[i] > maxx:
            maxx = arr[i]
            l = i
            
        # Index of rightmost 
        # smallest element
        if arr[i] <= minn:
            minn = arr[i]
            r = i
            
    if r < l: 
        print(l + (n - r - 2))
    else:
        print(l + (n - r - 1))
        
# Driver code
arr = [5, 6, 1, 3]

minSwaps(arr)

# This code is contributed 
# by Tuhin Patra

C#

// C# program to count Minimum 
// number of swaps so that the 
// largest element is at beginning
// and the smallest element is at last
using System;

class GFG 
{
    // Function performing calculations
    public static void minimumSwaps(int []a, 
                                    int n)
    {
        int maxx = -1, l = 0, 
            minn = a[0], r = 0; 
        for (int i = 0; i < n; i++) 
        {

            // Index of leftmost 
            // largest element
            if (a[i] > maxx) 
            {
                maxx = a[i];
                l = i;
            }

            // Index of rightmost
            // smallest element
            if (a[i] <= minn) 
            {
                minn = a[i];
                r = i;
            }
        }
        if (r < l)
            Console.WriteLine(l + (n - r - 2));
        else
            Console.WriteLine(l + (n - r - 1));
    }

    // Driver Code
    public static void Main() 
    {

        int []a = { 5, 6, 1, 3 };
        int n = a.Length;
        
        // Calling function
        minimumSwaps(a, n);
    }
}

// This code is contributed by anuj_67.

PHP

<?php
// PHP program to count Minimum 
// number of adjacent /swaps so 
// that the largest element is
// at beginning and the smallest 
// element is at last

// Function that returns
// the minimum swaps
function solve($a, $n)
{
    $maxx = -1; $minn = $a[0]; 
    $l = 0; $r = 0;
    for ($i = 0; $i < $n; $i++) 
    {

        // Index of leftmost
        // largest element
        if ($a[$i] > $maxx) 
        {
            $maxx = $a[$i];
            $l = $i;
        }

        // Index of rightmost 
        // smallest element
        if ($a[$i] <= $minn) 
        {
            $minn = $a[$i];
            $r = $i;
        }
    }
    
    if ($r < $l)
        echo $l + ($n - $r - 2);
    else
        echo $l + ($n - $r - 1);
}

// Driver Code
$a = array(5, 6, 1, 3);
$n = count($a);
solve($a, $n);

// This code is contributed 
// by anuj_67.
?>

Output:

2

Time Complexity: O(N)



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Improved By : vt_m, TuhinPatra

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