Minimum number of swaps required to minimize sum of absolute differences between adjacent array elements

Given an array arr[] consisting of N distinct positive integers, the task is to find the minimum number of elements required to be swapped to minimize the sum of absolute difference of each pair of adjacent elements. 

Examples:

Input: arr[] = {8, 50, 11, 2}
Output: 2
Explanation:
Operation 1: Swapping of elements 8 and 2, modifies the array arr[] to {2, 50, 11, 8}.
Operation 2: Swapping of elements 8 and 50, modifies the array arr[] to {2, 8, 11, 50}.
The sum of absolute difference of adjacent elements of the modified array is 48, which is minimum.
Therefore, the minimum number of swaps required is 2. 

Input: arr[] = {3, 4, 2, 5, 1}
Output: 2

Approach: The given problem can be solved based on the observation that the sum of the absolute difference of adjacent elements will be minimum if the array is either sorted in increasing or decreasing order. Follow the steps to solve the problem.

• Find the minimum number of swaps required to sort the array given array in ascending order using the approach discussed in this article. Let the count be S1.
• Find the minimum number of swaps required to sort the array given array in descending order using the approach discussed in this article. Let the count be S2.
• After completing the above steps, print the minimum of the value S1 and S2 as the resultant minimum number of swaps required.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Comparator to sort in the descending
// order
bool mycmp(pair<int, int> a,
           pair<int, int> b)
{
    return a.first > b.first;
}

// Function to find the minimum number
// of swaps required to sort the array
// in increasing order
int minSwapsAsc(vector<int> arr, int n)
{
    // Stores the array elements with
    // its index
    pair<int, int> arrPos[n];
    for (int i = 0; i < n; i++) {
        arrPos[i].first = arr[i];
        arrPos[i].second = i;
    }

    // Sort the array in the
    // increasing order
    sort(arrPos, arrPos + n);

    // Keeps the track of
    // visited elements
    vector<bool> vis(n, false);

    // Stores the count of swaps required
    int ans = 0;

    // Traverse array elements
    for (int i = 0; i < n; i++) {

        // If the element is already
        // swapped or at correct position
        if (vis[i] || arrPos[i].second == i)
            continue;

        // Find out the number of
        // nodes in this cycle
        int cycle_size = 0;

        // Update the value of j
        int j = i;

        while (!vis[j]) {
            vis[j] = 1;

            // Move to the next element
            j = arrPos[j].second;

            // Increment cycle_size
            cycle_size++;
        }

        // Update the ans by adding
        // current cycle
        if (cycle_size > 0) {
            ans += (cycle_size - 1);
        }
    }

    return ans;
}

// Function to find the minimum number
// of swaps required to sort the array
// in decreasing order
int minSwapsDes(vector<int> arr, int n)
{
    // Stores the array elements with
    // its index
    pair<int, int> arrPos[n];

    for (int i = 0; i < n; i++) {
        arrPos[i].first = arr[i];
        arrPos[i].second = i;
    }

    // Sort the array in the
    // descending order
    sort(arrPos, arrPos + n, mycmp);

    // Keeps track of visited elements
    vector<bool> vis(n, false);

    // Stores the count of resultant
    // swap required
    int ans = 0;

    // Traverse array elements
    for (int i = 0; i < n; i++) {

        // If the element is already
        // swapped or at correct
        // position
        if (vis[i] || arrPos[i].second == i)
            continue;

        // Find out the number of
        // node in this cycle
        int cycle_size = 0;

        // Update the value of j
        int j = i;

        while (!vis[j]) {

            vis[j] = 1;

            // Move to the next element
            j = arrPos[j].second;

            // Increment the cycle_size
            cycle_size++;
        }

        // Update the ans by adding
        // current cycle size
        if (cycle_size > 0) {
            ans += (cycle_size - 1);
        }
    }

    return ans;
}

// Function to find minimum number of
// swaps required to minimize the sum
// of absolute difference of adjacent
// elements
int minimumSwaps(vector<int> arr)
{
    // Sort in ascending order
    int S1 = minSwapsAsc(arr, arr.size());

    // Sort in descending order
    int S2 = minSwapsDes(arr, arr.size());

    // Return the minimum value
    return min(S1, S2);
}

// Drive Code
int main()
{
    vector<int> arr{ 3, 4, 2, 5, 1 };
    cout << minimumSwaps(arr);

    return 0;
}

Java

// Java program for the above approach
import java.lang.*;
import java.util.*;

// Pair class
class pair
{
    int first, second;
    pair(int first, int second)
    {
        this.first = first;
        this.second = second;
    }
}

class GFG{

// Function to find the minimum number
// of swaps required to sort the array
// in increasing order
static int minSwapsAsc(int[] arr, int n)
{

    // Stores the array elements with
    // its index
    pair[] arrPos = new pair[n];
    for(int i = 0; i < n; i++)
    {
        arrPos[i] = new pair(arr[i], i);
    }

    // Sort the array in the
    // increasing order
    Arrays.sort(arrPos, (a, b)-> a.first - b.first);

    // Keeps the track of
    // visited elements
    boolean[] vis= new boolean[n];

    // Stores the count of swaps required
    int ans = 0;

    // Traverse array elements
    for(int i = 0; i < n; i++)
    {

        // If the element is already
        // swapped or at correct position
        if (vis[i] || arrPos[i].second == i)
            continue;

        // Find out the number of
        // nodes in this cycle
        int cycle_size = 0;

        // Update the value of j
        int j = i;

        while (!vis[j])
        {
            vis[j] = true;

            // Move to the next element
            j = arrPos[j].second;

            // Increment cycle_size
            cycle_size++;
        }

        // Update the ans by adding
        // current cycle
        if (cycle_size > 0)
        {
            ans += (cycle_size - 1);
        }
    }
    return ans;
}

// Function to find the minimum number
// of swaps required to sort the array
// in decreasing order
static int minSwapsDes(int[] arr, int n)
{

    // Stores the array elements with
    // its index
    pair[] arrPos = new pair[n];

    for(int i = 0; i < n; i++)
    {
        arrPos[i] = new pair(arr[i], i);
    }

    // Sort the array in the
    // descending order
    Arrays.sort(arrPos, (a, b)-> b.first - a.first);

    // Keeps track of visited elements
    boolean[] vis = new boolean[n];

    // Stores the count of resultant
    // swap required
    int ans = 0;

    // Traverse array elements
    for(int i = 0; i < n; i++)
    {

        // If the element is already
        // swapped or at correct
        // position
        if (vis[i] || arrPos[i].second == i)
            continue;

        // Find out the number of
        // node in this cycle
        int cycle_size = 0;

        // Update the value of j
        int j = i;

        while (!vis[j])
        {

            vis[j] = true;

            // Move to the next element
            j = arrPos[j].second;

            // Increment the cycle_size
            cycle_size++;
        }

        // Update the ans by adding
        // current cycle size
        if (cycle_size > 0)
        {
            ans += (cycle_size - 1);
        }
    }
    return ans;
}

// Function to find minimum number of
// swaps required to minimize the sum
// of absolute difference of adjacent
// elements
static int minimumSwaps(int[] arr)
{

    // Sort in ascending order
    int S1 = minSwapsAsc(arr, arr.length);

    // Sort in descending order
    int S2 = minSwapsDes(arr, arr.length);

    // Return the minimum value
    return Math.min(S1, S2);
}

// Driver code
public static void main(String[] args)
{
    int[] arr = { 3, 4, 2, 5, 1 };
    System.out.println(minimumSwaps(arr));
}
}

// This code is contributed by offbeat

Python3

# Python3 program for the above approach

# Function to find the minimum number
# of swaps required to sort the array
# in increasing order
def minSwapsAsc(arr, n):

    # Stores the array elements with
    # its index
    arrPos = [[arr[i], i] for i in range(n)]

    # Sort the array in the
    # increasing order
    arrPos = sorted(arrPos)

    # Keeps the track of
    # visited elements
    vis = [False] * (n)

    # Stores the count of swaps required
    ans = 0

    # Traverse array elements
    for i in range(n):

        # If the element is already
        # swapped or at correct position
        if (vis[i] or arrPos[i][1] == i):
            continue

        # Find out the number of
        # nodes in this cycle
        cycle_size = 0

        # Update the value of j
        j = i

        while (not vis[j]):
            vis[j] = 1

            # Move to the next element
            j = arrPos[j][1]

            # Increment cycle_size
            cycle_size += 1

        # Update the ans by adding
        # current cycle
        if (cycle_size > 0):
            ans += (cycle_size - 1)

    return ans

# Function to find the minimum number
# of swaps required to sort the array
# in decreasing order
def minSwapsDes(arr, n):

    # Stores the array elements with
    # its index
    arrPos = [[0, 0] for i in range(n)]

    for i in range(n):
        arrPos[i][0] = arr[i]
        arrPos[i][1] = i

    # Sort the array in the
    # descending order
    arrPos = sorted(arrPos)[::-1]

    # Keeps track of visited elements
    vis = [False] * n

    # Stores the count of resultant
    # swap required
    ans = 0

    # Traverse array elements
    for i in range(n):

        # If the element is already
        # swapped or at correct
        # position
        if (vis[i] or arrPos[i][1] == i):
            continue

        # Find out the number of
        # node in this cycle
        cycle_size = 0

        # Update the value of j
        j = i

        while (not vis[j]):
            vis[j] = 1

            # Move to the next element
            j = arrPos[j][1]

            # Increment the cycle_size
            cycle_size += 1

        # Update the ans by adding
        # current cycle size
        if (cycle_size > 0):
            ans += (cycle_size - 1)

    return ans

# Function to find minimum number of
# swaps required to minimize the sum
# of absolute difference of adjacent
# elements
def minimumSwaps(arr):

    # Sort in ascending order
    S1 = minSwapsAsc(arr, len(arr))

    # Sort in descending order
    S2 = minSwapsDes(arr, len(arr))

    # Return the minimum value
    return min(S1, S2)

# Drive Code
if __name__ == '__main__':

    arr = [ 3, 4, 2, 5, 1 ]
    print (minimumSwaps(arr))

# This code is contributed by mohit kumar 29

C#

// C# code to implement the approach
using System;

// Pair class
class Pair {
  public int First
  {
    get;
    set;
  }
  public int Second
  {
    get;
    set;
  }
  public Pair(int first, int second)
  {
    First = first;
    Second = second;
  }
}

class GFG {

  // Function to find the minimum number
  // of swaps required to sort the array
  // in increasing order
  static int MinSwapsAsc(int[] arr, int n)
  {

    // Stores the array elements with
    // its index
    Pair[] arrPos = new Pair[n];
    for (int i = 0; i < n; i++) {
      arrPos[i] = new Pair(arr[i], i);
    }

    // Sort the array in the
    // increasing order
    Array.Sort(arrPos, (a, b) => a.First - b.First);

    // Keeps the track of
    // visited elements
    bool[] vis = new bool[n];

    // Stores the count of swaps required
    int ans = 0;

    for (int i = 0; i < n; i++) {

      // If the element is already
      // swapped or at correct position
      if (vis[i] || arrPos[i].Second == i) {
        continue;
      }

      // Find out the number of
      // nodes in this cycle
      int cycleSize = 0;

      // Update the value of j
      int j = i;

      while (!vis[j]) {
        vis[j] = true;
        j = arrPos[j].Second;
        cycleSize++;
      }

      // Update the ans by adding
      // current cycle
      if (cycleSize > 0) {
        ans += cycleSize - 1;
      }
    }

    return ans;
  }

  // Function to find the minimum number
  // of swaps required to sort the array
  // in decreasing order
  static int MinSwapsDes(int[] arr, int n)
  {

    // Stores the array elements with
    // its index
    Pair[] arrPos = new Pair[n];
    for (int i = 0; i < n; i++) {
      arrPos[i] = new Pair(arr[i], i);
    }

    // Sort the array in the
    // descending order
    Array.Sort(arrPos, (a, b) => b.First - a.First);

    bool[] vis = new bool[n];

    // Stores the count of resultant
    // swap required
    int ans = 0;

    // Traverse array elements
    for (int i = 0; i < n; i++) {

      // If the element is already
      // swapped or at correct
      // position
      if (vis[i] || arrPos[i].Second == i) {
        continue;
      }

      // Find out the number of
      // node in this cycle
      int cycleSize = 0;
      int j = i;

      while (!vis[j]) {
        vis[j] = true;
        j = arrPos[j].Second;

        // Increment the cycle_size
        cycleSize++;
      }

      // Update the ans by adding
      // current cycle size
      if (cycleSize > 0) {
        ans += cycleSize - 1;
      }
    }

    return ans;
  }

  // Function to find minimum number of
  // swaps required to minimize the sum
  // of absolute difference of adjacent
  // elements
  static int MinimumSwaps(int[] arr)
  {
    // Sort in ascending order
    int S1 = MinSwapsAsc(arr, arr.Length);

    // Sort in descending order
    int S2 = MinSwapsDes(arr, arr.Length);

    // Return the minimum value
    return Math.Min(S1, S2);
  }

  // Driver code
  static void Main(string[] args)
  {
    int[] arr = { 3, 4, 2, 5, 1 };
    Console.WriteLine(MinimumSwaps(arr));
  }
}

// This code is contributed by phasing17

Javascript

<script>
        // Javascript program for the above approach

        // Comparator to sort in the descending
        // order
        function mycmp(a, b) {
            return a[0] > b[0];
        }

        // Function to find the minimum number
        // of swaps required to sort the array
        // in increasing order
        function minSwapsAsc(arr, n) {
            // Stores the array elements with
            // its index
            let arrPos = new Array(n);
            for (let i = 0; i < n; i++)
                arrPos[i] = new Array(2);

            for (let i = 0; i < n; i++) {
                arrPos[i][0] = arr[i];
                arrPos[i][1] = i;
            }

            // Sort the array in the
            // increasing order
            arrPos.sort(function (a, b) { return a[0] - b[0] })

            // Keeps the track of
            // visited elements
            let vis = new Array(n);
            for (let i = 0; i < n; i++)
                vis[i] = false

            // Stores the count of swaps required
            let ans = 0;

            // Traverse array elements
            for (let i = 0; i < n; i++) {

                // If the element is already
                // swapped or at correct position
                if (vis[i] || arrPos[i][1] == i)
                    continue;

                // Find out the number of
                // nodes in this cycle
                let cycle_size = 0;

                // Update the value of j
                let j = i;

                while (!vis[j]) {
                    vis[j] = 1;

                    // Move to the next element
                    j = arrPos[j][1];

                    // Increment cycle_size
                    cycle_size++;
                }

                // Update the ans by adding
                // current cycle
                if (cycle_size > 0) {
                    ans += (cycle_size - 1);
                }
            }

            return ans;
        }

        // Function to find the minimum number
        // of swaps required to sort the array
        // in decreasing order
        function minSwapsDes(arr, n) {
            // Stores the array elements with
            // its index
            let arrPos = new Array(n);
            for (let i = 0; i < n; i++)
                arrPos[i] = new Array(2);

            for (let i = 0; i < n; i++) {
                arrPos[i][0] = arr[i];
                arrPos[i][1] = i;
            }

            // Sort the array in the
            // descending order
            arrPos.sort(function (a, b) { return b[0] - a[0] })

            // Keeps track of visited elements
            let vis = new Array(n);
            for (let i = 0; i < n; i++)
                vis[i] = false

            // Stores the count of resultant
            // swap required
            let ans = 0;

            // Traverse array elements
            for (let i = 0; i < n; i++) {

                // If the element is already
                // swapped or at correct
                // position
                if (vis[i] || arrPos[i][1] == i)
                    continue;

                // Find out the number of
                // node in this cycle
                let cycle_size = 0;

                // Update the value of j
                let j = i;

                while (!vis[j]) {

                    vis[j] = 1;

                    // Move to the next element
                    j = arrPos[j][1];

                    // Increment the cycle_size
                    cycle_size++;
                }

                // Update the ans by adding
                // current cycle size
                if (cycle_size > 0) {
                    ans += (cycle_size - 1);
                }
            }

            return ans;
        }

        // Function to find minimum number of
        // swaps required to minimize the sum
        // of absolute difference of adjacent
        // elements
        function minimumSwaps(arr) {
            // Sort in ascending order
            let S1 = minSwapsAsc(arr, arr.length);

            // Sort in descending order
            let S2 = minSwapsDes(arr, arr.length);

            // Return the minimum value
            return Math.min(S1, S2);
        }

        // Drive Code

        let arr = [ 3, 4, 2, 5, 1 ];
        document.write(minimumSwaps(arr));

        // This code is contributed by Hritik
    </script>

Output:

2

Time Complexity: O(N * log N), The time complexity of this code is O(Nl*ogN), where N is the size of the array. The reason behind this time complexity is due to the sorting algorithm used to sort the array. The code uses the built-in sort() function to sort the array elements in both ascending and descending order. The time complexity of the sort() function is O(N*logN), which dominates the overall time complexity of the code.

Auxiliary Space: O(N), The space complexity of this code is O(N). The reason behind this space complexity is due to the usage of two additional data structures – vector<bool> vis(n, false) and pair<int, int> arrPos[n]. The vis vector is used to keep track of visited elements during the traversal of the array, and arrPos is used to store the index and value of each element in the array. Both of these data structures have a space complexity of O(N). Therefore, the overall space complexity of the code is O(N).