Data Modelling & AI Data Structure & Algorithm

Minimize cost to sort an Array by swapping any pair of element (X, Y) with cost as (X + Y)

23 June 2025

0

Given an array arr[] consisting of N integers, the task is to find the minimum cost to sort the given array arr[] in ascending order by swapping any pair of elements (X, Y) such that the cost of swapping is (X + Y).

Examples:

Input: arr[] = {3, 2, 1}
Output: 4
Explanation:
Following are the swapping of array elements performed to sort the array:

Swapping the array elements at index 0 and 2 modifies the array to {1, 2, 3}. The cost of this swapping operation is (arr[0] + arr[2]) = (3 + 1) = 4.

After the above steps, the given array is sorted and the total cost is 4, which is the minimum among all possible combinations of swapping.

Input: arr[] = {7, 9, 15}
Output: 0

Approach: The given problem can be solved based on the following observations:

Form a directed graph by forming edges between every i^th element of the current array and the sorted array.
It can be observed that every component will always form a cycle as there every node will have in-degree and out-degree equal to 1.
Therefore, the idea is to sort the elements of each cycle separately.

Illustration:

Suppose the given array is {8, 4, 5, 3, 2, 7}. The sorted array will be equal to the {2, 3, 4, 5, 7, 8}.
For the above array, the graph will contain two components which are cycles.

If two elements are swapped in a cycle, of length K > 1 such that at least 1 element of this cycle will go to its destination. Then after the swap, it will be split into two cycles, one with length K – 1 and another one with length 1.
For the given array, If 2 and 8 are swapped of 2 ? 8 ? 7 ? 2 cycles. 2 will go to its destination, and 7 ? 8 ? 7 will form a smaller cycle.

Therefore, the minimum number of swaps needed to sort a cycle of size K is equal to the (K-1).
It can be observed that each swap will add two elements to the cost. So 2 × (K – 1) elements will be added to the cost and there are K elements. So some elements will be added multiple times to the cost.
Therefore, the idea is to, swap with the minimum value of a cycle with every other element to place them at the correct positions. Then in the final cost, every element will be added once and the minimum element will be added K – 1 time. So this is the optimal approach to solve a cycle.
For sorting a cycle there are two choices: either to use only the local minimum of the cycle or to use both local and overall minimum of the array.

Follow the steps below to solve the problem:

Initialize a variable res as 0 to store the total cost.
Copy every element of arr[] to another array, say copyArr[] and sort copyArr[] in ascending order.
Initialize a hashmap say place and store array elements and the correct position of it in the sorted array as key-value pair.
Also, Initialize an array, visited[] of size N, and mark all its entries as false.
Iterate over the array, arr[] using the variable i, and perform the following steps:
- If visited[i] is true then continue.
- If the element visited index is visited true and perform the following steps:
- If place[arr[i]] is equal to i then mark visited[i], true and continue.
- Initialize a variable say min_value as arr[i] and sum as 0, to store the minimum value of the current cycle and the sum of the elements of the cycle.
- Also, initialize a variable j as i to iterate over the cycle and a variable num as 0 to store the count of numbers in the cycle.
- Iterate until visited[j] is not true and in each iteration perform the following steps:
  - Increment sum by arr[j] and num by 1 and then mark the visited[j] as true.
  - Update min_value to min(min_value, arr[j]). And then assign value of place[arr[j]] to j.
- Decrement sum by the min_value.
- Now find the cost obtained by using the local minimum, min_value, and store it in a variable say Cost1.
  - Cost1 = sum + min_val*(num-1).
- Now find the cost obtained by using the global minimum, copyArr[0], and store it in a variable say Cost2.
  - Cost2 = copyArr[0] * (num + 1) + 2 * min_val+ sum
- Now increment res by min(Cost1, Cost2).
After completing the above steps, print the res as the total cost.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the minimum cost to
// sort the array
int findMinimumCost(int arr[], int N)
{
    // Stores the required result
    int res = 0;
 
    // Create 2 arrays
    int copyArr[N], visited[N];
 
    for (int i = 0; i < N; ++i) {
        copyArr[i] = arr[i];
        visited[i] = false;
    }
 
    // Sort the array, copyArr[] in
    // increasing order
    sort(copyArr, copyArr + N);
 
    // Map the numbers to their desired
    // place after sorting
    map<int, int> place;
 
    // Store the original places of the
    // elements in map
    for (int i = 0; i < N; ++i) {
        place[copyArr[i]] = i;
    }
 
    // Iterate in the range [0, N-1]
    for (int i = 0; i < N; ++i) {
 
        // If the ith index is not visited
        if (visited[i] == false) {
 
            // If the original place and
            // the place in sorted array
            // is same then only mark this
            // element as visited
            if (place[arr[i]] == i) {
                visited[i] = true;
                continue;
            }
 
            // Else a new cycle is present
            int min_val = arr[i], cost1, cost2;
            int num = 0;
            int sum = 0;
            int j = i;
 
            // Iterate while the nodes
            // in the current cycle is
            // not visited
            while (visited[j] == false) {
 
                // Increment sum by arr[j]
                sum += arr[j];
                num++;
 
                // Update the min_val value
                if (arr[j] < min_val) {
                    min_val = arr[j];
                }
 
                // Mark j as visited
                visited[j] = true;
 
                // Place j at its
                // original place
                j = place[arr[j]];
            }
 
            // Sum of all numbers of
            // cycle other than minimum
            sum -= min_val;
 
            // Cost from local minimum
            cost1 = sum + min_val * (num - 1);
 
            // Cost from overall minimum
            cost2 = copyArr[0] * (num + 1) + 2 * min_val
                    + sum;
 
            // Add the lower cost to
            // the final result
            if (cost1 < cost2) {
                res += cost1;
            }
            else {
                res += cost2;
            }
        }
    }
 
    // Print the minimum cost
    return res;
}
 
// Driver Code
int main()
{
    int arr[] = { 3, 2, 1 };
    int N = (sizeof(arr) / sizeof(int));
    cout << findMinimumCost(arr, N);
 
    return 0;
}

Java

import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;
 
//Java program for above approach
class GFG{
 
    // Function to find the minimum cost to
// sort the array
    static int findMinimumCost(int[] arr, int N)
    {
        // Stores the required result
        int res = 0;
 
        // Create 2 arrays
        int[] copyArr = new int[N];
        boolean[] visited = new boolean[N];
 
        for (int i = 0; i < N; ++i) {
            copyArr[i] = arr[i];
            visited[i] = false;
        }
 
        // Sort the array, copyArr[] in
        // increasing order
        Arrays.sort(copyArr);
 
        // Map the numbers to their desired
        // place after sorting
        Map<Integer, Integer> place = new HashMap<>();
 
        // Store the original places of the
        // elements in map
        for (int i = 0; i < N; ++i) {
            place.put(copyArr[i],i);
        }
 
        // Iterate in the range [0, N-1]
        for (int i = 0; i < N; ++i) {
 
            // If the ith index is not visited
            if (visited[i] == false) {
 
                // If the original place and
                // the place in sorted array
                // is same then only mark this
                // element as visited
                if (place.get(arr[i]) == i) {
                    visited[i] = true;
                    continue;
                }
 
                // Else a new cycle is present
                int min_val = arr[i], cost1, cost2;
                int num = 0;
                int sum = 0;
                int j = i;
 
                // Iterate while the nodes
                // in the current cycle is
                // not visited
                while (visited[j] == false) {
 
                    // Increment sum by arr[j]
                    sum += arr[j];
                    num++;
 
                    // Update the min_val value
                    if (arr[j] < min_val) {
                        min_val = arr[j];
                    }
 
                    // Mark j as visited
                    visited[j] = true;
 
                    // Place j at its
                    // original place
                    j = place.get(arr[j]);
                }
 
                // Sum of all numbers of
                // cycle other than minimum
                sum -= min_val;
 
                // Cost from local minimum
                cost1 = sum + min_val * (num - 1);
 
                // Cost from overall minimum
                cost2 = copyArr[0] * (num + 1) + 2 * min_val
                        + sum;
 
                // Add the lower cost to
                // the final result
                if (cost1 < cost2) {
                    res += cost1;
                }
                else {
                    res += cost2;
                }
            }
        }
 
        // Print the minimum cost
        return res;
    }
 
    // Driver Code
    public static void main(String[] args) {
        int[] arr = { 3, 2, 1 };
        int N = arr.length;
        System.out.println(findMinimumCost(arr, N));
 
    }
}
 
// This code is contributed by hritikrommie.

Python3

# Python program for the above approach
 
# Function to find the minimum cost to
# sort the array
def findMinimumCost(arr, N):
    # Stores the required result
    res = 0
 
    # Create 2 arrays
    copyArr = [0] * N
    visited = [0] * N
 
    for i in range(N):
        copyArr[i] = arr[i]
        visited[i] = False
 
    # Sort the array, copyArr[] in
    # increasing order
    copyArr.sort()
 
    # Map the numbers to their desired
    # place after sorting
    place = {}
 
    # Store the original places of the
    # elements in map
    for i in range(N):
        place[copyArr[i]] = i
 
    # Iterate in the range [0, N-1]
    for i in range(N):
 
        # If the ith index is not visited
        if (visited[i] == False):
 
            # If the original place and
            # the place in sorted array
            # is same then only mark this
            # element as visited
            if (place[arr[i]] == i):
                visited[i] = True
                continue
 
            # Else a new cycle is present
            min_val = arr[i]
            num = 0
            sum = 0
            j = i
 
            # Iterate while the nodes
            # in the current cycle is
            # not visited
            while (visited[j] == False):
 
                # Increment sum by arr[j]
                sum += arr[j]
                num += 1
 
                # Update the min_val value
                if (arr[j] < min_val):
                    min_val = arr[j]
 
                # Mark j as visited
                visited[j] = True
 
                # Place j at its
                # original place
                j = place[arr[j]]
 
            # Sum of all numbers of
            # cycle other than minimum
            sum -= min_val
 
            # Cost from local minimum
            cost1 = sum + min_val * (num - 1)
 
            # Cost from overall minimum
            cost2 = copyArr[0] * (num + 1) + 2 * min_val + sum
 
            # Add the lower cost to
            # the final result
            if (cost1 < cost2):
                res += cost1
            else:
                res += cost2
 
    # Print the minimum cost
    return res
 
 
# Driver Code
 
arr = [3, 2, 1]
N = len(arr)
print(findMinimumCost(arr, N))
 
 
# This code is contributed by gfgking

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
     
// Function to find the minimum cost to
// sort the array
static int findMinimumCost(int[] arr, int N)
{
     
    // Stores the required result
    int res = 0;
 
    // Create 2 arrays
    int[] copyArr = new int[N]; 
    int[] visited = new int[N]; 
 
    for(int i = 0; i < N; ++i) 
    {
        copyArr[i] = arr[i];
        visited[i] = 0;
    }
 
    // Sort the array, copyArr[] in
    // increasing order
    Array.Sort(copyArr);
 
    // Map the numbers to their desired
    // place after sorting
    Dictionary<int,
               int> place = new Dictionary<int,
                                           int>();
 
    // Store the original places of the
    // elements in map
    for(int i = 0; i < N; ++i)
    {
        place[copyArr[i]] = i;
    }
 
    // Iterate in the range [0, N-1]
    for(int i = 0; i < N; ++i)
    {
         
        // If the ith index is not visited
        if (visited[i] == 0)
        {
             
            // If the original place and
            // the place in sorted array
            // is same then only mark this
            // element as visited
            if (place[arr[i]] == i) 
            {
                visited[i] = 1;
                continue;
            }
 
            // Else a new cycle is present
            int min_val = arr[i], cost1, cost2;
            int num = 0;
            int sum = 0;
            int j = i;
 
            // Iterate while the nodes
            // in the current cycle is
            // not visited
            while (visited[j] == 0)
            {
                 
                // Increment sum by arr[j]
                sum += arr[j];
                num++;
 
                // Update the min_val value
                if (arr[j] < min_val) 
                {
                    min_val = arr[j];
                }
 
                // Mark j as visited
                visited[j] = 1;
 
                // Place j at its
                // original place
                j = place[arr[j]];
            }
 
            // Sum of all numbers of
            // cycle other than minimum
            sum -= min_val;
 
            // Cost from local minimum
            cost1 = sum + min_val * (num - 1);
 
            // Cost from overall minimum
            cost2 = copyArr[0] * (num + 1) + 
                             2 * min_val + sum;
 
            // Add the lower cost to
            // the final result
            if (cost1 < cost2) 
            {
                res += cost1;
            }
            else
            {
                res += cost2;
            }
        }
    }
 
    // Print the minimum cost
    return res;
}
     
// Driver code
public static void Main()
{
    int[] arr = { 3, 2, 1 };
    int N = arr.Length;
     
    Console.WriteLine(findMinimumCost(arr, N));
}
}
 
// This code is contributed by sanjoy_62

Javascript

<script>
 
       // JavaScript program for the above approach
 
 
       // Function to find the minimum cost to
       // sort the array
       function findMinimumCost(arr, N) {
           // Stores the required result
           let res = 0;
 
           // Create 2 arrays
           let copyArr = Array(N);
           let visited = Array(N);
 
           for (let i = 0; i < N; ++i) {
               copyArr[i] = arr[i];
               visited[i] = false;
           }
 
           // Sort the array, copyArr[] in
           // increasing order
           copyArr.sort(function (a, b) { return a - b; });
 
           // Map the numbers to their desired
           // place after sorting
           let place = new Map();
 
           // Store the original places of the
           // elements in map
           for (let i = 0; i < N; ++i) {
               place.set(copyArr[i], i);
           }
 
           // Iterate in the range [0, N-1]
           for (let i = 0; i < N; ++i) {
 
               // If the ith index is not visited
               if (visited[i] == false) {
 
                   // If the original place and
                   // the place in sorted array
                   // is same then only mark this
                   // element as visited
                   if (place.get(arr[i]) == i) {
                       visited[i] = true;
                       continue;
                   }
 
                   // Else a new cycle is present
                   let min_val = arr[i], cost1, cost2;
                   let num = 0;
                   let sum = 0;
                   let j = i;
 
                   // Iterate while the nodes
                   // in the current cycle is
                   // not visited
                   while (visited[j] == false) {
 
                       // Increment sum by arr[j]
                       sum += arr[j];
                       num++;
 
                       // Update the min_val value
                       if (arr[j] < min_val) {
                           min_val = arr[j];
                       }
 
                       // Mark j as visited
                       visited[j] = true;
 
                       // Place j at its
                       // original place
                       j = place.get(arr[j]);
                   }
 
                   // Sum of all numbers of
                   // cycle other than minimum
                   sum -= min_val;
 
                   // Cost from local minimum
                   cost1 = sum + min_val * (num - 1);
 
                   // Cost from overall minimum
                   cost2 = copyArr[0] * (num + 1) + 2 * min_val
                       + sum;
 
                   // Add the lower cost to
                   // the final result
                   if (cost1 < cost2) {
                       res += cost1;
                   }
                   else {
                       res += cost2;
                   }
               }
           }
 
           // Print the minimum cost
           return res;
       }
 
       // Driver Code
 
       let arr = [3, 2, 1];
       let N = arr.length;
       document.write(findMinimumCost(arr, N));
 
 
   // This code is contributed by Potta Lokesh
   </script>

Output:

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

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Minimize cost to sort an Array by swapping any pair of element (X, Y) with cost as (X + Y)

C++

Java

Python3

C#

Javascript

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