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Minimize cost to Swap two given Arrays

  • Last Updated : 26 Jul, 2021

Given two arrays A[] and B[] both of size N consisting of distinct elements, the task is to find the minimum cost to swap two given arrays. Cost of swapping two elements A[i] and B[j] is min(A[i], A[j]). The total cost is the cumulative sum of the costs of all swap operations. 

Note: Here, the order of elements can differ from the original arrays after swapping.

Examples:

Input: N = 3, A[] = {1, 4, 2}, B[] = {10, 6, 12} 
Output:
Explanation: 
Following swap operations will give the minimum cost: 
swap(A[0], B[2]): cost = min(A[0], B[2]) = 1, A[ ] = {12, 4, 2}, B[ ] = {10, 6, 1} 
swap(A[2], B[2]): cost = min(A[2], B[2]) = 1, A[ ] = {12, 4, 1}, B[ ] = {10, 6, 2} 
swap(A[2], B[0]): cost = min(A[2], B[0]) = 1, A[ ] = {12, 4, 10}, B[ ] = {1, 6, 2} 
swap(A[1], B[0]): cost = min(A[1], B[0]) = 1, A[ ] = {12, 1, 10}, B[ ] = {4, 6, 2} 
swap(A[1], B[1]): cost = min(A[1], B[1]) = 1, A[ ] = {12, 6, 10}, B[ ] = {4, 1, 2} 
Therefore, the minimum cost to swap two arrays = 1 + 1 + 1 + 1 + 1 = 5
Input: N = 2, A[] = {9, 12}, B[] = {3, 15} 
Output:
 

Approach: 
Follow the steps below to solve the problem:  



  • Traverse the arrays simultaneously and find the minimum element from them, say K.
  • Now, every element with K until the two arrays are swapped. Therefore, the number of swaps required is 2*N – 1.
  • Print K * (2 * N – 1) as the answer.

Below is the implementation of the above approach:

C++




// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate and return the
// minimum cost required to swap two arrays
int getMinCost(vector<int> A, vector<int> B,
               int N)
{
 
    int mini = INT_MAX;
    for (int i = 0; i < N; i++) {
        mini = min(mini, min(A[i], B[i]));
    }
 
    // Return the total minimum cost
    return mini * (2 * N - 1);
}
 
// Driver Code
int main()
{
    int N = 3;
 
    vector<int> A = { 1, 4, 2 };
    vector<int> B = { 10, 6, 12 };
 
    cout << getMinCost(A, B, N);
    return 0;
}

Java




// Java program to implement
// the above approach
class GFG{
 
// Function to calculate and return the
// minimum cost required to swap two arrays
static int getMinCost(int [] A, int [] B,
                                   int N)
{
    int mini = Integer.MAX_VALUE;
    for (int i = 0; i < N; i++)
    {
        mini = Math.min(mini,
               Math.min(A[i], B[i]));
    }
 
    // Return the total minimum cost
    return mini * (2 * N - 1);
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 3;
 
    int [] A = { 1, 4, 2 };
    int [] B = { 10, 6, 12 };
 
    System.out.print(getMinCost(A, B, N));
}
}
 
// This code is contributed by sapnasingh4991

Python3




# Python3 program to implement
# the above approach
import sys
 
# Function to calculate and return the
# minimum cost required to swap two arrays
def getMinCost(A, B, N):
 
    mini = sys.maxsize
    for i in range(N):
        mini = min(mini, min(A[i], B[i]))
 
    # Return the total minimum cost
    return mini * (2 * N - 1)
 
# Driver Code
N = 3
 
A = [ 1, 4, 2 ]
B = [ 10, 6, 12 ]
 
print(getMinCost(A, B, N))
 
# This code is contributed by chitranayal

C#




// C# program to implement
// the above approach
using System;
class GFG{
 
    // Function to calculate and return the
    // minimum cost required to swap two arrays
    static int getMinCost(int[] A, int[] B, int N)
    {
        int mini = int.MaxValue;
        for (int i = 0; i < N; i++)
        {
            mini = Math.Min(mini, Math.Min(A[i], B[i]));
        }
 
        // Return the total minimum cost
        return mini * (2 * N - 1);
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        int N = 3;
          int[] A = {1, 4, 2};
        int[] B = {10, 6, 12};
          Console.Write(getMinCost(A, B, N));
    }
}
 
// This code is contributed by shikhasingrajput

Javascript




<script>
// Java script program to implement
// the above approach
function  getMinCost(A,B,N)
{
    let mini = Number.MAX_VALUE;
    for (let i = 0; i < N; i++)
    {
        mini = Math.min(mini,
            Math.min(A[i], B[i]));
    }
 
    // Return the total minimum cost
    return mini * (2 * N - 1);
}
 
// Driver Code
 
    let N = 3;
 
    let A = [ 1, 4, 2 ];
    let B = [ 10, 6, 12 ];
 
    document.write(getMinCost(A, B, N));
 
 
// This code is contributed by manoj
</script>
Output: 
5

 

Time Complexity: O(N) 
Auxiliary Space: O(1) 
 

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