Minimum cost required to rearrange a given array to make it equal to another given array

Given two arrays A[] and B[] consisting of M and N integers respectively, and an integer C, the task is to find the minimum cost required to make the sequence A exactly the same as B(consists of distinct elements only) by performing the following operations on array A[]:

• Remove any element from the array with cost 0.
• Insert a new element anywhere in the array with cost C.

Examples:

Input: A[] = {1, 6, 3, 5, 10}, B[] = {3, 1, 5}, C = 2
Output: 2
Explanation:
Removing elements 1, 6, and 10 from the array costs 0. The array A[] becomes {3, 5}.
Add 1 in between 3 and 5, then the array arr[] becomes {3, 1, 5} which is the same as the array B[].
The cost of above operation is 2.

Input: A[] = {10, 5, 2, 4, 10, 5}, B[] = {5, 1, 2, 10, 4}, C = 3
Output: 6
Explanation:
Removing elements 10, 10, and 5 from the array costs 0. The array A[] becomes {5, 2, 4}.
Add element 1 and 10 in the array as {5, 1, 2, 10, 4} which is the same as the array B[].
The cost of above operation is 3*2 = 6.

Naive Approach: The simplest approach is to erase all the elements from array A[] which are not present in array B[] by using two for loops. After that generate all permutations of the remaining element in the array and for each sequence check for the minimum cost such that the array A[] is the same as the array B[]. Print the minimum cost of the same.

Time Complexity: O(N!*N*M), where N is the size of the array A[] and M is the size of the array B[].
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is to first find the length of the longest common subsequence of array A[] and B[] and subtract it from the size of array B[], which gives the number of elements to be added in array A[]. And add amount C into the cost for every new element added. Therefore, the total cost is given by:

Cost = C*(N – LCS(A, B))
where, 
LCS is the longest common subsequence of the arrays A[] and B[], 
N is the length of the array A[], and
C is the cost of adding each element in the array B[].

Follow the steps below to solve the problem:

• Create a new array say index[] and initialize it with -1 and an array nums[].
• Map each element of the array B[] to its corresponding index in the array index[].
• Traverse the given array A[] and insert the values with its mapped values i.e., index number into the array nums[] array and if the index number is not -1.
• Now find the Longest Increasing Subsequence of the array nums[] to obtain the length of the longest common subsequence of the two given arrays.
• After finding the LCS in the above steps, find the value of cost using the formula discussed above.

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 length of the
// longest common subsequence
int findLCS(int* nums, int N)
{
    int k = 0;
    for (int i = 0; i < N; i++) {

        // Find position where element
        // is to be inserted
        int pos = lower_bound(nums, nums + k,
                              nums[i])
                  - nums;
        nums[pos] = nums[i];
        if (k == pos) {
            k = pos + 1;
        }
    }

    // Return the length of LCS
    return k;
}

// Function to find the minimum cost
// required to convert the sequence A
// exactly same as B
int minimumCost(int* A, int* B, int M,
                int N, int C)
{
    // Auxiliary array
    int nums[1000000];

    // Stores positions of elements of A[]
    int index[1000000];

    // Initialize index array with -1
    memset(index, -1, sizeof(index));

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

        // Update the index array with
        // index of corresponding
        // elements of B
        index[B[i]] = i;
    }

    int k = 0;

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

        // Place only A's array values
        // with its mapped values
        // into nums array
        if (index[A[i]] != -1) {
            nums[k++] = index[A[i]];
        }
    }

    // Find LCS
    int lcs_length = findLCS(nums, k);

    // No of elements to be added
    // in array A[]
    int elements_to_be_added
        = N - lcs_length;

    // Stores minimum cost
    int min_cost
        = elements_to_be_added * C;

    // Print the minimum cost
    cout << min_cost;
}

// Driver Code
int main()
{
    // Given array A[]
    int A[] = { 1, 6, 3, 5, 10 };
    int B[] = { 3, 1, 5 };

    // Given C
    int C = 2;

    // Size of arr A
    int M = sizeof(A) / sizeof(A[0]);

    // Size of arr B
    int N = sizeof(B) / sizeof(B[0]);

    // Function Call
    minimumCost(A, B, M, N, C);

    return 0;
}

Java

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

class GFG{

// Function to find lower_bound
static int LowerBound(int a[], int k,
                      int x)
{
    int l = -1;
    int r = k;

    while (l + 1 < r)
    {
        int m = (l + r) >>> 1;
        if (a[m] >= x)
        {
            r = m;
        }
        else
        {
            l = m;
        }
    }
    return r;
}

// Function to find length of the
// longest common subsequence
static int findLCS(int[] nums, int N)
{
    int k = 0;
    for(int i = 0; i < N; i++)
    {

        // Find position where element
        // is to be inserted
        int pos = LowerBound(nums, k,
                             nums[i]);
        nums[pos] = nums[i];
        if (k == pos)
        {
            k = pos + 1;
        }
    }

    // Return the length of LCS
    return k;
}

// Function to find the minimum cost
// required to convert the sequence A
// exactly same as B
static int  minimumCost(int[] A, int[] B,
                        int M, int N, int C)
{

    // Auxiliary array
    int[] nums = new int[100000];

    // Stores positions of elements of A[]
    int[] index = new int[100000];

    // Initialize index array with -1
    for(int i = 0; i < 100000; i++)
        index[i] = -1;

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

        // Update the index array with
        // index of corresponding
        // elements of B
        index[B[i]] = i;
    }

    int k = 0;

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

        // Place only A's array values
        // with its mapped values
        // into nums array
        if (index[A[i]] != -1)
        {
            nums[k++] = index[A[i]];
        }
    }

    // Find LCS
    int lcs_length = findLCS(nums, k);

    // No of elements to be added
    // in array A[]
    int elements_to_be_added = N - lcs_length;

    // Stores minimum cost
    int min_cost = elements_to_be_added * C;

    // Print the minimum cost
    System.out.println( min_cost);
    return 0;
}

// Driver code
public static void main(String[] args)
{

    // Given array A[]
    int[] A = { 1, 6, 3, 5, 10 };
    int[] B = { 3, 1, 5 };

    // Given C
    int C = 2;

    // Size of arr A
    int M = A.length;

    // Size of arr B
    int N = B.length;

    // Function call
    minimumCost(A, B, M, N, C);
}
}

// This code is contributed by sallagondaavinashreddy7

Python3

# Python3 program for the above approach

# Function to find lower_bound
def LowerBound(a, k, x):

    l = -1
    r = k

    while (l + 1 < r):
        m = (l + r) >> 1

        if (a[m] >= x):
            r = m
        else:
            l = m

    return r

# Function to find length of the
# longest common subsequence
def findLCS(nums, N):

    k = 0
    for i in range(N):

        # Find position where element
        # is to be inserted
        pos = LowerBound(nums, k, nums[i])
        nums[pos] = nums[i]

        if (k == pos):
            k = pos + 1

    # Return the length of LCS
    return k

# Function to find the minimum cost
# required to convert the sequence A
# exactly same as B
def  minimumCost(A, B, M, N, C):

    # Auxiliary array
    nums = [0] * 100000

    # Stores positions of elements of A[]
    # Initialize index array with -1
    index = [-1] * 100000

    for i in range(N):

        # Update the index array with
        # index of corresponding
        # elements of B
        index[B[i]] = i

    k = 0

    for i in range(M):

        # Place only A's array values
        # with its mapped values
        # into nums array
        if (index[A[i]] != -1):
            k += 1
            nums[k] = index[A[i]]

    # Find LCS
    lcs_length = findLCS(nums, k)

    # No of elements to be added
    # in array A[]
    elements_to_be_added = N - lcs_length

    # Stores minimum cost
    min_cost = elements_to_be_added * C

    # Print the minimum cost
    print( min_cost)

# Driver Code

# Given array A[]
A = [ 1, 6, 3, 5, 10 ]
B = [ 3, 1, 5 ]

# Given C
C = 2

# Size of arr A
M = len(A)

# Size of arr B
N = len(B)

# Function call
minimumCost(A, B, M, N, C)

# This code is contributed by divyeshrabadiya07

C#

// C# program for the above approach
using System;

class GFG{

// Function to find lower_bound
static int LowerBound(int[] a, int k,
                      int x)
{
    int l = -1;
    int r = k;

    while (l + 1 < r)
    {
        int m = (l + r) >> 1;
        if (a[m] >= x)
        {
            r = m;
        }
        else
        {
            l = m;
        }
    }
    return r;
}

// Function to find length of the
// longest common subsequence
static int findLCS(int[] nums, int N)
{
    int k = 0;

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

        // Find position where element
        // is to be inserted
        int pos = LowerBound(nums, k,
                             nums[i]);
        nums[pos] = nums[i];

        if (k == pos)
        {
            k = pos + 1;
        }
    }

    // Return the length of LCS
    return k;
}

// Function to find the minimum cost
// required to convert the sequence A
// exactly same as B
static int  minimumCost(int[] A, int[] B,
                        int M, int N, int C)
{

    // Auxiliary array
    int[] nums = new int[100000];

    // Stores positions of elements of A[]
    int[] index = new int[100000];

    // Initialize index array with -1
    for(int i = 0; i < 100000; i++)
        index[i] = -1;

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

        // Update the index array with
        // index of corresponding
        // elements of B
        index[B[i]] = i;
    }

    int k = 0;

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

        // Place only A's array values
        // with its mapped values
        // into nums array
        if (index[A[i]] != -1)
        {
            nums[k++] = index[A[i]];
        }
    }

    // Find LCS
    int lcs_length = findLCS(nums, k);

    // No of elements to be added
    // in array A[]
    int elements_to_be_added = N - lcs_length;

    // Stores minimum cost
    int min_cost = elements_to_be_added * C;

    // Print the minimum cost
    Console.WriteLine(min_cost);
    return 0;
}

// Driver code
public static void Main()
{

    // Given array A[]
    int[] A = { 1, 6, 3, 5, 10 };
    int[] B = { 3, 1, 5 };

    // Given C
    int C = 2;

    // Size of arr A
    int M = A.Length;

    // Size of arr B
    int N = B.Length;

    // Function call
    minimumCost(A, B, M, N, C);
}
}

// This code is contributed by code_hunt

Javascript

<script>
// javascript program for the above approach

    // Function to find lower_bound
    function LowerBound(a , k , x) {
        var l = -1;
        var r = k;

        while (l + 1 < r) {
            var m = (l + r) >>> 1;
            if (a[m] >= x) {
                r = m;
            } else {
                l = m;
            }
        }
        return r;
    }

    // Function to find length of the
    // longest common subsequence
    function findLCS(nums , N) {
        var k = 0;
        for (i = 0; i < N; i++) {

            // Find position where element
            // is to be inserted
            var pos = LowerBound(nums, k, nums[i]);
            nums[pos] = nums[i];
            if (k == pos) {
                k = pos + 1;
            }
        }

        // Return the length of LCS
        return k;
    }

    // Function to find the minimum cost
    // required to convert the sequence A
    // exactly same as B
    function minimumCost(A,  B , M , N , C) {

        // Auxiliary array
        var nums = Array(100000).fill(0);

        // Stores positions of elements of A
        var index = Array(100000).fill(0);

        // Initialize index array with -1
        for (i = 0; i < 100000; i++)
            index[i] = -1;

        for (i = 0; i < N; i++) {

            // Update the index array with
            // index of corresponding
            // elements of B
            index[B[i]] = i;
        }

        var k = 0;

        for (i = 0; i < M; i++) {

            // Place only A's array values
            // with its mapped values
            // into nums array
            if (index[A[i]] != -1) {
                nums[k++] = index[A[i]];
            }
        }

        // Find LCS
        var lcs_length = findLCS(nums, k);

        // No of elements to be added
        // in array A
        var elements_to_be_added = N - lcs_length;

        // Stores minimum cost
        var min_cost = elements_to_be_added * C;

        // Print the minimum cost
        document.write(min_cost);
        return 0;
    }

    // Driver code


        // Given array A
        var A = [ 1, 6, 3, 5, 10 ];
        var B = [ 3, 1, 5 ];

        // Given C
        var C = 2;

        // Size of arr A
        var M = A.length;

        // Size of arr B
        var N = B.length;

        // Function call
        minimumCost(A, B, M, N, C);

// This code contributed by umadevi9616
</script>

Output:

2

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