Minimum total sum from the given two arrays

Given two arrays A[] and B[] of N positive integers and a cost C. We can choose any one element from each index of the given arrays i.e., for any index i we can choose only element A[i] or B[i]. The task is to find the minimum total sum of selecting N elements from the given two arrays and if we are selecting any element from A[] to B[] or vice-versa in the next iteration then the cost C is also added to the sum. 

Note: Choose element in increasing order of index i.e., 0 ≤ i < N.

Examples:

Input: N = 9, A[] = {7, 6, 18, 6, 16, 18, 1, 17, 17}, B[] = {6, 9, 3, 10, 9, 1, 10, 1, 5}, C = 2 
Output: 49 
Explanation:
On taking the 1st element from array A, sum = 7 
On taking the 2nd element from array A, sum = 7 + 6 = 13 
On taking the 3rd element from array B, as we are entering from array A to array B, sum = 13 + 3 + 2 = 18 
On taking the 4th element from array A, as we are entering from array B to array A, sum = 18 + 6 + 2 = 26 
On taking the 5th element from array B, as we are entering from array A to array B, sum = 26 + 9 + 2 = 37 
On taking the 6th element form array B, sum = 37 + 1 = 38 
On taking the 7th element from array A, as we are entering from array B to array A, sum = 38 + 1 + 2 = 41 
On taking the 8th element form array B, as we are entering from array A to array B, sum = 41 + 1 + 2 = 44 
On taking the 9th element from array B, sum = 44 + 5 = 49.

Input: N = 9, A = {3, 2, 3, 1, 3, 3, 1, 4, 1}, B = {1, 2, 3, 4, 4, 1, 2, 1, 3}, C = 1 
Output: 18 
Explanation:
On taking the 1st element from array B, sum = 1 
On taking the 2nd element from array A, sum = 1 + 2 = 3 
On taking the 3rd element from array A, sum = 3 + 3 = 6 
On taking the 4th element from array A, sum = 6 + 1 = 7 
On taking the 5th element from array A, sum = 7 + 3 = 10 
On taking the 6th element form array B, as we are entering from array A to array B, sum = 10 + 1 + 1 = 12 
On taking the 7th element from array A, as we are entering from array B to array A, sum = 12 + 1 + 1 = 14 
On taking the 8th element form array B, as we are entering from array A to array B, sum = 14 + 1 + 1 = 16 
On taking the 9th element from array A, as we are entering from array B to array A, sum = 16 + 1 + 1 = 18. 

Approach: We will use Dynamic Programming to solve this problem. Below are the steps:  

1. Create a 2D array dp[][] of N rows and two columns and initialize all elements of dp to infinity.
2. There can be 4 possible cases of adding the elements from both the arrays: 
   • Adding an element from array a[] when the previously added element is from array a[].
   • Adding an element from array a[] when the previously added element is from array b[]. In this case there is a penalty of adding the integer C with the result.
   • Adding an element from array b[] when the previously added element is from array b[].
   • Adding an element from array b[] when the previously added element is from array a[]. In this case there is a penalty of adding the integer C with the result.
3. Update the dp array each time with the minimum value of the above four conditions.
4. The minimum of dp[n-1][0] and dp[n-1][1] is the total minimum sum of selecting N elements.

Below is the implementation of the above approach:  

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function that prints minimum sum
// after selecting N elements
void minimumSum(int a[], int b[],
                int c, int n)
{

    // Initialise the dp array
    vector<vector<int> > dp(n,
                            vector<int>(2,
                                        1e6));

    // Base Case
    dp[0][0] = a[0];
    dp[0][1] = b[0];

    for (int i = 1; i < n; i++) {

        // Adding the element of array a if
        // previous element is also from array a
        dp[i][0] = min(dp[i][0],
                    dp[i - 1][0] + a[i]);

        // Adding the element of array a if
        // previous element is from array b
        dp[i][0] = min(dp[i][0],
                    dp[i - 1][1] + a[i] + c);

        // Adding the element of array b if
        // previous element is from array a
        // with an extra penalty of integer C
        dp[i][1] = min(dp[i][1],
                    dp[i - 1][0] + b[i] + c);

        // Adding the element of array b if
        // previous element is also from array b
        dp[i][1] = min(dp[i][1],
                    dp[i - 1][1] + b[i]);
    }

    // Print the minimum sum
    cout << min(dp[n - 1][0],
                dp[n - 1][1])
        << "\n";
}

// Driver Code
int main()
{
    // Given array arr1[] and arr2[]
    int arr1[] = { 7, 6, 18, 6, 16,
                18, 1, 17, 17 };

    int arr2[] = { 6, 9, 3, 10, 9,
                1, 10, 1, 5 };

    // Given cost
    int C = 2;

    int N = sizeof(arr1) / sizeof(arr1[0]);

    // Function Call
    minimumSum(arr1, arr2, C, N);

    return 0;
}

Java

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

class GFG{

// Function that prints minimum sum
// after selecting N elements
static void minimumSum(int a[], int b[],
                       int c, int n)
{

    // Initialise the dp array
    int [][]dp = new int[n][2];
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < 2; j++)
        {
            dp[i][j] = (int) 1e6;
        }
    }

    // Base Case
    dp[0][0] = a[0];
    dp[0][1] = b[0];

    for(int i = 1; i < n; i++)
    {

        // Adding the element of array a if
        // previous element is also from array a
        dp[i][0] = Math.min(dp[i][0],
                            dp[i - 1][0] + a[i]);

        // Adding the element of array a if
        // previous element is from array b
        dp[i][0] = Math.min(dp[i][0],
                            dp[i - 1][1] + a[i] + c);

        // Adding the element of array b if
        // previous element is from array a
        // with an extra penalty of integer C
        dp[i][1] = Math.min(dp[i][1],
                            dp[i - 1][0] + b[i] + c);

        // Adding the element of array b if
        // previous element is also from array b
        dp[i][1] = Math.min(dp[i][1],
                            dp[i - 1][1] + b[i]);
    }

    // Print the minimum sum
    System.out.print(Math.min(dp[n - 1][0],
                              dp[n - 1][1]) + "\n");
}

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

    // Given array arr1[] and arr2[]
    int arr1[] = { 7, 6, 18, 6, 16,
                   18, 1, 17, 17 };

    int arr2[] = { 6, 9, 3, 10, 9,
                   1, 10, 1, 5 };

    // Given cost
    int C = 2;

    int N = arr1.length;

    // Function call
    minimumSum(arr1, arr2, C, N);
}
}

// This code is contributed by gauravrajput1

Python3

# Python3 program for the above approach

# Function that prints minimum sum
# after selecting N elements
def minimumSum(a, b, c, n):

    # Initialise the dp array
    dp = [[1e6 for i in range(2)]
            for j in range(n)]

    # Base Case
    dp[0][0] = a[0]
    dp[0][1] = b[0]

    for i in range(1, n):

        # Adding the element of array a if
        # previous element is also from array a
        dp[i][0] = min(dp[i][0],
                    dp[i - 1][0] + a[i])

        # Adding the element of array a if
        # previous element is from array b
        dp[i][0] = min(dp[i][0],
                    dp[i - 1][1] + a[i] + c)

        # Adding the element of array b if
        # previous element is from array a
        # with an extra penalty of integer C
        dp[i][1] = min(dp[i][1],
                    dp[i - 1][0] + b[i] + c)

        # Adding the element of array b if
        # previous element is also from array b
        dp[i][1] = min(dp[i][1],
                    dp[i - 1][1] + b[i])

    # Print the minimum sum
    print(min(dp[n - 1][0], dp[n - 1][1]))

# Driver code
if __name__ == '__main__':

    # Given array arr[]
    arr1 = [ 7, 6, 18, 6, 16,
            18, 1, 17, 17 ]

    arr2 = [ 6, 9, 3, 10, 9,
            1, 10, 1, 5 ]

    # Given cost
    C = 2

    N = len(arr1)

    # Function Call
    minimumSum(arr1, arr2, C, N)

# This code is contributed by Shivam Singh

C#

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

class GFG{

// Function that prints minimum sum
// after selecting N elements
static void minimumSum(int []a, int []b,
                       int c, int n)
{

    // Initialise the dp array
    int [,]dp = new int[n, 2];
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < 2; j++)
        {
            dp[i, j] = (int)1e6;
        }
    }

    // Base Case
    dp[0, 0] = a[0];
    dp[0, 1] = b[0];

    for(int i = 1; i < n; i++)
    {

        // Adding the element of array a if
        // previous element is also from array a
        dp[i, 0] = Math.Min(dp[i, 0],
                            dp[i - 1, 0] + a[i]);

        // Adding the element of array a if
        // previous element is from array b
        dp[i, 0] = Math.Min(dp[i, 0],
                            dp[i - 1, 1] + a[i] + c);

        // Adding the element of array b if
        // previous element is from array a
        // with an extra penalty of integer C
        dp[i, 1] = Math.Min(dp[i, 1],
                            dp[i - 1, 0] + b[i] + c);

        // Adding the element of array b if
        // previous element is also from array b
        dp[i, 1] = Math.Min(dp[i, 1],
                            dp[i - 1, 1] + b[i]);
    }

    // Print the minimum sum
    Console.Write(Math.Min(dp[n - 1, 0],
                           dp[n - 1, 1]) + "\n");
}

// Driver Code
public static void Main(String[] args)
{

    // Given array arr1[] and arr2[]
    int []arr1 = { 7, 6, 18, 6, 16,
                   18, 1, 17, 17 };

    int []arr2 = { 6, 9, 3, 10, 9,
                   1, 10, 1, 5 };

    // Given cost
    int C = 2;

    int N = arr1.Length;

    // Function call
    minimumSum(arr1, arr2, C, N);
}
}

// This code is contributed by Rajput-Ji

Javascript

<script>

// JavaScript program for the above approach

// Function that prints minimum sum
// after selecting N elements
function minimumSum(a, b, c, n)
{

    // Initialise the dp array
    let dp =  new Array(n);
    for (var i = 0; i < dp.length; i++) {
    dp[i] = new Array(2);
    }

    for(let i = 0; i < n; i++)
    {
        for(let j = 0; j < 2; j++)
        {
            dp[i][j] = 1e6;
        }
    }

    // Base Case
    dp[0][0] = a[0];
    dp[0][1] = b[0];

    for(let i = 1; i < n; i++)
    {

        // Adding the element of array a if
        // previous element is also from array a
        dp[i][0] = Math.min(dp[i][0],
                            dp[i - 1][0] + a[i]);

        // Adding the element of array a if
        // previous element is from array b
        dp[i][0] = Math.min(dp[i][0],
                            dp[i - 1][1] + a[i] + c);

        // Adding the element of array b if
        // previous element is from array a
        // with an extra penalty of integer C
        dp[i][1] = Math.min(dp[i][1],
                            dp[i - 1][0] + b[i] + c);

        // Adding the element of array b if
        // previous element is also from array b
        dp[i][1] = Math.min(dp[i][1],
                            dp[i - 1][1] + b[i]);
    }

    // Print the minimum sum
    document.write(Math.min(dp[n - 1][0],
                              dp[n - 1][1]) + "<br/>");
}

// Driver Code

    // Given array arr1[] and arr2[]
    let arr1 = [ 7, 6, 18, 6, 16,
                   18, 1, 17, 17 ];

    let arr2 = [ 6, 9, 3, 10, 9,
                   1, 10, 1, 5 ];

    // Given cost
    let C = 2;

    let N = arr1.length;

    // Function call
    minimumSum(arr1, arr2, C, N);

</script>

Output:

49

Time Complexity: O(N), where N is the size of the given array.
Auxiliary Space: O(N), for creating additional dp array.