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# 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] and dp[n-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 ` `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 > dp(n, ` `                            ``vector<``int``>(2, ` `                                        ``1e6)); `   `    ``// Base Case ` `    ``dp = a; ` `    ``dp = b; `   `    ``for` `(``int` `i = 1; i < n; i++) { `   `        ``// Adding the element of array a if ` `        ``// previous element is also from array a ` `        ``dp[i] = min(dp[i], ` `                    ``dp[i - 1] + a[i]); `   `        ``// Adding the element of array a if ` `        ``// previous element is from array b ` `        ``dp[i] = min(dp[i], ` `                    ``dp[i - 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] = min(dp[i], ` `                    ``dp[i - 1] + b[i] + c); `   `        ``// Adding the element of array b if ` `        ``// previous element is also from array b ` `        ``dp[i] = min(dp[i], ` `                    ``dp[i - 1] + b[i]); ` `    ``} `   `    ``// Print the minimum sum ` `    ``cout << min(dp[n - 1], ` `                ``dp[n - 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); `   `    ``// 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;` `    ``dp[0, 1] = b;`   `    ``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

 ``

Output:

`49`

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

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