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Minimum total sum from the given two arrays
• Last Updated : 07 Aug, 2020

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`
Output:
```49
```

Time Complexity: O(N)

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