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# Maximum sum by picking elements from two arrays in order | Set 2

• Last Updated : 29 Apr, 2021

Given two arrays A[] and B[], each of size N, and two integers X and Y denoting the maximum number of elements that can be picked from A[] and B[] respectively, the task is to find the maximum possible sum by selecting N elements in such a way that for any index i, either A[i] or B[i] can be chosen.
Note: It is guaranteed that (X + Y) >= N.
Examples:

Input: A[] = {1, 2, 3, 4, 5}, B[] = {5, 4, 3, 2, 1}, X = 3, Y = 2
Output: 21
i = 0 -> 5 picked
i = 1 -> 4 picked
i = 2 -> 3 picked
i = 3 -> 4 picked
i = 4 -> 5 picked
5 + 4 + 3 + 4 + 5 = 21
Input: A[] = {1, 4, 3, 2, 7, 5, 9, 6}, B[] = {1, 2, 3, 6, 5, 4, 9, 8}, X = 4, Y = 4
Output: 43

Greedy Approach: The greedy approach to solve this problem has already been discussed in this post.
Dynamic Programming Approach: In this article, we are discussing a Dynamic Programming based solution.
Follow the steps below to solve the problem:

1. Utmost min (N, X) elements can be selected from A[]and min(N, Y) elements can be selected from B[].
2. Initialize a 2D array dp[][] such that dp[i][j] contains the maximum sum possible by selecting i elements from A[] and j elements from B[] where i and j range within min (N, X) and min (N, X) respectively.
3. Initialize a variable max_sum to store the maximum sum possible.
4. Traverse the array and for every array, element do the following:
• The (i + j)th element can either be A[i + j – 1] or B[i + j – 1].
• If the (i + j)th element is selected from A[], then the cost of (i + 1)th element in A[] will be added to the total cost. Hence, the cost of selecting (i + 1)th element is dp[i][j] = dp[ i – 1 ][ j ] + A[ i + j – 1 ] for this case.
• If the (i + j)th element is selected from B[], then the cost of selecting (i + 1)th element is dp[i][j] = dp[ i – 1 ][ j ] + B[ i + j – 1 ].
5. Now, the target is to maximize the cost. Hence, the recurrence relation is:

dp[i][j] = max(dp[ i – 1 ][ j ] + A[ i + j – 1 ], dp[ i – 1 ][ j ] + B[ i + j – 1 ])

1. Keep updating max_sum after every iteration. After complete traversal of the array print the final value of max_sum.

Below is the implementation of the above approach :

## C++

 `// C++ program to find maximum sum``// possible by selecting an element``// from one of two arrays for every index` `#include ``using` `namespace` `std;` `// Function to calculate maximum sum``int` `maximumSum(``int` `A[], ``int` `B[],``               ``int` `length,``               ``int` `X, ``int` `Y)``{``    ``int` `l = length;``    ``// Maximum elements that can be``    ``// chosen from array A``    ``int` `l1 = min(length, X);` `    ``// Maximum elements that can be``    ``// chosen from array B``    ``int` `l2 = min(length, Y);` `    ``int` `dp[l1 + 1][l2 + 1];``    ``memset``(dp, 0, ``sizeof``(dp));``    ``dp = 0;` `    ``// Stores the maximum``    ``// sum possible``    ``int` `max_sum = INT_MIN;` `    ``// Fill the dp[][] for base case when``    ``// all elements are selected from A[]``    ``for` `(``int` `i = 1; i <= l1; i++) {``        ``dp[i] = dp[i - 1] + A[i - 1];``        ``max_sum = max(max_sum, dp[i]);``    ``}` `    ``// Fill the dp[][] for base case when``    ``// all elements are selected from B[]``    ``for` `(``int` `i = 1; i <= l2; i++) {``        ``dp[i] = dp[i - 1] + B[i - 1];``        ``max_sum = max(max_sum, dp[i]);``    ``}` `    ``for` `(``int` `i = 1; i <= l1; i++) {``        ``for` `(``int` `j = 1; j <= l2; j++) {``            ``if` `(i + j <= l)``                ``dp[i][j]``                    ``= max(dp[i - 1][j]``                              ``+ A[i + j - 1],``                          ``dp[i][j - 1]``                              ``+ B[i + j - 1]);``            ``max_sum = max(dp[i][j],``                          ``max_sum);``        ``}``    ``}` `    ``// Return the final answer``    ``return` `max_sum;``}` `// Driver Program``int` `main()``{``    ``int` `A[] = { 1, 2, 3, 4, 5 };``    ``int` `B[] = { 5, 4, 3, 2, 1 };``    ``int` `X = 3, Y = 2;` `    ``int` `N = ``sizeof``(A) / ``sizeof``(A);` `    ``cout << maximumSum(A, B, N, X, Y);` `    ``return` `0;``}`

## Java

 `// Java program to find maximum sum``// possible by selecting an element``// from one of two arrays for every index``class` `GFG{``    ` `// Function to calculate maximum sum``static` `int` `maximumSum(``int` `A[], ``int` `B[],``                      ``int` `length, ``int` `X,``                      ``int` `Y)``{``    ``int` `l = length;``    ` `    ``// Maximum elements that can be``    ``// chosen from array A``    ``int` `l1 = Math.min(length, X);` `    ``// Maximum elements that can be``    ``// chosen from array B``    ``int` `l2 = Math.min(length, Y);` `    ``int` `dp[][] = ``new` `int` `[l1 + ``1``][l2 + ``1``];` `    ``// Stores the maximum``    ``// sum possible``    ``int` `max_sum = Integer.MIN_VALUE;` `    ``// Fill the dp[][] for base case when``    ``// all elements are selected from A[]``    ``for``(``int` `i = ``1``; i <= l1; i++)``    ``{``        ``dp[i][``0``] = dp[i - ``1``][``0``] + A[i - ``1``];``        ``max_sum = Math.max(max_sum, dp[i][``0``]);``    ``}` `    ``// Fill the dp[][] for base case when``    ``// all elements are selected from B[]``    ``for``(``int` `i = ``1``; i <= l2; i++)``    ``{``        ``dp[``0``][i] = dp[``0``][i - ``1``] + B[i - ``1``];``        ``max_sum = Math.max(max_sum, dp[``0``][i]);``    ``}` `    ``for``(``int` `i = ``1``; i <= l1; i++)``    ``{``        ``for``(``int` `j = ``1``; j <= l2; j++)``        ``{``            ``if` `(i + j <= l)``                ``dp[i][j] = Math.max(dp[i - ``1``][j] +``                                     ``A[i + j - ``1``],``                                    ``dp[i][j - ``1``] +``                                     ``B[i + j - ``1``]);``            ``max_sum = Math.max(dp[i][j], max_sum);``        ``}``    ``}``    ` `    ``// Return the final answer``    ``return` `max_sum;``}``    ` `// Driver code``public` `static` `void` `main (String[] args)``{``    ``int` `A[] = ``new` `int``[]{ ``1``, ``2``, ``3``, ``4``, ``5` `};``    ``int` `B[] = ``new` `int``[]{ ``5``, ``4``, ``3``, ``2``, ``1` `};``    ` `    ``int` `X = ``3``, Y = ``2``;``    ``int` `N = A.length;` `    ``System.out.println(maximumSum(A, B, N, X, Y));``}``}` `// This code is contributed by Pratima Pandey`

## Python

 `# Python3 program to find maximum sum``# possible by selecting an element``# from one of two arrays for every index``# Function to calculate maximum sum``def` `maximumSum(A, B, length, X, Y):``    ``l ``=` `length``    ` `    ``# Maximum elements that can be``    ``# chosen from array A``    ``l1 ``=` `min``(length, X)` `    ``# Maximum elements that can be``    ``# chosen from array B``    ``l2 ``=` `min``(length, Y)` `    ``dp``=``[[ ``0` `for` `i ``in` `range``(l2 ``+` `1``)]``        ``for` `i ``in` `range``(l1 ``+` `1``)]``    ``dp[``0``][``0``] ``=` `0` `    ``# Stores the maximum``    ``# sum possible``    ``max_sum ``=` `-``10` `*` `9` `    ``# Fill the dp[]for base case when``    ``# all elements are selected from A[]``    ``for` `i ``in` `range``(``1``, l1 ``+` `1``):``        ``dp[i][``0``] ``=` `dp[i ``-` `1``][``0``] ``+` `A[i ``-` `1``]``        ``max_sum ``=` `max``(max_sum, dp[i][``0``])`  `    ``# Fill the dp[]for base case when``    ``# all elements are selected from B[]``    ``for` `i ``in` `range``(``1``, l2 ``+` `1``):``        ``dp[``0``][i] ``=` `dp[``0``][i ``-` `1``] ``+` `B[i ``-` `1``]``        ``max_sum ``=` `max``(max_sum, dp[``0``][i])` `    ``for` `i ``in` `range``(``1``, l1 ``+` `1``):``        ``for` `j ``in` `range``(``1``, l2 ``+` `1``):``            ``if` `(i ``+` `j <``=` `l):``                ``dp[i][j]``=` `max``(dp[i ``-` `1``][j] ``+` `A[i ``+` `j ``-` `1``],``                              ``dp[i][j ``-` `1``] ``+` `B[i ``+` `j ``-` `1``])``            ``max_sum ``=` `max``(dp[i][j], max_sum)` `    ``# Return the final answer``    ``return` `max_sum` `# Driver Program``if` `__name__ ``=``=` `'__main__'``:``    ``A``=`  `[``1``, ``2``, ``3``, ``4``, ``5``]``    ``B``=`  `[``5``, ``4``, ``3``, ``2``, ``1``]``    ``X ``=` `3``    ``Y ``=` `2``    ``N ``=` `len``(A)``    ``print``(maximumSum(A, B, N, X, Y))` `# This code is contributed by Mohit Kumar 29`

## C#

 `// C# program to find maximum sum``// possible by selecting an element``// from one of two arrays for every index``using` `System;` `class` `GFG{``    ` `// Function to calculate maximum sum``static` `int` `maximumSum(``int` `[]A, ``int` `[]B,``                      ``int` `length, ``int` `X,``                      ``int` `Y)``{``    ``int` `l = length;``    ` `    ``// Maximum elements that can be``    ``// chosen from array A``    ``int` `l1 = Math.Min(length, X);` `    ``// Maximum elements that can be``    ``// chosen from array B``    ``int` `l2 = Math.Min(length, Y);` `    ``int` `[,]dp = ``new` `int` `[l1 + 1, l2 + 1];` `    ``// Stores the maximum``    ``// sum possible``    ``int` `max_sum = ``int``.MinValue;` `    ``// Fill the [,]dp for base case when``    ``// all elements are selected from []A``    ``for``(``int` `i = 1; i <= l1; i++)``    ``{``        ``dp[i, 0] = dp[i - 1, 0] + A[i - 1];``        ``max_sum = Math.Max(max_sum, dp[i, 0]);``    ``}` `    ``// Fill the [,]dp for base case when``    ``// all elements are selected from []B``    ``for``(``int` `i = 1; i <= l2; i++)``    ``{``        ``dp[0, i] = dp[0, i - 1] + B[i - 1];``        ``max_sum = Math.Max(max_sum, dp[0, i]);``    ``}` `    ``for``(``int` `i = 1; i <= l1; i++)``    ``{``        ``for``(``int` `j = 1; j <= l2; j++)``        ``{``            ``if` `(i + j <= l)``                ``dp[i, j] = Math.Max(dp[i - 1, j] +``                                     ``A[i + j - 1],``                                    ``dp[i, j - 1] +``                                     ``B[i + j - 1]);``            ``max_sum = Math.Max(dp[i, j], max_sum);``        ``}``    ``}``    ` `    ``// Return the readonly answer``    ``return` `max_sum;``}``    ` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``int` `[]A = ``new` `int``[]{ 1, 2, 3, 4, 5 };``    ``int` `[]B = ``new` `int``[]{ 5, 4, 3, 2, 1 };``    ` `    ``int` `X = 3, Y = 2;``    ``int` `N = A.Length;` `    ``Console.WriteLine(maximumSum(A, B, N, X, Y));``}``}` `// This code is contributed by Amit Katiyar`

## Javascript

 ``
Output:
`21`

Time Complexity: O(N2)
Auxiliary Space: O(N2)

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