Given two arrays arr1[] and arr2[] each of size N. The task is to choose some elements from both the arrays such that no two elements have the same index and no two consecutive numbers can be selected from a single array. Find the maximum sum possible of above-chosen numbers.
Examples:
Input : arr1[] = {9, 3, 5, 7, 3}, arr2[] = {5, 8, 1, 4, 5}
Output : 29
Select first, third and fivth element from the first array.
Select the second and fourth element from the second array.Input : arr1[] = {1, 2, 9}, arr2[] = {10, 1, 1}
Output : 19
Select last element from the first array and first element from the second array.
Approach :
This problem is based on dynamic programming.
.
Recursion realtions are :
dp(i, 1)=max(dp (i – 1, 2) + arr(i, 1), dp(i – 1, 3) + arr(i, 1), arr(i, 1) );
dp(i, 2)=max(dp(i – 1, 1) + arr(i, 2 ), dp(i – 1, 3) + arr (i, 2), arr(i, 2));
dp(i, 3)=max(dp(i- 1, 1), dp( i-1, 2) ).
We don’t actually need dp( i, 3), if we update dp(i, 1) as max(dp(i, 1), dp(i-1, 1)) and dp(i, 2) as max(dp(i, 2), dp(i-1, 2)).
Thus, dp(i, j) is the maximum total sum of the elements that are selected if the last element was taken from the position (i-1, 1) or less. The same with dp(i, 2). Therefore the answer to the above problem is max(dp(n, 1), dp(n, 2)).
Below is the implementation of the above approach :
C++
// CPP program to maximum sum // combination from two arrays #include <bits/stdc++.h> using namespace std; // Function to maximum sum // combination from two arrays int Max_Sum(int arr1[], int arr2[], int n) { // To store dp value int dp[n][2]; // For loop to calculate the value of dp for (int i = 0; i < n; i++) { if(i==0) { dp[i][0] = arr1[i]; dp[i][1] = arr2[i]; continue; } dp[i][0] = max(dp[i - 1][0], dp[i - 1][1] + arr1[i]); dp[i][1] = max(dp[i - 1][1], dp[i - 1][0] + arr2[i]); } // Return the required answer return max(dp[n-1][0], dp[n-1][1]); } // Driver code int main() { int arr1[] = {9, 3, 5, 7, 3}; int arr2[] = {5, 8, 1, 4, 5}; int n = sizeof(arr1) / sizeof(arr1[0]); // Function call cout << Max_Sum(arr1, arr2, n); return 0; } |
Java
// Java program to maximum sum // combination from two arrays class GFG { // Function to maximum sum // combination from two arrays static int Max_Sum(int arr1[], int arr2[], int n) { // To store dp value int [][]dp = new int[n][2]; // For loop to calculate the value of dp for (int i = 0; i < n; i++) { if(i == 0) { dp[i][0] = arr1[i]; dp[i][1] = arr2[i]; continue; } dp[i][0] = Math.max(dp[i - 1][0], dp[i - 1][1] + arr1[i]); dp[i][1] = Math.max(dp[i - 1][1], dp[i - 1][0] + arr2[i]); } // Return the required answer return Math.max(dp[n - 1][0], dp[n - 1][1]); } // Driver code public static void main(String[] args) { int arr1[] = {9, 3, 5, 7, 3}; int arr2[] = {5, 8, 1, 4, 5}; int n = arr1.length; // Function call System.out.println(Max_Sum(arr1, arr2, n)); } } // This code is contributed // by PrinciRaj1992 |
Python3
# Python3 program to maximum sum # combination from two arrays # Function to maximum sum # combination from two arrays def Max_Sum(arr1, arr2, n): # To store dp value dp = [[0 for i in range(2)] for j in range(n)] # For loop to calculate the value of dp for i in range(n): if(i == 0): dp[i][0] = arr1[i] dp[i][1] = arr2[i] continue else: dp[i][0] = max(dp[i - 1][0], dp[i - 1][1] + arr1[i]) dp[i][1] = max(dp[i - 1][1], dp[i - 1][0] + arr2[i]) # Return the required answer return max(dp[n - 1][0], dp[n - 1][1]) # Driver code if __name__ == '__main__': arr1 = [9, 3, 5, 7, 3] arr2 = [5, 8, 1, 4, 5] n = len(arr1) # Function call print(Max_Sum(arr1, arr2, n)) # This code is contributed by # Surendra_Gangwar |
C#
// C# program to maximum sum // combination from two arrays using System; class GFG { // Function to maximum sum // combination from two arrays static int Max_Sum(int []arr1, int []arr2, int n) { // To store dp value int [,]dp = new int[n, 2]; // For loop to calculate the value of dp for (int i = 0; i < n; i++) { if(i == 0) { dp[i, 0] = arr1[i]; dp[i, 1] = arr2[i]; continue; } dp[i, 0] = Math.Max(dp[i - 1, 0], dp[i - 1, 1] + arr1[i]); dp[i, 1] = Math.Max(dp[i - 1, 1], dp[i - 1, 0] + arr2[i]); } // Return the required answer return Math.Max(dp[n - 1, 0], dp[n - 1, 1]); } // Driver code public static void Main() { int []arr1 = {9, 3, 5, 7, 3}; int []arr2 = {5, 8, 1, 4, 5}; int n = arr1.Length; // Function call Console.WriteLine(Max_Sum(arr1, arr2, n)); } } // This code is contributed // by anuj_67.. |
29
Time Complexity: O(N)
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