Given two arrays arr1[] and arr2[] each of size N. The task is to choose some elements from both 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 the above-chosen numbers.
Examples:
Input : arr1[] = {9, 3, 5, 7, 3}, arr2[] = {5, 8, 1, 4, 5}
Output : 29
Select first, third and fifth 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.
- Let dp(i, 1) be the maximum sum of the newly selected elements if the last element was taken from the position(i-1, 1).
- dp(i, 2) is the same but the last element taken has the position (i-1, 2)
- dp(i, 3) the same but we didn’t take any element from position i-1
Recursion relations 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 arraysint 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 codeint 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 arraysclass GFG{// Function to maximum sum // combination from two arraysstatic 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 codepublic 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 arraysdef 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 codeif __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 arraysusing System;class GFG{// Function to maximum sum // combination from two arraysstatic 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 codepublic 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.. |
Javascript
<script> // Javascript program to maximum sum combination from two arrays // Function to maximum sum // combination from two arrays function Max_Sum(arr1, arr2, n) { // To store dp value let dp = new Array(n); for (let i = 0; i < n; i++) { dp[i] = new Array(2); for (let j = 0; j < 2; j++) { dp[i][j] = 0; } } // For loop to calculate the value of dp for (let 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]); } let arr1 = [9, 3, 5, 7, 3]; let arr2 = [5, 8, 1, 4, 5]; let n = arr1.length; // Function call document.write(Max_Sum(arr1, arr2, n));</script> |
Output:
29
Time Complexity: O(N), where N is the length of the given arrays.
Auxiliary Space: O(N)
Efficient approach : Space optimization O(1)
To optimize the space complexity since we only need to access the values of dp[i] and dp[i-1], we can just use variables to store these values instead of an entire array. This way, the space complexity will be reduced from O(N) to O(1)
Implementation Steps:
- Initialize prev1 and prev2 with the first elements of arr1 and arr2 respectively.
- Create two variables curr1 and curr2.
- Use a loop to iterate over the arrays from index 1 to n-1.
- Update prev1 and prev2 to curr1 and curr2 respectively for further iterations.
- Return the maximum of prev1 and prev2 as the maximum sum combination from the two arrays.
Implementation :
C++
#include <bits/stdc++.h>using namespace std;// Function to maximum sum combination from two arraysint Max_Sum(int arr1[], int arr2[], int n){ // To store dp value int prev1 = arr1[0], prev2 = arr2[0]; int curr1, curr2; // For loop to calculate the value of dp for (int i = 1; i < n; i++) { curr1 = max(prev1, prev2 + arr1[i]); curr2 = max(prev2, prev1 + arr2[i]); // assigning values for further iteration prev1 = curr1; prev2 = curr2; } // Return the required answer return max(prev1, prev2);}// Driver codeint 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
import java.util.*;public class Main { // Function to maximum sum combination from two arrays static int Max_Sum(int[] arr1, int[] arr2, int n) { // To store dp value int prev1 = arr1[0], prev2 = arr2[0]; int curr1, curr2; // For loop to calculate the value of dp for (int i = 1; i < n; i++) { curr1 = Math.max(prev1, prev2 + arr1[i]); curr2 = Math.max(prev2, prev1 + arr2[i]); // assigning values for further iteration prev1 = curr1; prev2 = curr2; } // Return the required answer return Math.max(prev1, prev2); } // 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)); }} |
Python3
def Max_Sum(arr1, arr2, n): # To store dp value prev1 = arr1[0] prev2 = arr2[0] curr1 = 0 curr2 = 0 # For loop to calculate the value of dp for i in range(1, n): curr1 = max(prev1, prev2 + arr1[i]) curr2 = max(prev2, prev1 + arr2[i]) # assigning values for further iteration prev1 = curr1 prev2 = curr2 # Return the required answer return max(prev1, prev2)# Driver codearr1 = [9, 3, 5, 7, 3]arr2 = [5, 8, 1, 4, 5]n = len(arr1)# Function callprint(Max_Sum(arr1, arr2, n)) |
C#
using System;class MainClass { // Function to maximum sum combination from two arrays public static int Max_Sum(int[] arr1, int[] arr2, int n) { // To store dp value int prev1 = arr1[0], prev2 = arr2[0]; int curr1, curr2; // For loop to calculate the value of dp for (int i = 1; i < n; i++) { curr1 = Math.Max(prev1, prev2 + arr1[i]); curr2 = Math.Max(prev2, prev1 + arr2[i]); // assigning values for further iteration prev1 = curr1; prev2 = curr2; } // Return the required answer return Math.Max(prev1, prev2); } // 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)); }} |
Javascript
// Function to maximum sum combination from two arraysfunction Max_Sum(arr1, arr2, n) { // To store dp value let prev1 = arr1[0], prev2 = arr2[0]; let curr1, curr2; // For loop to calculate the value of dp for (let i = 1; i < n; i++) { curr1 = Math.max(prev1, prev2 + arr1[i]); curr2 = Math.max(prev2, prev1 + arr2[i]); // assigning values for further iteration prev1 = curr1; prev2 = curr2; } // Return the required answer return Math.max(prev1, prev2);}// Driver codelet arr1 = [9, 3, 5, 7, 3];let arr2 = [5, 8, 1, 4, 5];let n = arr1.length;// Function callconsole.log(Max_Sum(arr1, arr2, n));//This code is contributed by sarojmcy2e |
Output
29
Time Complexity: O(N), where N is the length of the given arrays.
Auxiliary Space: O(1)
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