Maximum sum combination from two arrays
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++
#include <bits/stdc++.h>
using namespace std;
int Max_Sum( int arr1[], int arr2[], int n)
{
int dp[n][2];
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 max(dp[n-1][0], dp[n-1][1]);
}
int main()
{
int arr1[] = {9, 3, 5, 7, 3};
int arr2[] = {5, 8, 1, 4, 5};
int n = sizeof (arr1) / sizeof (arr1[0]);
cout << Max_Sum(arr1, arr2, n);
return 0;
}
|
Java
class GFG
{
static int Max_Sum( int arr1[],
int arr2[], int n)
{
int [][]dp = new int [n][ 2 ];
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 Math.max(dp[n - 1 ][ 0 ],
dp[n - 1 ][ 1 ]);
}
public static void main(String[] args)
{
int arr1[] = { 9 , 3 , 5 , 7 , 3 };
int arr2[] = { 5 , 8 , 1 , 4 , 5 };
int n = arr1.length;
System.out.println(Max_Sum(arr1, arr2, n));
}
}
|
Python3
def Max_Sum(arr1, arr2, n):
dp = [[ 0 for i in range ( 2 )]
for j in range (n)]
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 max (dp[n - 1 ][ 0 ],
dp[n - 1 ][ 1 ])
if __name__ = = '__main__' :
arr1 = [ 9 , 3 , 5 , 7 , 3 ]
arr2 = [ 5 , 8 , 1 , 4 , 5 ]
n = len (arr1)
print (Max_Sum(arr1, arr2, n))
|
C#
using System;
class GFG
{
static int Max_Sum( int []arr1,
int []arr2, int n)
{
int [,]dp = new int [n, 2];
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 Math.Max(dp[n - 1, 0],
dp[n - 1, 1]);
}
public static void Main()
{
int []arr1 = {9, 3, 5, 7, 3};
int []arr2 = {5, 8, 1, 4, 5};
int n = arr1.Length;
Console.WriteLine(Max_Sum(arr1, arr2, n));
}
}
|
Javascript
<script>
function Max_Sum(arr1, arr2, n)
{
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 (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 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;
document.write(Max_Sum(arr1, arr2, n));
</script>
|
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;
int Max_Sum( int arr1[], int arr2[], int n)
{
int prev1 = arr1[0], prev2 = arr2[0];
int curr1, curr2;
for ( int i = 1; i < n; i++)
{
curr1 = max(prev1, prev2 + arr1[i]);
curr2 = max(prev2, prev1 + arr2[i]);
prev1 = curr1;
prev2 = curr2;
}
return max(prev1, prev2);
}
int main()
{
int arr1[] = {9, 3, 5, 7, 3};
int arr2[] = {5, 8, 1, 4, 5};
int n = sizeof (arr1) / sizeof (arr1[0]);
cout << Max_Sum(arr1, arr2, n);
return 0;
}
|
Java
import java.util.*;
public class Main
{
static int Max_Sum( int [] arr1, int [] arr2, int n)
{
int prev1 = arr1[ 0 ], prev2 = arr2[ 0 ];
int curr1, curr2;
for ( int i = 1 ; i < n; i++) {
curr1 = Math.max(prev1, prev2 + arr1[i]);
curr2 = Math.max(prev2, prev1 + arr2[i]);
prev1 = curr1;
prev2 = curr2;
}
return Math.max(prev1, prev2);
}
public static void main(String[] args)
{
int [] arr1 = { 9 , 3 , 5 , 7 , 3 };
int [] arr2 = { 5 , 8 , 1 , 4 , 5 };
int n = arr1.length;
System.out.println(Max_Sum(arr1, arr2, n));
}
}
|
Python3
def Max_Sum(arr1, arr2, n):
prev1 = arr1[ 0 ]
prev2 = arr2[ 0 ]
curr1 = 0
curr2 = 0
for i in range ( 1 , n):
curr1 = max (prev1, prev2 + arr1[i])
curr2 = max (prev2, prev1 + arr2[i])
prev1 = curr1
prev2 = curr2
return max (prev1, prev2)
arr1 = [ 9 , 3 , 5 , 7 , 3 ]
arr2 = [ 5 , 8 , 1 , 4 , 5 ]
n = len (arr1)
print (Max_Sum(arr1, arr2, n))
|
C#
using System;
class MainClass {
public static int Max_Sum( int [] arr1, int [] arr2, int n)
{
int prev1 = arr1[0], prev2 = arr2[0];
int curr1, curr2;
for ( int i = 1; i < n; i++) {
curr1 = Math.Max(prev1, prev2 + arr1[i]);
curr2 = Math.Max(prev2, prev1 + arr2[i]);
prev1 = curr1;
prev2 = curr2;
}
return Math.Max(prev1, prev2);
}
public static void Main()
{
int [] arr1 = { 9, 3, 5, 7, 3 };
int [] arr2 = { 5, 8, 1, 4, 5 };
int n = arr1.Length;
Console.WriteLine(Max_Sum(arr1, arr2, n));
}
}
|
Javascript
function Max_Sum(arr1, arr2, n) {
let prev1 = arr1[0],
prev2 = arr2[0];
let curr1, curr2;
for (let i = 1; i < n; i++) {
curr1 = Math.max(prev1, prev2 + arr1[i]);
curr2 = Math.max(prev2, prev1 + arr2[i]);
prev1 = curr1;
prev2 = curr2;
}
return Math.max(prev1, prev2);
}
let arr1 = [9, 3, 5, 7, 3];
let arr2 = [5, 8, 1, 4, 5];
let n = arr1.length;
console.log(Max_Sum(arr1, arr2, n));
|
Output
29
Time Complexity: O(N), where N is the length of the given arrays.
Auxiliary Space: O(1)
Last Updated :
19 Apr, 2023
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