Maximum sum from three arrays such that picking elements consecutively from same is not allowed

Difficulty Level : Medium
Last Updated : 12 Sep, 2021

Given three arrays A[], B[] and C[] of N integers. We can choose N elements from this array such that for every index i only one element can be chosen from these array i.e. either A[i], B[i] or C[i] and no two consecutive elements can be chosen from the same array. The task is to print the maximum sum of numbers that we can make by choosing elements from these arrays.

Examples:

Input: a[] = {10, 20, 30}, b[] = {40, 50, 60}, c[] = {70, 80, 90}
Output: 210
70 + 50 + 90 = 210
Input: a[] = {6, 8, 2, 7, 4, 2, 7}, b[] = {7, 8, 5, 8, 6, 3, 5}, c[] = {8, 3, 2, 6, 8, 4, 1}
Output: 46
Choose elements from C, A, B, A, C, B and A.

Approach: The above problem can be solved using Dynamic Programming. Let dp[i][j] be considered the maximum sum till i if element from j-th array is chosen. We can select element from any array for the first index, but later on recursively we can choose an element only from the rest two arrays for the next step. The maximum sum returned by all of the combinations will be our answer. Use memoization to avoid repetitive and multiple same function calls.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
const int N = 3;
 
// Function to return the maximum sum
int FindMaximumSum(int ind, int kon, int a[], int b[],
                   int c[], int n, int dp[][N])
{
 
    // Base case
    if (ind == n)
        return 0;
 
    // Already visited
    if (dp[ind][kon] != -1)
        return dp[ind][kon];
    int ans = -1e9 + 5;
 
    // If the element has been taken
    // from first array in previous step
    if (kon == 0) {
        ans = max(ans, b[ind] + FindMaximumSum(ind + 1,
                                               1, a, b,
                                               c, n, dp));
        ans = max(ans, c[ind] + FindMaximumSum(ind + 1,
                                               2, a, b,
                                               c, n, dp));
    }
 
    // If the element has been taken
    // from second array in previous step
    else if (kon == 1) {
        ans = max(ans, a[ind] + FindMaximumSum(ind + 1,
                                               0, a, b,
                                               c, n, dp));
        ans = max(ans, c[ind] + FindMaximumSum(ind + 1,
                                               2, a, b,
                                               c, n, dp));
    }
 
    // If the element has been taken
    // from third array in previous step
    else if (kon == 2) {
        ans = max(ans, a[ind] + FindMaximumSum(ind + 1,
                                               1, a, b,
                                               c, n, dp));
        ans = max(ans, b[ind] + FindMaximumSum(ind + 1,
                                               0, a, b,
                                               c, n, dp));
    }
 
    return dp[ind][kon] = ans;
}
 
// Driver code
int main()
{
    int a[] = { 6, 8, 2, 7, 4, 2, 7 };
    int b[] = { 7, 8, 5, 8, 6, 3, 5 };
    int c[] = { 8, 3, 2, 6, 8, 4, 1 };
    int n = sizeof(a) / sizeof(a[0]);
    int dp[n][N];
    memset(dp, -1, sizeof dp);
 
    // Pick element from first array
    int x = FindMaximumSum(0, 0, a, b, c, n, dp);
 
    // Pick element from second array
    int y = FindMaximumSum(0, 1, a, b, c, n, dp);
 
    // Pick element from third array
    int z = FindMaximumSum(0, 2, a, b, c, n, dp);
 
    // Print the maximum of them
    cout << max(x, max(y, z));
 
    return 0;
}

Java

// Java program for the above approach
 
class GFG {
     
static int N = 3;
  
// Function to return the maximum sum
static int FindMaximumSum(int ind, int kon, int a[],
               int b[], int c[], int n, int dp[][])
                    
{
    // Base case
    if (ind == n)
        return 0;
  
    // Already visited
    if (dp[ind][kon] != -1)
        return dp[ind][kon];
    int ans = (int) (-1e9 + 5);
  
    // If the element has been taken
    // from first array in previous step
    if (kon == 0) {
        ans = Math.max(ans, b[ind] +
              FindMaximumSum(ind + 1,
                  1, a, b,c, n, dp));
                                                
                                                
        ans = Math.max(ans, c[ind] +
              FindMaximumSum(ind + 1,
                 2, a, b,c, n, dp));
                                                                                        
    }
  
    // If the element has been taken
    // from second array in previous step
    else if (kon == 1) {
        ans = Math.max(ans, a[ind] +
              FindMaximumSum(ind + 1,
                 0, a, b, c, n, dp));
                                                                                 
        ans = Math.max(ans, c[ind] +
              FindMaximumSum(ind + 1,
                 2, a, b, c, n, dp));
                                                                                           
    }
  
    // If the element has been taken
    // from third array in previous step
    else if (kon == 2) {
        ans = Math.max(ans, a[ind] +
              FindMaximumSum(ind + 1,
                 1, a, b, c, n, dp));
                                                
                                                
        ans = Math.max(ans, b[ind] +
              FindMaximumSum(ind + 1,
                 0, a, b, c, n, dp));
                                                
                                                
    }
  
    return dp[ind][kon] = ans;
}
  
// Driver code
public static void main(String[] args) {
 
    int a[] = { 6, 8, 2, 7, 4, 2, 7 };
    int b[] = { 7, 8, 5, 8, 6, 3, 5 };
    int c[] = { 8, 3, 2, 6, 8, 4, 1 };
    int n = a.length;
    int dp[][] = new int[n][N];
 
    for(int i = 0; i < n; i++) {
 
        for(int j = 0; j < N; j++) {
            dp[i][j] =- 1;
        }
    }
  
    // Pick element from first array
    int x = FindMaximumSum(0, 0, a, b, c, n, dp);
  
    // Pick element from second array
    int y = FindMaximumSum(0, 1, a, b, c, n, dp);
  
    // Pick element from third array
    int z = FindMaximumSum(0, 2, a, b, c, n, dp);
  
    // Print the maximum of them
    System.out.println(Math.max(x, Math.max(y, z)));
  
    }
}
// This code has been contributed
// by 29AjayKumar

Python3

# Python3 implementation of the approach
 
# Function to return the maximum sum
def FindMaximumSum(ind, kon, a, b, c, n, dp):
 
    # Base case
    if ind == n:
        return 0
 
    # Already visited
    if dp[ind][kon] != -1:
        return dp[ind][kon]
     
    ans = -10 ** 9 + 5
 
    # If the element has been taken
    # from first array in previous step
    if kon == 0:
        ans = max(ans, b[ind] +
                  FindMaximumSum(ind + 1, 1,
                                 a, b, c, n, dp))
        ans = max(ans, c[ind] +
                  FindMaximumSum(ind + 1, 2,
                                 a, b, c, n, dp))
     
    # If the element has been taken
    # from second array in previous step
    elif kon == 1:
        ans = max(ans, a[ind] +
                  FindMaximumSum(ind + 1, 0,
                                 a, b, c, n, dp))
        ans = max(ans, c[ind] +
                  FindMaximumSum(ind + 1, 2,
                                 a, b, c, n, dp))
     
    # If the element has been taken
    # from third array in previous step
    elif kon == 2:
        ans = max(ans, a[ind] +
                  FindMaximumSum(ind + 1, 1,
                                 a, b, c, n, dp))
        ans = max(ans, b[ind] +
                  FindMaximumSum(ind + 1, 0,
                                 a, b, c, n, dp))
     
    dp[ind][kon] = ans
    return ans
 
# Driver code
if __name__ == "__main__":
     
    N = 3
    a = [6, 8, 2, 7, 4, 2, 7]
    b = [7, 8, 5, 8, 6, 3, 5]
    c = [8, 3, 2, 6, 8, 4, 1]
    n = len(a)
     
    dp = [[-1 for i in range(N)]
              for j in range(n)]
 
    # Pick element from first array
    x = FindMaximumSum(0, 0, a, b, c, n, dp)
 
    # Pick element from second array
    y = FindMaximumSum(0, 1, a, b, c, n, dp)
 
    # Pick element from third array
    z = FindMaximumSum(0, 2, a, b, c, n, dp)
 
    # Print the maximum of them
    print(max(x, y, z))
 
# This code is contributed
# by Rituraj Jain

C#

// C# program for the above approach
using System;
 
class GFG
{
     
    static int N = 3;
     
    // Function to return the maximum sum
    static int FindMaximumSum(int ind, int kon, int []a,
                int []b, int []c, int n, int [,]dp)
                         
    {
        // Base case
        if (ind == n)
            return 0;
     
        // Already visited
        if (dp[ind,kon] != -1)
            return dp[ind,kon];
        int ans = (int) (-1e9 + 5);
     
        // If the element has been taken
        // from first array in previous step
        if (kon == 0)
        {
            ans = Math.Max(ans, b[ind] +
                FindMaximumSum(ind + 1,
                    1, a, b,c, n, dp));
                                                     
                                                     
            ans = Math.Max(ans, c[ind] +
                FindMaximumSum(ind + 1,
                    2, a, b,c, n, dp));
                                                                                             
        }
     
        // If the element has been taken
        // from second array in previous step
        else if (kon == 1)
        {
            ans = Math.Max(ans, a[ind] +
                FindMaximumSum(ind + 1,
                    0, a, b, c, n, dp));
                                                                                     
            ans = Math.Max(ans, c[ind] +
                FindMaximumSum(ind + 1,
                    2, a, b, c, n, dp));
                                                                                                 
        }
     
        // If the element has been taken
        // from third array in previous step
        else if (kon == 2)
        {
            ans = Math.Max(ans, a[ind] +
                FindMaximumSum(ind + 1,
                    1, a, b, c, n, dp));
                                                     
                                                     
            ans = Math.Max(ans, b[ind] +
                FindMaximumSum(ind + 1,
                    0, a, b, c, n, dp));
                                                     
                                                     
        }
     
        return dp[ind,kon] = ans;
    }
     
    // Driver code
    public static void Main()
    {
     
        int []a = { 6, 8, 2, 7, 4, 2, 7 };
        int []b = { 7, 8, 5, 8, 6, 3, 5 };
        int []c = { 8, 3, 2, 6, 8, 4, 1 };
        int n = a.Length;
        int [,]dp = new int[n,N];
     
        for(int i = 0; i < n; i++)
        {
     
            for(int j = 0; j < N; j++)
            {
                dp[i,j] =- 1;
            }
        }
     
        // Pick element from first array
        int x = FindMaximumSum(0, 0, a, b, c, n, dp);
     
        // Pick element from second array
        int y = FindMaximumSum(0, 1, a, b, c, n, dp);
     
        // Pick element from third array
        int z = FindMaximumSum(0, 2, a, b, c, n, dp);
     
        // Print the maximum of them
        Console.WriteLine(Math.Max(x, Math.Max(y, z)));
     
        }
}
 
// This code has been contributed by Ryuga

PHP

<?php
// PHP implementation of the approach
$N = 3;
 
// Function to return the maximum sum
function FindMaximumSum($ind, $kon, $a,
                        $b, $c, $n, $dp)
{
    global $N;
     
    // Base case
    if ($ind == $n)
        return 0;
 
    // Already visited
    if ($dp[$ind][$kon] != -1)
        return $dp[$ind][$kon];
    $ans = -1e9 + 5;
 
    // If the element has been taken
    // from first array in previous step
    if ($kon == 0)
    {
        $ans = max($ans, $b[$ind] +
                   FindMaximumSum($ind + 1, 1, $a,
                                  $b, $c, $n, $dp));
        $ans = max($ans, $c[$ind] +
                   FindMaximumSum($ind + 1, 2, $a,
                                  $b, $c, $n, $dp));
    }
 
    // If the element has been taken
    // from second array in previous step
    else if ($kon == 1)
    {
        $ans = max($ans, $a[$ind] +
                   FindMaximumSum($ind + 1, 0,
                                  $a, $b, $c, $n, $dp));
        $ans = max($ans, $c[$ind] +
                   FindMaximumSum($ind + 1, 2,
                                  $a, $b, $c, $n, $dp));
    }
 
    // If the element has been taken
    // from third array in previous step
    else if ($kon == 2)
    {
        $ans = max($ans, $a[$ind] +
                   FindMaximumSum($ind + 1, 1,
                                  $a, $b, $c, $n, $dp));
        $ans = max($ans, $b[$ind] +
                   FindMaximumSum($ind + 1, 0,
                                  $a, $b, $c, $n, $dp));
    }
 
    return $dp[$ind][$kon] = $ans;
}
 
// Driver code
$a = array( 6, 8, 2, 7, 4, 2, 7 );
$b = array( 7, 8, 5, 8, 6, 3, 5 );
$c = array( 8, 3, 2, 6, 8, 4, 1 );
$n = count($a);
$dp = array_fill(0, $n,
      array_fill(0, $N, -1));
 
// Pick element from first array
$x = FindMaximumSum(0, 0, $a, $b,
                      $c, $n, $dp);
 
// Pick element from second array
$y = FindMaximumSum(0, 1, $a, $b,
                      $c, $n, $dp);
 
// Pick element from third array
$z = FindMaximumSum(0, 2, $a, $b,
                      $c, $n, $dp);
 
// Print the maximum of them
print(max($x, max($y, $z)));
 
// This code is contributed by mits
?>

Javascript

<script>
 
// JavaScript implementation of the approach
var N = 3;
 
// Function to return the maximum sum
function FindMaximumSum(ind, kon, a, b, c, n, dp)
{
 
    // Base case
    if (ind == n)
        return 0;
 
    // Already visited
    if (dp[ind][kon] != -1)
        return dp[ind][kon];
    var ans = -1000000005;
 
    // If the element has been taken
    // from first array in previous step
    if (kon == 0) {
        ans = Math.max(ans, b[ind] + FindMaximumSum(ind + 1,
                                               1, a, b,
                                               c, n, dp));
        ans = Math.max(ans, c[ind] + FindMaximumSum(ind + 1,
                                               2, a, b,
                                               c, n, dp));
    }
 
    // If the element has been taken
    // from second array in previous step
    else if (kon == 1) {
        ans = Math.max(ans, a[ind] + FindMaximumSum(ind + 1,
                                               0, a, b,
                                               c, n, dp));
        ans = Math.max(ans, c[ind] + FindMaximumSum(ind + 1,
                                               2, a, b,
                                               c, n, dp));
    }
 
    // If the element has been taken
    // from third array in previous step
    else if (kon == 2) {
        ans = Math.max(ans, a[ind] + FindMaximumSum(ind + 1,
                                               1, a, b,
                                               c, n, dp));
        ans = Math.max(ans, b[ind] + FindMaximumSum(ind + 1,
                                               0, a, b,
                                               c, n, dp));
    }
 
    return dp[ind][kon] = ans;
}
 
// Driver code
var a = [ 6, 8, 2, 7, 4, 2, 7 ];
var b = [ 7, 8, 5, 8, 6, 3, 5 ];
var c = [ 8, 3, 2, 6, 8, 4, 1 ];
var n = a.length;
var dp = Array.from(Array(n), ()=> Array(n).fill(-1));
// Pick element from first array
var x = FindMaximumSum(0, 0, a, b, c, n, dp);
// Pick element from second array
var y = FindMaximumSum(0, 1, a, b, c, n, dp);
// Pick element from third array
var z = FindMaximumSum(0, 2, a, b, c, n, dp);
// Print the maximum of them
document.write( Math.max(x, Math.max(y, z)));
 
</script>

Output:

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

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