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Maximize sum of pairwise products generated from the given Arrays

Last Updated : 15 Jul, 2025
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Given three arrays arr1[], arr2[] and arr3[] of length N1, N2, and N3 respectively, the task is to find the maximum sum possible by adding the products of pairs taken from different arrays.

Note: Each array element can be a part of a single pair.

Examples:

Input: arr1[] = {3, 5}, arr2[] = {2, 1}, arr3[] = {4, 3, 5}
Output : 43
Explanation
After sorting the arrays in descending order, following modifications are obtained: arr1[] = {5, 3}, arr2[] = {2, 1}, arr3[] = {5, 4, 3}. 
Therefore, maximized product = (arr1[0] * arr3[0]) + (arr1[1] * arr3[1]) + (arr2[0] * arr3[2]) = (5*5 + 3*4 + 2*3) = 43

Input: arr1[] = {3, 5, 9, 8, 7}, arr2[] = {6}, arr3[] = {3, 5}
Output : 115
Explanation
Sort the arrays in descending order, the following modifications are obtained: arr1[] = {9, 8, 7, 5, 3}, arr2[] = {6}, arr3[] = {5, 3}. 
Therefore, maximized product = (arr1[0] * arr2[0]) + (arr1[1] * arr3[0]) + (arr1[2] * arr3[1]) = (9*6 + 8*5 + 7*3) = 155

Approach: The given problem can be solved using a 3D Memoization table to store the maximum sums for all possible combinations of pairs. Suppose i, j, k are the number of elements taken from the arrays arr1[], arr2[] and arr3[] respectively to form pairs, then the memorization table dp[][][] will store the maximum possible sum of products generated from this combination of elements in dp[i][j][k]. 
Follow the below steps to solve the problem-

  • Sort the given arrays in descending order.
  • Initialize a dp table dp[][][], where dp[i][j][k] store the maximum sum obtained by taking i largest numbers from the first array, j largest numbers from the second array, and k largest numbers from the third array.
  • For every i, j, and kth element from the three arrays respectively, check for all possible pairs and calculate the maximum sum possible by considering each pair and memoize the maximum sum obtained for further computation.
  • Finally, print the maximum possible sum returned by the dp[][][] matrix.

Below is the implementation of the above approach:

C++ 
// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;

#define maxN 201

// Variables which represent
// the size of the array
int n1, n2, n3;

// Stores the results
int dp[maxN][maxN][maxN];

// Function to return the
// maximum possible sum
int getMaxSum(int i, int j,
              int k, int arr1[],
              int arr2[], int arr3[])
{
    // Stores the count of
    // arrays processed
    int cnt = 0;

    if (i >= n1)
        cnt++;

    if (j >= n2)
        cnt++;

    if (k >= n3)
        cnt++;

    // If more than two arrays
    // have been processed
    if (cnt >= 2)
        return 0;

    // If an already computed
    // subproblem occurred
    if (dp[i][j][k] != -1)
        return dp[i][j][k];

    int ans = 0;

    // Explore all the possible pairs
    if (i < n1 && j < n2)

        // Recursive function call
        ans = max(ans,
                  getMaxSum(i + 1, j + 1, k,
                            arr1, arr2, arr3)
                      + arr1[i] * arr2[j]);

    if (i < n1 && k < n3)
        ans = max(ans,
                  getMaxSum(i + 1, j, k + 1,
                            arr1, arr2, arr3)
                      + arr1[i] * arr3[k]);

    if (j < n2 && k < n3)
        ans = max(ans,
                  getMaxSum(i, j + 1, k + 1,
                            arr1, arr2, arr3)
                      + arr2[j] * arr3[k]);

    // Memoize the maximum
    dp[i][j][k] = ans;

    // Returning the value
    return dp[i][j][k];
}

// Function to return the maximum sum of
// products of pairs possible
int maxProductSum(int arr1[], int arr2[],
                  int arr3[])
{
    // Initialising the dp array to -1
    memset(dp, -1, sizeof(dp));

    // Sort the arrays in descending order
    sort(arr1, arr1 + n1);
    reverse(arr1, arr1 + n1);

    sort(arr2, arr2 + n2);
    reverse(arr2, arr2 + n2);

    sort(arr3, arr3 + n3);
    reverse(arr3, arr3 + n3);

    return getMaxSum(0, 0, 0,
                     arr1, arr2, arr3);
}

// Driver Code
int main()
{
    n1 = 2;
    int arr1[] = { 3, 5 };

    n2 = 2;
    int arr2[] = { 2, 1 };

    n3 = 3;
    int arr3[] = { 4, 3, 5 };

    cout << maxProductSum(arr1, arr2, arr3);

    return 0;
}
Java 
// Java program for above approach
import java.util.*;
import java.lang.*;

class GFG{

static final int maxN = 201; 

// Variables which represent 
// the size of the array 
static int n1, n2, n3; 

// Stores the results 
static int[][][] dp = new int[maxN][maxN][maxN]; 

// Function to return the 
// maximum possible sum 
static int getMaxSum(int i, int j, 
                     int k, int arr1[], 
                     int arr2[], int arr3[]) 
{ 
    
    // Stores the count of 
    // arrays processed 
    int cnt = 0; 

    if (i >= n1) 
        cnt++; 

    if (j >= n2) 
        cnt++; 

    if (k >= n3) 
        cnt++; 

    // If more than two arrays 
    // have been processed 
    if (cnt >= 2) 
        return 0; 

    // If an already computed 
    // subproblem occurred 
    if (dp[i][j][k] != -1) 
        return dp[i][j][k]; 

    int ans = 0; 

    // Explore all the possible pairs 
    if (i < n1 && j < n2) 

        // Recursive function call 
        ans = Math.max(ans, 
                       getMaxSum(i + 1, j + 1, k, 
                                arr1, arr2, arr3) + 
                                arr1[i] * arr2[j]); 

    if (i < n1 && k < n3) 
        ans = Math.max(ans, 
                       getMaxSum(i + 1, j, k + 1, 
                                 arr1, arr2, arr3) +
                                 arr1[i] * arr3[k]); 

    if (j < n2 && k < n3) 
        ans = Math.max(ans, 
                       getMaxSum(i, j + 1, k + 1, 
                                 arr1, arr2, arr3) +
                                 arr2[j] * arr3[k]); 

    // Memoize the maximum 
    dp[i][j][k] = ans; 

    // Returning the value 
    return dp[i][j][k]; 
} 

static void reverse(int[] tmp)
{
    int i, k, t;
    int n = tmp.length;
    
        for(i = 0; i < n/ 2; i++) 
        { 
            t = tmp[i]; 
            tmp[i] = tmp[n - i - 1]; 
            tmp[n - i - 1] = t; 
        } 
}

// Function to return the maximum sum of 
// products of pairs possible 
static int maxProductSum(int arr1[], int arr2[], 
                         int arr3[]) 
{ 
    
    // Initialising the dp array to -1 
    for(int i = 0; i < dp.length; i++)
        for(int j = 0; j < dp[0].length; j++)
            for(int k = 0; k < dp[j][0].length; k++)
                dp[i][j][k] = -1;

    // Sort the arrays in descending order 
    Arrays.sort(arr1);
    reverse(arr1);
    
    Arrays.sort(arr2);
    reverse(arr2);
    
    Arrays.sort(arr3);
    reverse(arr3);
    
    return getMaxSum(0, 0, 0, 
                     arr1, arr2, arr3); 
} 

// Driver Code
public static void main (String[] args) 
{
    n1 = 2; 
    int arr1[] = { 3, 5 }; 
    
    n2 = 2; 
    int arr2[] = { 2, 1 }; 
    
    n3 = 3; 
    int arr3[] = { 4, 3, 5 }; 
    
    System.out.println(maxProductSum(arr1, arr2, arr3)); 
}
}

// This code is contributed by offbeat
Python3 
# Python3 program for 
# the above approach
maxN = 201;

# Variables which represent
# the size of the array
n1, n2, n3 = 0, 0, 0;

# Stores the results
dp = [[[0 for i in range(maxN)] 
          for j in range(maxN)] 
          for j in range(maxN)];

# Function to return the
# maximum possible sum
def getMaxSum(i, j, k, 
              arr1, arr2, arr3):
  
    # Stores the count of
    # arrays processed
    cnt = 0;

    if (i >= n1):
        cnt += 1;

    if (j >= n2):
        cnt += 1;

    if (k >= n3):
        cnt += 1;

    # If more than two arrays
    # have been processed
    if (cnt >= 2):
        return 0;

    # If an already computed
    # subproblem occurred
    if (dp[i][j][k] != -1):
        return dp[i][j][k];

    ans = 0;

    # Explore all the possible pairs
    if (i < n1 and j < n2):

        # Recursive function call
        ans = max(ans, getMaxSum(i + 1, j + 1, 
                                 k, arr1,
                                 arr2, arr3) + 
                       arr1[i] * arr2[j]);

    if (i < n1 and k < n3):
        ans = max(ans, getMaxSum(i + 1, j, 
                                 k + 1, arr1, 
                                 arr2, arr3) + 
                       arr1[i] * arr3[k]);

    if (j < n2 and k < n3):
        ans = max(ans, getMaxSum(i, j + 1, 
                                 k + 1, arr1, 
                                 arr2, arr3) + 
                       arr2[j] * arr3[k]);

    # Memoize the maximum
    dp[i][j][k] = ans;

    # Returning the value
    return dp[i][j][k];


def reverse(tmp):
    i, k, t = 0, 0, 0;
    n = len(tmp);

    for i in range(n // 2):
        t = tmp[i];
        tmp[i] = tmp[n - i - 1];
        tmp[n - i - 1] = t;

# Function to return the maximum sum of
# products of pairs possible
def maxProductSum(arr1, arr2, arr3):
    # Initialising the dp array to -1
    for i in range(len(dp)):
        for j in range(len(dp[0])):
            for k in range(len(dp[j][0])):
                dp[i][j][k] = -1;

    # Sort the arrays in descending order
    arr1.sort();
    reverse(arr1);

    arr2.sort();
    reverse(arr2);

    arr3.sort();
    reverse(arr3);

    return getMaxSum(0, 0, 0, 
                     arr1, arr2, arr3);

# Driver Code
if __name__ == '__main__':
  n1 = 2;
  arr1 = [3, 5];

  n2 = 2;
  arr2 = [2, 1];

  n3 = 3;
  arr3 = [4, 3, 5];

  print(maxProductSum(arr1, arr2, arr3));

# This code is contributed by Rajput-Ji
C# 
// C# program for above approach 
using System; 

class GFG{ 

const int maxN = 201; 

// Variables which represent 
// the size of the array 
static int n1, n2, n3; 

// Stores the results 
static int[,,] dp = new int[maxN, maxN, maxN]; 

// Function to return the 
// maximum possible sum 
static int getMaxSum(int i, int j, 
                     int k, int []arr1, 
                     int []arr2, int []arr3) 
{ 
    
    // Stores the count of 
    // arrays processed 
    int cnt = 0; 

    if (i >= n1) 
        cnt++; 

    if (j >= n2) 
        cnt++; 

    if (k >= n3) 
        cnt++; 

    // If more than two arrays 
    // have been processed 
    if (cnt >= 2) 
        return 0; 

    // If an already computed 
    // subproblem occurred 
    if (dp[i, j, k] != -1) 
        return dp[i, j, k]; 

    int ans = 0; 

    // Explore all the possible pairs 
    if (i < n1 && j < n2) 

        // Recursive function call 
        ans = Math.Max(ans, 
                       getMaxSum(i + 1, j + 1, k, 
                                 arr1, arr2, arr3) + 
                                 arr1[i] * arr2[j]); 

    if (i < n1 && k < n3) 
        ans = Math.Max(ans, 
                       getMaxSum(i + 1, j, k + 1, 
                                 arr1, arr2, arr3) + 
                                 arr1[i] * arr3[k]); 

    if (j < n2 && k < n3) 
        ans = Math.Max(ans, 
                       getMaxSum(i, j + 1, k + 1, 
                                 arr1, arr2, arr3) + 
                                 arr2[j] * arr3[k]); 
    
    // Memoize the maximum 
    dp[i, j, k] = ans; 

    // Returning the value 
    return dp[i, j, k]; 
} 

static void reverse(int[] tmp) 
{ 
    int i, t; 
    int n = tmp.Length; 

    for(i = 0; i < n / 2; i++) 
    { 
        t = tmp[i]; 
        tmp[i] = tmp[n - i - 1]; 
        tmp[n - i - 1] = t; 
    } 
} 

// Function to return the maximum sum of 
// products of pairs possible 
static int maxProductSum(int []arr1, int []arr2, 
                         int []arr3) 
{ 
    
    // Initialising the dp array to -1 
    for(int i = 0; i < maxN; i++) 
        for(int j = 0; j < maxN; j++) 
            for(int k = 0; k < maxN; k++) 
                dp[i, j, k] = -1; 

    // Sort the arrays in descending order 
    Array.Sort(arr1); 
    reverse(arr1); 
    
    Array.Sort(arr2); 
    reverse(arr2); 
    
    Array.Sort(arr3); 
    reverse(arr3); 
    
    return getMaxSum(0, 0, 0, 
                     arr1, arr2, arr3); 
} 

// Driver Code 
public static void Main (string[] args) 
{ 
    n1 = 2; 
    int []arr1 = { 3, 5 }; 
    
    n2 = 2; 
    int []arr2 = { 2, 1 }; 
    
    n3 = 3; 
    int []arr3 = { 4, 3, 5 }; 
    
    Console.Write(maxProductSum(arr1, arr2, arr3)); 
} 
} 

// This code is contributed by rutvik_56
JavaScript 
<script>

// JavaScript Program to implement
// the above approach

var maxN = 201;

// Variables which represent
// the size of the array
var n1, n2, n3;

// Stores the results
var dp = Array.from(Array(maxN), ()=>Array(maxN));
for(var i =0; i<maxN; i++)
        for(var j =0; j<maxN; j++)
            dp[i][j] = new Array(maxN).fill(-1);

// Function to return the
// maximum possible sum
function getMaxSum(i, j, k, arr1, arr2, arr3)
{
    // Stores the count of
    // arrays processed
    var cnt = 0;

    if (i >= n1)
        cnt++;

    if (j >= n2)
        cnt++;

    if (k >= n3)
        cnt++;

    // If more than two arrays
    // have been processed
    if (cnt >= 2)
        return 0;

    // If an already computed
    // subproblem occurred
    if (dp[i][j][k] != -1)
        return dp[i][j][k];

    var ans = 0;

    // Explore all the possible pairs
    if (i < n1 && j < n2)

        // Recursive function call
        ans = Math.max(ans,
                  getMaxSum(i + 1, j + 1, k,
                            arr1, arr2, arr3)
                      + arr1[i] * arr2[j]);

    if (i < n1 && k < n3)
        ans = Math.max(ans,
                  getMaxSum(i + 1, j, k + 1,
                            arr1, arr2, arr3)
                      + arr1[i] * arr3[k]);

    if (j < n2 && k < n3)
        ans = Math.max(ans,
                  getMaxSum(i, j + 1, k + 1,
                            arr1, arr2, arr3)
                      + arr2[j] * arr3[k]);

    // Memoize the maximum
    dp[i][j][k] = ans;

    // Returning the value
    return dp[i][j][k];
}

// Function to return the maximum sum of
// products of pairs possible
function maxProductSum(arr1, arr2, arr3)
{

    // Sort the arrays in descending order
    arr1.sort();
    arr1.reverse();
    arr2.sort();
    arr2.reverse();
    arr3.sort();
    arr3.reverse();

    return getMaxSum(0, 0, 0,
                     arr1, arr2, arr3);
}

// Driver Code
n1 = 2;
var arr1 = [3, 5];
n2 = 2;
var arr2 = [2, 1];
n3 = 3;
var arr3 = [4, 3, 5];
document.write( maxProductSum(arr1, arr2, arr3));

</script>

Output: 
43

Time Complexity: O((N1 * N2 * N3)) 
Auxiliary Space: O(N1 * N2 * N3)


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