Maximize product of same-indexed elements of same size subsequences

Given two integer arrays a[] and b[], the task is to find the maximum possible product of the same indexed elements of two equal length subsequences from the two given arrays.

Examples:

Input: a[] = {-3, 1, -12, 7}, b[] = {3, 2, -6, 7} 
Output: 124 
Explanation: The subsequence [1, -12, 7] from a[] and subsequence [3, -6, 7] from b[] gives the maximum scaler product, (1*3 + (-12)*(-6) + 7*7) = 124.

Input: a[] = {-2, 6, -2, -5}, b[] = {-3, 4, -2, 8} 
Output: 54 
Explanation: The subsequence [-2, 6] from a[] and subsequence [-3, 8] from b[] gives the maximum scaler product, ((-2)*(-3) + 6*8) = 54. 

Approach: The problem is similar to the Longest Common Subsequence problem of Dynamic Programming. A Recursive and Memoization based Top-Down method is explained below: 

• Let us define a function F(X, Y), which returns the maximum scalar product between the arrays a[0-X] and b[0-Y].
• Following is the recursive definition:

F(X, Y) = max(a[X] * b[Y] + F(X – 1, Y – 1); Use the last elements from both arrays and check further 
a[X] * b[Y]; Only use the last element as the maximum product
F(X – 1, Y); Ignore the last element from the first array 
F(X, Y – 1)); Ignore the last element from the second element 

• Use a 2-D array for memoization and to avoid recomputation of same sub-problems.

Below is the implementation of the above approach: 

C++

// C++ implementation to maximize
// product of same-indexed elements
// of same size subsequences

#include <bits/stdc++.h>
using namespace std;

#define INF 10000000

// Utility function to find the maximum
int maximum(int A, int B, int C, int D)
{
    return max(max(A, B), max(C, D));
}

// Utility function to find
// the maximum scalar product
// from the equal length sub-sequences
// taken from the two given array
int maxProductUtil(
    int X, int Y,
    int* A, int* B,
    vector<vector<int> >& dp)
{
    // Return a very small number
    // if index is invalid
    if (X < 0 or Y < 0)
        return -INF;

    // If the sub-problem is already
    // evaluated, then return it
    if (dp[X][Y] != -1)
        return dp[X][Y];

    // Take the maximum of all
    // the recursive cases
    dp[X][Y]
        = maximum(
            A[X] * B[Y]
                + maxProductUtil(
                      X - 1, Y - 1, A, B, dp),
            A[X] * B[Y],
            maxProductUtil(
                X - 1, Y, A, B, dp),
            maxProductUtil(
                X, Y - 1, A, B, dp));

    return dp[X][Y];
}

// Function to find maximum scalar
// product from same size sub-sequences
// taken from the two given array
int maxProduct(int A[], int N,
               int B[], int M)
{
    // Initialize a 2-D array
    // for memoization
    vector<vector<int> >
    dp(N, vector<int>(M, -1));

    return maxProductUtil(
        N - 1, M - 1,
        A, B, dp);
}

// Driver Code
int main()
{
    int a[] = { -2, 6, -2, -5 };
    int b[] = { -3, 4, -2, 8 };

    int n = sizeof(a) / sizeof(a[0]);
    int m = sizeof(b) / sizeof(b[0]);

    cout << maxProduct(a, n, b, m);
}

Java

// Java implementation to maximize
// product of same-indexed elements
// of same size subsequences
class GFG{

static final int INF = 10000000;

// Utility function to find the maximum
static int maximum(int A, int B,
                   int C, int D)
{
    return Math.max(Math.max(A, B),
                    Math.max(C, D));
}

// Utility function to find the
// maximum scalar product from
// the equal length sub-sequences
// taken from the two given array
static int maxProductUtil(int X, int Y,
                          int []A, int[] B,
                          int [][]dp)
{

    // Return a very small number
    // if index is invalid
    if (X < 0 || Y < 0)
        return -INF;

    // If the sub-problem is already
    // evaluated, then return it
    if (dp[X][Y] != -1)
        return dp[X][Y];

    // Take the maximum of all
    // the recursive cases
    dp[X][Y] = maximum(A[X] * B[Y] +
                       maxProductUtil(X - 1, Y - 1,
                                      A, B, dp),
                       A[X] * B[Y],
                       maxProductUtil(X - 1, Y,
                                      A, B, dp),
                       maxProductUtil(X, Y - 1,
                                      A, B, dp));

    return dp[X][Y];
}

// Function to find maximum scalar
// product from same size sub-sequences
// taken from the two given array
static int maxProduct(int A[], int N,
                      int B[], int M)
{

    // Initialize a 2-D array
    // for memoization
    int [][]dp = new int[N][M];
    for(int i = 0; i < N; i++)
    {
       for(int j = 0; j < M; j++)
       {
          dp[i][j] = -1;
       }
    }

    return maxProductUtil(N - 1, M - 1,
                          A, B, dp);
}

// Driver Code
public static void main(String[] args)
{
    int a[] = { -2, 6, -2, -5 };
    int b[] = { -3, 4, -2, 8 };

    int n = a.length;
    int m = b.length;

    System.out.print(maxProduct(a, n, b, m));
}
}

// This code is contributed by Amal Kumar Choubey

Python3

# Python3 implementation to maximize
# product of same-indexed elements
# of same size subsequences
INF = 10000000

# Utility function to find the maximum
def maximum(A, B, C, D):

    return max(max(A, B), max(C, D))

# Utility function to find
# the maximum scalar product
# from the equal length sub-sequences
# taken from the two given array
def maxProductUtil(X, Y, A, B, dp):

    # Return a very small number
    # if index is invalid
    if (X < 0 or Y < 0):
        return -INF

    # If the sub-problem is already
    # evaluated, then return it
    if (dp[X][Y] != -1):
        return dp[X][Y]

    # Take the maximum of all
    # the recursive cases
    dp[X][Y]= maximum(A[X] * B[Y] +
                      maxProductUtil(X - 1, Y - 1,
                                     A, B, dp),
                      A[X] * B[Y],
                      maxProductUtil(X - 1, Y, A,
                                     B, dp),
                      maxProductUtil(X, Y - 1, A,
                                     B, dp))

    return dp[X][Y]

# Function to find maximum scalar
# product from same size sub-sequences
# taken from the two given array
def maxProduct(A, N, B, M):

    # Initialize a 2-D array
    # for memoization
    dp = [[-1 for i in range(m)]
              for i in range(n)]

    return maxProductUtil(N - 1, M - 1,
                          A, B, dp)

# Driver Code
a = [ -2, 6, -2, -5 ]
b = [ -3, 4, -2, 8 ]
n = len(a)
m = len(b)

print(maxProduct(a, n, b, m))

# This code is contributed by avanitrachhadiya2155

C#

// C# implementation to maximize
// product of same-indexed elements
// of same size subsequences
using System;

class GFG{

static readonly int INF = 10000000;

// Utility function to find the maximum
static int maximum(int A, int B,
                   int C, int D)
{
    return Math.Max(Math.Max(A, B),
                    Math.Max(C, D));
}

// Utility function to find the
// maximum scalar product from
// the equal length sub-sequences
// taken from the two given array
static int maxProductUtil(int X, int Y,
                          int []A, int[] B,
                          int [,]dp)
{

    // Return a very small number
    // if index is invalid
    if (X < 0 || Y < 0)
        return -INF;

    // If the sub-problem is already
    // evaluated, then return it
    if (dp[X, Y] != -1)
        return dp[X, Y];

    // Take the maximum of all
    // the recursive cases
    dp[X, Y] = maximum(A[X] * B[Y] +
                       maxProductUtil(X - 1, Y - 1,
                                        A, B, dp),
                       A[X] * B[Y],
                       maxProductUtil(X - 1, Y,
                                      A, B, dp),
                       maxProductUtil(X, Y - 1,
                                      A, B, dp));

    return dp[X, Y];
}

// Function to find maximum scalar
// product from same size sub-sequences
// taken from the two given array
static int maxProduct(int []A, int N,
                      int []B, int M)
{

    // Initialize a 2-D array
    // for memoization
    int [,]dp = new int[N, M];
    for(int i = 0; i < N; i++)
    {
       for(int j = 0; j < M; j++)
       {
          dp[i, j] = -1;
       }
    }

    return maxProductUtil(N - 1, M - 1,
                          A, B, dp);
}

// Driver Code
public static void Main(String[] args)
{
    int []a = { -2, 6, -2, -5 };
    int []b = { -3, 4, -2, 8 };

    int n = a.Length;
    int m = b.Length;

    Console.Write(maxProduct(a, n, b, m));
}
}

// This code is contributed by amal kumar choubey

Javascript

<script>
// Javascript implementation to maximize
// product of same-indexed elements
// of same size subsequences

let INF = 10000000;

// Utility function to find the maximum
function maximum(A, B, C, D)
{
    return Math.max(Math.max(A, B),
                    Math.max(C, D));
}

// Utility function to find the
// maximum scalar product from
// the equal length sub-sequences
// taken from the two given array
function maxProductUtil(X, Y, A, B, dp)
{

    // Return a very small number
    // if index is invalid
    if (X < 0 || Y < 0)
        return -INF;

    // If the sub-problem is already
    // evaluated, then return it
    if (dp[X][Y] != -1)
        return dp[X][Y];

    // Take the maximum of all
    // the recursive cases
    dp[X][Y] = maximum(A[X] * B[Y] +
                       maxProductUtil(X - 1, Y - 1,
                                      A, B, dp),
                       A[X] * B[Y],
                       maxProductUtil(X - 1, Y,
                                      A, B, dp),
                       maxProductUtil(X, Y - 1,
                                      A, B, dp));

    return dp[X][Y];
}

// Function to find maximum scalar
// product from same size sub-sequences
// taken from the two given array
function maxProduct(A, N, B, M)
{

    // Initialize a 2-D array
    // for memoization
    let dp =new Array(N);
    for (var i = 0; i < dp.length; i++) {
    dp[i] = new Array(2);
    }
    for(let i = 0; i < N; i++)
    {
       for(let j = 0; j < M; j++)
       {
          dp[i][j] = -1;
       }
    }

    return maxProductUtil(N - 1, M - 1,
                          A, B, dp);
}

  // Driver Code

    let a = [ -2, 6, -2, -5 ];
    let b = [ -3, 4, -2, 8 ];

    let n = a.length;
    let m = b.length;

    document.write(maxProduct(a, n, b, m));

</script>

Output:

54