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Find Maximum dot product of two arrays with insertion of 0’s

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  • Difficulty Level : Hard
  • Last Updated : 06 Jul, 2022
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Given two arrays of positive integers of size m and n where m > n. We need to maximize the dot product by inserting zeros in the second array but we cannot disturb the order of elements.

Examples: 

Input : A[] = {2, 3 , 1, 7, 8} 
        B[] = {3, 6, 7}        
Output : 107
Explanation : We get maximum dot product after
inserting 0 at first and third positions in 
second array.
Maximum Dot Product : = A[i] * B[j] 
2*0 + 3*3 + 1*0 + 7*6 + 8*7 = 107

Input : A[] = {1, 2, 3, 6, 1, 4}
        B[] = {4, 5, 1}
Output : 46

Asked in: Directi Interview

Recommended Practice

Another way to look at this problem is, for every pair of elements element A[i] and B[j] where j >= i , we have two choices:  

  1. We multiply A[i] and B[j] and add to product (We include A[i]).
  2. We exclude A[i] from product (In other words, we insert 0 at current position in B[])

The idea is to use Dynamic programming

1) Given Array A[] of size 'm' and B[] of size 'n'

2) Create 2D matrix 'DP[n + 1][m + 1]' initialize it
with '0'

3) Run loop outer loop for i = 1 to n
     Inner loop j = i to m

       // Two cases arise
       // 1) Include A[j]
       // 2) Exclude A[j] (insert 0 in B[])      
       dp[i][j]  = max(dp[i-1][j-1] + A[j-1] * B[i -1],
                       dp[i][j-1]) 
      
    // Last return maximum dot product that is 
    return dp[n][m]

Below is the implementation of above idea.  

C++




// C++ program to find maximum dot product of two array
#include<bits/stdc++.h>
using namespace std;
 
// Function compute Maximum Dot Product and
// return it
long long int MaxDotProduct(int A[], int B[],
                            int m, int n)
{
    // Create 2D Matrix that stores dot product
    // dp[i+1][j+1] stores product considering B[0..i]
    // and A[0...j]. Note that since all m > n, we fill
    // values in upper diagonal of dp[][]
    long long int dp[n+1][m+1];
    memset(dp, 0, sizeof(dp));
 
    // Traverse through all elements of B[]
    for (int i=1; i<=n; i++)
 
        // Consider all values of A[] with indexes greater
        // than or equal to i and compute dp[i][j]
        for (int j=i; j<=m; j++)
 
            // Two cases arise
            // 1) Include A[j]
            // 2) Exclude A[j] (insert 0 in B[])
            dp[i][j] = max((dp[i-1][j-1] + (A[j-1]*B[i-1])) ,
                            dp[i][j-1]);
 
    // return Maximum Dot Product
    return dp[n][m] ;
}
 
// Driver program to test above function
int main()
{
    int A[] = { 2, 3 , 1, 7, 8 } ;
    int B[] = { 3, 6, 7 } ;
    int m = sizeof(A)/sizeof(A[0]);
    int n = sizeof(B)/sizeof(B[0]);
    cout << MaxDotProduct(A, B, m, n);
    return 0;
}


Java




// Java program to find maximum
// dot product of two array
import java.util.*;
 
class GFG
{
// Function to compute Maximum
// Dot Product and return it
static int MaxDotProduct(int A[], int B[], int m, int n)
{
    // Create 2D Matrix that stores dot product
    // dp[i+1][j+1] stores product considering B[0..i]
    // and A[0...j]. Note that since all m > n, we fill
    // values in upper diagonal of dp[][]
    int dp[][] = new int[n + 1][m + 1];
    for (int[] row : dp)
    Arrays.fill(row, 0);
 
    // Traverse through all elements of B[]
    for (int i = 1; i <= n; i++)
 
    // Consider all values of A[] with indexes greater
    // than or equal to i and compute dp[i][j]
    for (int j = i; j <= m; j++)
 
        // Two cases arise
        // 1) Include A[j]
        // 2) Exclude A[j] (insert 0 in B[])
        dp[i][j] =
            Math.max((dp[i - 1][j - 1] +
                    (A[j - 1] * B[i - 1])), dp[i][j - 1]);
 
    // return Maximum Dot Product
    return dp[n][m];
}
 
// Driver code
public static void main(String[] args) {
    int A[] = {2, 3, 1, 7, 8};
    int B[] = {3, 6, 7};
    int m = A.length;
    int n = B.length;
    System.out.print(MaxDotProduct(A, B, m, n));
}
}
 
// This code is contributed by Anant Agarwal.


Python3




# Python 3 program to find maximum dot
# product of two array
 
# Function compute Maximum Dot Product
# and return it
def MaxDotProduct(A, B, m, n):
     
    # Create 2D Matrix that stores dot product
    # dp[i+1][j+1] stores product considering
    # B[0..i] and A[0...j]. Note that since
    # all m > n, we fill values in upper
    # diagonal of dp[][]
    dp = [[0 for i in range(m + 1)]
             for j in range(n + 1)]
 
    # Traverse through all elements of B[]
    for i in range(1, n + 1, 1):
         
        # Consider all values of A[] with indexes
        # greater than or equal to i and compute
        # dp[i][j]
        for j in range(i, m + 1, 1):
             
            # Two cases arise
            # 1) Include A[j]
            # 2) Exclude A[j] (insert 0 in B[])
            dp[i][j] = max((dp[i - 1][j - 1] +
                            (A[j - 1] * B[i - 1])) ,
                            dp[i][j - 1])
 
    # return Maximum Dot Product
    return dp[n][m]
 
# Driver Code
if __name__ == '__main__':
    A = [2, 3 , 1, 7, 8]
    B = [3, 6, 7]
    m = len(A)
    n = len(B)
    print(MaxDotProduct(A, B, m, n))
 
# This code is contributed by
# Sanjit_Prasad


C#




// C# program to find maximum
// dot product of two array
using System;
 
public class GFG{
 
    // Function to compute Maximum
    // Dot Product and return it
    static int MaxDotProduct(int []A, int []B, int m, int n)
    {
        // Create 2D Matrix that stores dot product
        // dp[i+1][j+1] stores product considering B[0..i]
        // and A[0...j]. Note that since all m > n, we fill
        // values in upper diagonal of dp[][]
        int [,]dp = new int[n + 1,m + 1];
 
        // Traverse through all elements of B[]
        for (int i = 1; i <= n; i++)
 
        // Consider all values of A[] with indexes greater
        // than or equal to i and compute dp[i][j]
        for (int j = i; j <= m; j++)
 
            // Two cases arise
            // 1) Include A[j]
            // 2) Exclude A[j] (insert 0 in B[])
            dp[i,j] =
                Math.Max((dp[i - 1,j - 1] +
                        (A[j - 1] * B[i - 1])), dp[i,j - 1]);
 
        // return Maximum Dot Product
        return dp[n,m];
    }
 
    // Driver code
    public static void Main() {
        int []A = {2, 3, 1, 7, 8};
        int []B = {3, 6, 7};
        int m = A.Length;
        int n = B.Length;
        Console.Write(MaxDotProduct(A, B, m, n));
    }
}
 
/*This code is contributed by 29AjayKumar*/


Javascript




<script>
// Javascript program to find maximum
// dot product of two array
 
// Function to compute Maximum
// Dot Product and return it
function MaxDotProduct(A, B, m, n)
{
    // Create 2D Matrix that stores dot product
    // dp[i+1][j+1] stores product considering B[0..i]
    // and A[0...j]. Note that since all m > n, we fill
    // values in upper diagonal of dp[][]
    let dp = new Array(n + 1);
    for(let i = 0; i < (n + 1); i++)
    {
        dp[i] = new Array(m+1);
        for(let j = 0; j < m + 1; j++)
        {
            dp[i][j] = 0;
        }
    }
   
    // Traverse through all elements of B[]
    for (let i = 1; i <= n; i++)
   
    // Consider all values of A[] with indexes greater
    // than or equal to i and compute dp[i][j]
    for (let j = i; j <= m; j++)
   
        // Two cases arise
        // 1) Include A[j]
        // 2) Exclude A[j] (insert 0 in B[])
        dp[i][j] =
            Math.max((dp[i - 1][j - 1] +
                    (A[j - 1] * B[i - 1])), dp[i][j - 1]);
   
    // return Maximum Dot Product
    return dp[n][m];
}
 
// Driver code
let A = [2, 3, 1, 7, 8];
let B = [3, 6, 7];
let m = A.length;
let n = B.length;
document.write(MaxDotProduct(A, B, m, n));
 
// This code is contributed by rag2127
</script>


Output

107

Time Complexity : O(nm)

This article is contributed by Nishant Singh. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks. 


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