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

3.7

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

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 programing .

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 = 1 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 c++ implementation of above idea

// 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;
}

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 contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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