Find Maximum dot product of two arrays with insertion of 0's - GeeksforGeeks

31 Aug, 2021
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:  
We multiply A[i] and B[j] and add to product (We include A[i]).We exclude A[i] from product (In other words, we insert 0 at current position in B[])
We multiply A[i] and B[j] and add to product (We include A[i]).
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++
Java
Python3
C#
Javascript
// C++ program to find maximum dot product of two array#include<bits/stdc++.h>using namespace std; // Function compute Maximum Dot Product and// return itlong 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 functionint 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 program to find maximum// dot product of two arrayimport java.util.*; class GFG{// Function to compute Maximum// Dot Product and return itstatic 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 codepublic 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.
# Python 3 program to find maximum dot# product of two array # Function compute Maximum Dot Product# and return itdef 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 Codeif __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# program to find maximum// dot product of two arrayusing 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*/
<script>// Javascript program to find maximum// dot product of two array // Function to compute Maximum// Dot Product and return itfunction 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 codelet 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)
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thechosenone
29AjayKumar
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Directi
Arrays
Dynamic Programming
Directi
Arrays
Dynamic Programming
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