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 = 107Input : 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 the product (We include A[i]).
- We exclude A[i] from the product (In other words, we insert 0 at the 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 the 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 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
// 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. |
Python3
# 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#
// 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*/ |
Javascript
<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(n*m)
Auxiliary Space: O(n*m)
Efficient approach : Space optimization
In previous approach the current value dp[i][j] is only depend upon the current and previous row values of DP. So to optimize the space complexity we use a single 1D array to store the computations.
Implementation steps:
- Create an array dp to store the maximum dot product values.
- Initialize all elements of dp to 0.
- Iterate over the A array and process each element.
- For each element in A, iterate over the B array in reverse order and process each element.
- Update dp[j] with the maximum value between the current dp[j] and dp[j – 1] + (A[i – 1] * B[j – 1]).
- After the loops, dp[n] will represent the maximum dot product achievable.
- Return dp[n] as the result.
Implementation:
C++
#include<bits/stdc++.h>using namespace std;long long int MaxDotProduct(int A[], int B[], int m, int n){ long long int dp[n + 1]; memset(dp, 0, sizeof(dp)); for (int i = 1; i <= m; i++) { for (int j = n; j >= 1; j--) { dp[j] = max(dp[j], dp[j - 1] + (A[i - 1] * B[j - 1])); // Calculate the maximum dot product at position j // by choosing between not including B[j-1] and including B[j-1] // Update dp[j] with the maximum value } } return dp[n]; // Return the maximum dot product}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
import java.util.Arrays;public class GFG { // Function to find the maximum dot product of two // arrays static long MaxDotProduct(int[] A, int[] B, int m, int n) { // Initialize an array to store the maximum dot // products at each position in B long[] dp = new long[n + 1]; Arrays.fill(dp, 0); // Initialize the array with zeros // Iterate through each element in array A for (int i = 1; i <= m; i++) { // Iterate through each position in array B in // reverse order for (int j = n; j >= 1; j--) { // Calculate the maximum dot product at // position j by choosing between not // including B[j-1] (dp[j]) and including // B[j-1] (dp[j - 1] + (A[i - 1] * B[j - // 1])) dp[j] = Math.max( dp[j], dp[j - 1] + (A[i - 1] * B[j - 1])); // dp[j] represents the maximum dot product // of A[0...i] and B[0...j] } } // The final result is stored in dp[n], which // represents the maximum dot product of A and B return dp[n]; } 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; // Call the MaxDotProduct function and print the // result System.out.println(MaxDotProduct(A, B, m, n)); }} |
Python3
def MaxDotProduct(A, B, m, n): dp = [0] * (n + 1) for i in range(1, m + 1): for j in range(n, 0, -1): dp[j] = max(dp[j], dp[j - 1] + (A[i - 1] * B[j - 1])) # Calculate the maximum dot product at position j # by choosing between not including B[j-1] and # including B[j-1] # Update dp[j] with the maximum value return dp[n] # Return the maximum dot productA = [2, 3, 1, 7, 8]B = [3, 6, 7]m = len(A)n = len(B)print(MaxDotProduct(A, B, m, n)) |
C#
using System;class GFG { // Function to calculate the maximum dot product between // two arrays A and B static long MaxDotProduct(int[] A, int[] B, int m, int n) { // Create an array to store the maximum dot products long[] dp = new long[n + 1]; // Calculate the maximum dot product for each // position j in array B for (int i = 1; i <= m; i++) { for (int j = n; j >= 1; j--) { // Calculate the maximum dot product at // position j by choosing between not // including B[j-1] and including B[j-1]. // Update dp[j] with the maximum value. dp[j] = Math.Max( dp[j], dp[j - 1] + (A[i - 1] * B[j - 1])); } } return dp[n]; // Return the maximum dot product } static void Main() { int[] A = { 2, 3, 1, 7, 8 }; int[] B = { 3, 6, 7 }; int m = A.Length; int n = B.Length; // Calculate and print the maximum dot product // between arrays A and B Console.WriteLine(MaxDotProduct(A, B, m, n)); }} |
Javascript
function MaxDotProduct(A, B, m, n) { // Create an array to store the maximum dot products const dp = new Array(n + 1).fill(0); for (let i = 1; i <= m; i++) { for (let j = n; j >= 1; j--) { // Calculate the maximum dot product at position j // by choosing between not including B[j-1] and including B[j-1] dp[j] = Math.max(dp[j], dp[j - 1] + (A[i - 1] * B[j - 1])); } } return dp[n]; // Return the maximum dot product}// Driver codeconst A = [2, 3, 1, 7, 8];const B = [3, 6, 7];// Calculate the lengths of arrays A and Bconst m = A.length;const n = B.length;// Call the MaxDotProduct function and output the resultconsole.log(MaxDotProduct(A, B, m, n)); |
Output
107
Time Complexity : O(n*m)
Auxiliary Space: O(n)
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