Find Maximum dot product of two arrays with insertion of 0’s
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[])
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 = 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; } |
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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. |
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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 |
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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*/ |
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Output:
107
Time Complexity : O(nm)
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