Maximize sum of products at corresponding indices in two arrays by reversing at most one subarray in first Array

Given two arrays A and B of size N, the task is to maximize the sum of A[i]*B[i] across all values of i from 0 to N-1 by reversing at most one subarray of array A.

Examples:

Input: N = 4, A = [5, 1, 2, 3], B = [1, 4, 3, 2]
Output: 33
Explanation: Array A after reversing the subarray A[0, 1] will become [1, 5, 2, 3]. Sum of A[i]*B[i] after the reversal becomes 1*1+5*4+2*3+3*2 = 33.

Input: N = 3, A = [6, 7, 3], B = [5, 1, 7]
Output: 82

Naive Approach: One simple way to solve this problem is to check for all possible subarrays of A and reverse them one by one to find the maximum possible value of the sum.

Below is the implementation of the above approach:

C++

// C++ program of the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to find the maximum sum of A[i]*B[i]
// across all values of i from 0 to N-1 by reversing
// at most one subarray of array A
int maxSum(vector<int>& A, vector<int>& B)
{
    int N = A.size();

    // Initialising maximum possible sum variable
    int maxPosSum = 0;

    // Iterating for all subarrays
    for (int L = 0; L < N - 1; L++) {
        for (int R = L; R < N; R++) {

            // Variable for storing the sum after reversing
            // The subarray from L to R
            int curSum = 0;
            for (int i = 0; i < N; i++) {

                // Checking if the current index is in the
                // reversed subarray
                if (i >= L && i <= R)
                    curSum += A[L + R - i] * B[i];
                else
                    curSum += A[i] * B[i];
            }

            // Updating the answer
            maxPosSum = max(maxPosSum, curSum);
        }
    }

    // Returning the Maximum Possible Sum of product
    return maxPosSum;
}

// Driver Code
int main()
{
    // Given Input
    int N = 4;
    vector<int> A = { 5, 1, 2, 3 }, B = { 1, 4, 3, 2 };

    cout << maxSum(A, B);
}

Java

// Java code to implement above approach
import java.util.*;
public class GFG {

  // Function to find the maximum sum of A[i]*B[i]
  // across all values of i from 0 to N-1 by reversing
  // at most one subarray of array A
  static int maxSum(int []A, int []B)
  {
    int N = A.length;

    // Initialising maximum possible sum variable
    int maxPosSum = 0;

    // Iterating for all subarrays
    for (int L = 0; L < N - 1; L++) {
      for (int R = L; R < N; R++) {

        // Variable for storing the sum after reversing
        // The subarray from L to R
        int curSum = 0;
        for (int i = 0; i < N; i++) {

          // Checking if the current index is in the
          // reversed subarray
          if (i >= L && i <= R)
            curSum += A[L + R - i] * B[i];
          else
            curSum += A[i] * B[i];
        }

        // Updating the answer
        maxPosSum = Math.max(maxPosSum, curSum);
      }
    }

    // Returning the Maximum Possible Sum of product
    return maxPosSum;
  }

  // Driver code
  public static void main(String args[])
  {

    // Given Input
    int N = 4;
    int []A = { 5, 1, 2, 3 };
    int []B = { 1, 4, 3, 2 };
    System.out.println(maxSum(A, B));

  }
}

// This code is contributed by Samim Hossain Mondal.

Python3

# Python code for the above approach

# Function to find the maximum sum of A[i]*B[i]
# across all values of i from 0 to N-1 by reversing
# at most one subarray of array A
def maxSum(A, B):
    N = len(A)

    # Initialising maximum possible sum variable
    maxPosSum = 0

    # Iterating for all subarrays
    for L in range(N - 1):
        for R in range(L, N):

            # Variable for storing the sum after reversing
            # The subarray from L to R
            curSum = 0
            for i in range(N):
                # Checking if the current index is in the
                # reversed subarray
                if (i >= L and i <= R):
                    curSum += A[L + R - i] * B[i]
                else:
                    curSum += A[i] * B[i]

            # Updating the answer
            maxPosSum = max(maxPosSum, curSum)

    # Returning the Maximum Possible Sum of product
    return maxPosSum

# Driver Code

# Given Input
N = 4
A = [5, 1, 2, 3]
B = [1, 4, 3, 2]
print(maxSum(A, B))

# This code is contributed by gfgking

C#

// C# code to implement above approach
using System;
public class GFG {

  // Function to find the maximum sum of A[i]*B[i]
  // across all values of i from 0 to N-1 by reversing
  // at most one subarray of array A
  static int maxSum(int []A, int []B)
  {
    int N = A.Length;

    // Initialising maximum possible sum variable
    int maxPosSum = 0;

    // Iterating for all subarrays
    for (int L = 0; L < N - 1; L++) {
      for (int R = L; R < N; R++) {

        // Variable for storing the sum after reversing
        // The subarray from L to R
        int curSum = 0;
        for (int i = 0; i < N; i++) {

          // Checking if the current index is in the
          // reversed subarray
          if (i >= L && i <= R)
            curSum += A[L + R - i] * B[i];
          else
            curSum += A[i] * B[i];
        }

        // Updating the answer
        maxPosSum = Math.Max(maxPosSum, curSum);
      }
    }

    // Returning the Maximum Possible Sum of product
    return maxPosSum;
  }

  // Driver code
  public static void Main()
  {

    // Given Input
    int []A = { 5, 1, 2, 3 };
    int []B = { 1, 4, 3, 2 };
    Console.Write(maxSum(A, B));

  }
}

// This code is contributed by Saurabh Jaiswal

Javascript

<script>
   // JavaScript code for the above approach

   // Function to find the maximum sum of A[i]*B[i]
   // across all values of i from 0 to N-1 by reversing
   // at most one subarray of array A
   function maxSum(A, B)
   {
     let N = A.length;

     // Initialising maximum possible sum variable
     let maxPosSum = 0;

     // Iterating for all subarrays
     for (let L = 0; L < N - 1; L++) {
       for (let R = L; R < N; R++) {

         // Variable for storing the sum after reversing
         // The subarray from L to R
         let curSum = 0;
         for (let i = 0; i < N; i++) {

           // Checking if the current index is in the
           // reversed subarray
           if (i >= L && i <= R)
             curSum += A[L + R - i] * B[i];
           else
             curSum += A[i] * B[i];
         }

         // Updating the answer
         maxPosSum = Math.max(maxPosSum, curSum);
       }
     }

     // Returning the Maximum Possible Sum of product
     return maxPosSum;
   }

   // Driver Code

   // Given Input
   let N = 4;
   let A = [5, 1, 2, 3], B = [1, 4, 3, 2];

   document.write(maxSum(A, B));

 // This code is contributed by Potta Lokesh
 </script>

Output

33

Time Complexity: O(N³)
Auxiliary Space: O(1)

Efficient Approach: The above problem can be solved with the use of dynamic programming. Follow the below steps to solve this problem:

• Let dp[L][[R] represent the sum of the product of A[i] and B[i] after reversing subarray A[L, R].
• Observe that if the subarray A[L, R] is reversed, the subarray A[L+1, R-1] is also reversed. Therefore, dp[L][R] can be calculated by the use of dp[L+1][R-1] and by using the formula:

 dp[L][R] = dp[L+1][R-1]+ (A[R] * B[L]) – (A[L] * B[L]) + (A[L] * B[R]) – (A[R] * B[R])

• Because In the calculation of dp[L+1][R-1], A[L]*B[L] and A[R]*B[R] is added whereas for the calculation dp[L][R], A[R]*B[L] and A[L]*B[R ]is added.
• Hence we need to subtract A[L]*B[L] and A[R]*B[R] and add A[R]*B[L] and A[L]*B[R] to dp[L+1][R-1] for calculation of dp[L][R].
• Print the answer according to the above observation.

Below is the implementation of the dynamic programming approach.

C++

// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the maximum sum of A[i]*B[i]
// across all values of i from 0 to N-1 by reversing
// at most one subarray of array A
int maxSum(vector<int>& A, vector<int>& B)
{

    int N = A.size();

    // Initialising maximum possible sum variable
    int maxPosSum = 0;
    int dp[N][N];

    // Initialising the dp array
    memset(dp, 0, sizeof(dp));

    // Value of maxPosSum when no subarray is reversed
    for (int i = 0; i < N; i++)
        maxPosSum += A[i] * B[i];

    // Initialising dp for subarray of length 1
    for (int i = 0; i < N; i++)
        dp[i][i] = maxPosSum;

    // Initialising dp for subarray of length 2
    for (int i = 0; i < N - 1; i++) {
        int R = i + 1;
        int L = i;
        dp[L][R] = maxPosSum + (A[R] * B[L]) - (A[L] * B[L])
                   + (A[L] * B[R]) - (A[R] * B[R]);
    }

    // Calculating the complete dp array
    for (int R = 0; R < N; R++) {
        for (int L = 0; L < N; L++) {

            // If length of subarray is less 3, then
            // continuing
            if (R - L + 1 < 3)
                continue;
            dp[L][R] = dp[L + 1][R - 1] + (A[R] * B[L])
                       - (A[L] * B[L]) + (A[L] * B[R])
                       - (A[R] * B[R]);
        }
    }

    // Updating the maxPosSum variable
    for (int L = 0; L < N; L++) {
        for (int R = L; R < N; R++) {
            maxPosSum = max(maxPosSum, dp[L][R]);
        }
    }

    // Returning the maximum possible sum of product
    return maxPosSum;
}

// Driver Code
int main()
{
    // Given Input
    vector<int> A = { 5, 1, 2, 3 }, B = { 1, 4, 3, 2 };

    cout << maxSum(A, B);
    return 0;
}

Java

// Java implementation of the above approach
import java.util.*;

class GFG{

// Function to find the maximum sum of A[i]*B[i]
// across all values of i from 0 to N-1 by reversing
// at most one subarray of array A
static int maxSum(int[] A, int[] B)
{

    int N = A.length;

    // Initialising maximum possible sum variable
    int maxPosSum = 0;
    int [][]dp = new int[N][N];


    // Value of maxPosSum when no subarray is reversed
    for (int i = 0; i < N; i++)
        maxPosSum += A[i] * B[i];

    // Initialising dp for subarray of length 1
    for (int i = 0; i < N; i++)
        dp[i][i] = maxPosSum;

    // Initialising dp for subarray of length 2
    for (int i = 0; i < N - 1; i++) {
        int R = i + 1;
        int L = i;
        dp[L][R] = maxPosSum + (A[R] * B[L]) - (A[L] * B[L])
                   + (A[L] * B[R]) - (A[R] * B[R]);
    }

    // Calculating the complete dp array
    for (int R = 0; R < N; R++) {
        for (int L = 0; L < N; L++) {

            // If length of subarray is less 3, then
            // continuing
            if (R - L + 1 < 3)
                continue;
            dp[L][R] = dp[L + 1][R - 1] + (A[R] * B[L])
                       - (A[L] * B[L]) + (A[L] * B[R])
                       - (A[R] * B[R]);
        }
    }

    // Updating the maxPosSum variable
    for (int L = 0; L < N; L++) {
        for (int R = L; R < N; R++) {
            maxPosSum = Math.max(maxPosSum, dp[L][R]);
        }
    }

    // Returning the maximum possible sum of product
    return maxPosSum;
}

// Driver Code
public static void main(String[] args)
{
    // Given Input
    int[] A = { 5, 1, 2, 3 }, B = { 1, 4, 3, 2 };

    System.out.print(maxSum(A, B));
}
}

// This code is contributed by 29AjayKumar

Python

# Python implementation of the above approach

# Function to find the maximum sum of A[i]*B[i]
# across all values of i from 0 to N-1 by reversing
# at most one subarray of array A
def maxSum(A, B):

    N = len(A)

    # Initialising maximum possible sum variable
    maxPosSum = 0

    # Initialising the dp array
    dp = ([[0 for i in range(N)]
           for i in range(N)])

    # Value of maxPosSum when no subarray is reversed
    for i in range(0, N):
        maxPosSum = maxPosSum + (A[i] * B[i])

    # Initialising dp for subarray of length 1
    for i in range(0, N):
        dp[i][i] = maxPosSum

    # Initialising dp for subarray of length 2
    for i in range(0, N - 1):
        R = i + 1
        L = i
        dp[L][R] = maxPosSum + (A[R] * B[L]) - \
            (A[L] * B[L]) + (A[L] * B[R]) - (A[R] * B[R])

    # Calculating the complete dp array
    for R in range(0, N):
        for L in range(0, N):

            # If length of subarray is less 3, then
            # continuing
            if (R - L + 1 < 3):
                continue
            dp[L][R] = dp[L + 1][R - 1] + \
                (A[R] * B[L]) - (A[L] * B[L]) + (A[L] * B[R]) - (A[R] * B[R])

    # Updating the maxPosSum variable
    for R in range(0, N):
        for L in range(0, N):
            maxPosSum = max(maxPosSum, dp[L][R])

    # Returning the maximum possible sum of product
    return maxPosSum

# Driver Code
# Given Input
A = [5, 1, 2, 3]
B = [1, 4, 3, 2]

print(maxSum(A, B))

# This code is contributed by Samim Hossain Mondal.

C#

// C# implementation of the above approach
using System;
class GFG {

    // Function to find the maximum sum of A[i]*B[i]
    // across all values of i from 0 to N-1 by reversing
    // at most one subarray of array A
    static int maxSum(int[] A, int[] B)
    {

        int N = A.Length;

        // Initialising maximum possible sum variable
        int maxPosSum = 0;
        int[, ] dp = new int[N, N];

        // Initialising the dp array
        // memset(dp, 0, sizeof(dp));

        // Value of maxPosSum when no subarray is reversed
        for (int i = 0; i < N; i++)
            maxPosSum += A[i] * B[i];

        // Initialising dp for subarray of length 1
        for (int i = 0; i < N; i++)
            dp[i, i] = maxPosSum;

        // Initialising dp for subarray of length 2
        for (int i = 0; i < N - 1; i++) {
            int R = i + 1;
            int L = i;
            dp[L, R] = maxPosSum + (A[R] * B[L])
                       - (A[L] * B[L]) + (A[L] * B[R])
                       - (A[R] * B[R]);
        }

        // Calculating the complete dp array
        for (int R = 0; R < N; R++) {
            for (int L = 0; L < N; L++) {

                // If length of subarray is less 3, then
                // continuing
                if (R - L + 1 < 3)
                    continue;
                dp[L, R] = dp[L + 1, R - 1] + (A[R] * B[L])
                           - (A[L] * B[L]) + (A[L] * B[R])
                           - (A[R] * B[R]);
            }
        }

        // Updating the maxPosSum variable
        for (int L = 0; L < N; L++) {
            for (int R = L; R < N; R++) {
                maxPosSum = Math.Max(maxPosSum, dp[L, R]);
            }
        }

        // Returning the maximum possible sum of product
        return maxPosSum;
    }

    // Driver Code
    public static void Main()
    {
        // Given Input
        int[] A = { 5, 1, 2, 3 }, B = { 1, 4, 3, 2 };

        Console.WriteLine(maxSum(A, B));
    }
}

// This code is contributed by ukasp.

Javascript

<script>
// javascript implementation of the above approach

// Function to find the maximum sum of A[i]*B[i]
// across all values of i from 0 to N-1 by reversing
// at most one subarray of array A
function maxSum(A, B)
{

    var N = A.length;

    // Initialising maximum possible sum variable
    var maxPosSum = 0;
    var dp = Array(N).fill(0).map(x => Array(N).fill(0));

    // Value of maxPosSum when no subarray is reversed
    for (var i = 0; i < N; i++)
        maxPosSum += A[i] * B[i];

    // Initialising dp for subarray of length 1
    for (var i = 0; i < N; i++)
        dp[i][i] = maxPosSum;

    // Initialising dp for subarray of length 2
    for (var i = 0; i < N - 1; i++) {
        var R = i + 1;
        var L = i;
        dp[L][R] = maxPosSum + (A[R] * B[L]) - (A[L] * B[L])
                   + (A[L] * B[R]) - (A[R] * B[R]);
    }

    // Calculating the complete dp array
    for (var R = 0; R < N; R++) {
        for (var L = 0; L < N; L++) {

            // If length of subarray is less 3, then
            // continuing
            if (R - L + 1 < 3)
                continue;
            dp[L][R] = dp[L + 1][R - 1] + (A[R] * B[L])
                       - (A[L] * B[L]) + (A[L] * B[R])
                       - (A[R] * B[R]);
        }
    }

    // Updating the maxPosSum variable
    for (var L = 0; L < N; L++) {
        for (var R = L; R < N; R++) {
            maxPosSum = Math.max(maxPosSum, dp[L][R]);
        }
    }

    // Returning the maximum possible sum of product
    return maxPosSum;
}

// Driver Code
// Given Input
var A = [ 5, 1, 2, 3 ], B = [ 1, 4, 3, 2 ];

document.write(maxSum(A, B));

// This code is contributed by 29AjayKumar
</script>

Output

33

Time Complexity: O(N²)
Auxiliary Space: O(N²)