Maximize the value of expression [i.j - K.(Ai | Aj)] over all pairs (i, j) in given Array

Given an array A[] of length N and an integer K, the task is to maximize the value of expression [i.j - K.(A_i | A_j)] over all pairs (i, j) in given Array, where (1 ≤ i < j ≤ N) and | denotes Bitwise OR operator.

Examples:

Input: A[] = {5, 20, 1, 0, 8, 11}, K = 10
Output: 2
Explanation: The maximum value of the expression f(i, j) = i.j - K.(A[i] | A[j]) can be found for below pair:
f(3, 4) = 3.4 - 10.(1 | 0) = 2

Input: A[] = {1, 5, 6, 7, 8, 19}, K = 3
Output: -9

Approach: The following observations have to be made:

Let f(i, j) = i.j - K.(A_i | A_j). 

=> Notice that for f(i, j) to be maximum, i.j should be maximum and K.(A_i | A_j) should be minimum.
=> Thus, in the best case, f(i, j) will be maximum if 
      => K.(A_i | A_j) is equal to 0
      => i = (N-1) and j = N.
=> So, the maximum value of the expression can be f(i, j) = (N-1)*N.

=> Also, the minimum value will be obtained by subtracting the maximum value of K.(A_i | A_j) from (N-1)*N. 

=> It is known that 
a | b < 2*max(a, b) where | is the bitwise OR operator

         From the above property and the given constraints, it can be inferred:

• The maximum value of (A_i | A_j) can be 2*N. because (0 ≤ A_i ≤ N)
• The maximum value of K can be 100 because (1 ≤ K ≤ min(N, 100)).
• So, the minimum value of f(i, j) will be:

f(i, j) = (N-1)*N - K*(A_i | Aj)
       = (N-1)*N - 100*2*N
       = N*(N - 201)

• It can be easily observed that the resultant answer will always lie between:

N*(N-201) <= ans <= (N-1)*N

• Also, notice that for i = N - 201 and j = N,
  • the maximum value of f(i, j) will be N*(N - 201),
  • which in turn is the minimum value of f(i, j) for i = N - 1 and j = N.
  • Thus, the maximum value of the expression has to be checked from i = N - 201 to i = N and j = i+1 to j = N.

Below is the implementation of the above approach:

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

// Function to find the maximum
// value of the given expression
long long int maxValue(int N, int K,
                       long long int A[])
{
    // Stores the maximum value of
    // the given expression
    long long int ans = LLONG_MIN;

    // Nested loops to find the maximum
    // value of the given expression
    for (long long int i = max(0, N - 201);
         i < N; ++i) {
        for (long long int j = i + 1;
             j < N; ++j) {
            ans = max(ans, (i + 1) * (j + 1)
                               - K * (A[i] | A[j]));
        }
    }

    // Return the answer
    return ans;
}

// Driver Code
int main()
{
    // Given input
    int N = 6, K = 10;
    long long int A[N]
        = { 5, 20, 1, 0, 8, 11 };

    // Function Call
    cout << maxValue(N, K, A);
    return 0;
}

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;

class GFG {

  // Function to find the maximum
  // value of the given expression
  static int maxValue(int N, int K,
                      int A[])
  {
    // Stores the maximum value of
    // the given expression
    int ans = Integer.MIN_VALUE;

    // Nested loops to find the maximum
    // value of the given expression
    for (int i = Math.max(0, N - 201);
         i < N; ++i) {
      for (int j = i + 1;
           j < N; ++j) {
        ans = Math.max(ans, (i + 1) * (j + 1)
                       - K * (A[i] | A[j]));
      }
    }

    // Return the answer
    return ans;
  }

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

    // Given input
    int N = 6, K = 10;
    int  A[]  = { 5, 20, 1, 0, 8, 11 };

    // Function Call
    System.out.println( maxValue(N, K, A));
  }
}

// This code is contributed by hrithikgarg03188.

# Python3 program for the above approach
LLONG_MIN = -9223372036854775808

# Function to find the maximum
# value of the given expression
def maxValue(N, K, A):

    # Stores the maximum value of
    # the given expression
    ans = LLONG_MIN

    # Nested loops to find the maximum
    # value of the given expression
    for i in range(max(0, N - 201), N):
        for j in range(i + 1, N):
            ans = max(ans, (i + 1) * (j + 1)
                      - K * (A[i] | A[j]))

    # Return the answer
    return ans

# Driver Code
if __name__ == "__main__":

    # Given input
    N, K = 6, 10
    A = [5, 20, 1, 0, 8, 11]

    # Function Call
    print(maxValue(N, K, A))

# This code is contributed by rakeshsahni

// C# program for the above approach
using System;

class GFG {

  // Function to find the maximum
  // value of the given expression
  static int maxValue(int N, int K,
                      int []A)
  {
    
    // Stores the maximum value of
    // the given expression
    int ans = Int32.MinValue;

    // Nested loops to find the maximum
    // value of the given expression
    for (int i = Math.Max(0, N - 201);
         i < N; ++i) {
      for (int j = i + 1;
           j < N; ++j) {
        ans = Math.Max(ans, (i + 1) * (j + 1)
                       - K * (A[i] | A[j]));
      }
    }

    // Return the answer
    return ans;
  }

  // Driver Code
  public static void Main () 
  {

    // Given input
    int N = 6, K = 10;
    int  []A  = { 5, 20, 1, 0, 8, 11 };

    // Function Call
    Console.WriteLine(maxValue(N, K, A));
  }
}

// This code is contributed by Samim Hossain Mondal.

<script>
        // JavaScript code for the above approach

        // Function to find the maximum
        // value of the given expression
        function maxValue(N, K, A)
        {
        
            // Stores the maximum value of
            // the given expression
            let ans = Number.MIN_VALUE;

            // Nested loops to find the maximum
            // value of the given expression
            for (let i = Math.max(0, N - 201);
                i < N; ++i) {
                for (let j = i + 1;
                    j < N; ++j) {
                    ans = Math.max(ans, (i + 1) * (j + 1)
                        - K * (A[i] | A[j]));
                }
            }

            // Return the answer
            return ans;
        }

        // Driver Code

        // Given input
        let N = 6, K = 10;
        let A
            = [5, 20, 1, 0, 8, 11];

        // Function Call
        document.write(maxValue(N, K, A));

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

2

Time Complexity: O(N²)
Auxiliary Space: O(1)