Open In App
Related Articles

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

Improve Article
Improve
Save Article
Save
Like Article
Like

Given an array A[] of length N and an integer K, the task is to maximize the value of expression [i.j – K.(Ai | Aj)] 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.(Ai | Aj)

=> Notice that for f(i, j) to be maximum, i.j should be maximum and K.(Ai | Aj) should be minimum.
=> Thus, in the best case, f(i, j) will be maximum if 
      => K.(Ai | Aj) 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.(Ai | Aj) 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 (Ai | Aj) can be 2*N. because (0 Ai 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*(Ai | 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++




// 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




// 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




# 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#




// 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.


Javascript




<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>


 
 

Output

2

 

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

 


Feeling lost in the world of random DSA topics, wasting time without progress? It's time for a change! Join our DSA course, where we'll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 geeks!

Last Updated : 21 Mar, 2022
Like Article
Save Article
Previous
Next
Similar Reads
Complete Tutorials