Count pairs with bitwise OR less than Max

Given an array arr[] of N integers, the task is to count the pairs of indices (i, j) such that 0 ? i < j ? N and arr[i] | arr[j] ? max(arr[i], arr[j]) where | is the bitwise OR.
Examples:

Input: arr[] = {1, 2, 3} 
Output: 2 
(1, 3) and (2, 3) are the only valid pairs.
Input: arr[] = {11, 45, 12, 14, 5} 
Output: 3 

Approach: Run two nested loops and check for every possible pair. If arr[i] | arr[j] <= max(arr[i], arr[j]) then increment the count.
Below is the implementation of the above approach: 

C++

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

// Function to return the count of valid pairs
int countPairs(int arr[], int n)
{
    int cnt = 0;

    // Check all possible pairs
    for (int i = 0; i < n - 1; i++)
        for (int j = i + 1; j < n; j++)
            if ((arr[i] | arr[j]) <= max(arr[i], arr[j]))
                cnt++;

    return cnt;
}

// Driver code
int main()
{
    int arr[] = { 1, 2, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << countPairs(arr, n);

    return 0;
}

Java

// Java implementation of the approach

class GFG
{

// Function to return the count of valid pairs
static int countPairs(int arr[], int n)
{
    int cnt = 0;

    // Check all possible pairs
    for (int i = 0; i < n - 1; i++)
        for (int j = i + 1; j < n; j++)
            if ((arr[i] | arr[j]) <= Math.max(arr[i], arr[j]))
                cnt++;

    return cnt;
}

// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3 };
    int n = arr.length;
    System.out.println(countPairs(arr, n));
}
}

// This code is Contributed by Code_Mech.

Python3

# Python 3 implementation of the approach

# Function to return the count
# of valid pairs
def countPairs(arr, n):
    cnt = 0

    # Check all possible pairs
    for i in range(n - 1):
        for j in range(i + 1, n, 1):
            if ((arr[i] | arr[j]) <= max(arr[i],
                                         arr[j])):
                cnt += 1

    return cnt

# Driver code
if __name__ == '__main__':
    arr = [1, 2, 3]
    n = len(arr)
    print(countPairs(arr, n))

# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of the approach
using System;

class GFG
{

// Function to return the count of valid pairs
static int countPairs(int []arr, int n)
{
    int cnt = 0;

    // Check all possible pairs
    for (int i = 0; i < n - 1; i++)
        for (int j = i + 1; j < n; j++)
            if ((arr[i] | arr[j]) <= Math.Max(arr[i], arr[j]))
                cnt++;

    return cnt;
}

// Driver code
static void Main()
{
    int []arr = { 1, 2, 3 };
    int n = arr.Length;
    Console.WriteLine(countPairs(arr, n));
}
}

// This code is Contributed by mits.

PHP

<?php
// PHP implementation of the approach

// Function to return the count of valid pairs
function countPairs($arr, $n)
{
    $cnt = 0;

    // Check all possible pairs
    for ($i = 0; $i < $n - 1; $i++)
        for ($j = $i + 1; $j < $n; $j++)
            if (($arr[$i] |
                 $arr[$j]) <= max($arr[$i],
                                  $arr[$j]))
                $cnt++;

    return $cnt;
}

// Driver code
$arr = array(1, 2, 3);
$n = sizeof($arr);
echo countPairs($arr, $n);

// This code is contributed
// by Akanksha Rai
?>

Javascript

<script>

// JavaScript implementation of the approach


// Function to return the count of valid pairs
function countPairs(arr, n) {
    let cnt = 0;

    // Check all possible pairs
    for (let i = 0; i < n - 1; i++)
        for (let j = i + 1; j < n; j++)
            if ((arr[i] | arr[j]) <= Math.max(arr[i], arr[j]))
                cnt++;

    return cnt;
}

// Driver code

let arr = [1, 2, 3];
let n = arr.length;
document.write(countPairs(arr, n));

// This code is contributed by _saurabh_jaiswal

</script>

Output:

2

Time Complexity: O(n²), where n is the size of the given array.
Auxiliary Space: O(1), no extra space is required, so it is a constant.