Remove minimum numbers from the array to get minimum OR value

Given an array arr[] of N positive integers, the task is to find the minimum number of elements to be deleted from the array so that the bitwise OR of the array elements get minimized. You are not allowed to remove all the elements i.e. at least one element must remain in the array.
Examples:

Input: arr[] = {1, 2, 3} 
Output: 2 
All possible subsets and there OR values are: 
a) {1, 2, 3} = 3 
b) {1, 2} = 3 
c) {2, 3} = 3 
d) {1, 3} = 3 
e) {1} = 1 
f) {2} = 2 
g) {3} = 3 
The minimum possible OR will be 1 from the subset {1}. 
So, we will need to remove 2 elements.
Input: arr[] = {3, 3, 3} 
Output: 0 

Naive approach: Generate all possible sub-sequences and test which one gives the minimum ‘OR’ value. Let the length of the largest sub-sequence with minimum possible OR be L then the answer will N – L. This will take exponential time.
Better approach: The minimum value will always be equal to the smallest number present in the array. If this number get bitwise ORed with any other number other than itself then the value of the OR will change and it won’t stay minimum anymore. Thus, we need to remove all the elements that are not equal to this minimum element. 

• Find the smallest number in the array.
• Find the frequency of this element in the array say cnt.
• The final answer will be N – cnt.

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 minimum
// deletions to get minimum OR
int findMinDel(int* arr, int n)
{

    // To store the minimum element
    int min_num = INT_MAX;

    // Find the minimum element
    // from the array
    for (int i = 0; i < n; i++)
        min_num = min(arr[i], min_num);

    // To store the frequency of
    // the minimum element
    int cnt = 0;

    // Find the frequency of the
    // minimum element
    for (int i = 0; i < n; i++)
        if (arr[i] == min_num)
            cnt++;

    // Return the final answer
    return n - cnt;
}

// Driver code
int main()
{
    int arr[] = { 3, 3, 2 };
    int n = sizeof(arr) / sizeof(int);

    cout << findMinDel(arr, n);

    return 0;
}

Java

// Java implementation of the approach
class GFG
{

// Function to return the minimum
// deletions to get minimum OR
static int findMinDel(int []arr, int n)
{

    // To store the minimum element
    int min_num = Integer.MAX_VALUE;

    // Find the minimum element
    // from the array
    for (int i = 0; i < n; i++)
        min_num = Math.min(arr[i], min_num);

    // To store the frequency of
    // the minimum element
    int cnt = 0;

    // Find the frequency of the
    // minimum element
    for (int i = 0; i < n; i++)
        if (arr[i] == min_num)
            cnt++;

    // Return the final answer
    return n - cnt;
}

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

    System.out.print(findMinDel(arr, n));
}
}

// This code is contributed by PrinciRaj1992

Python3

# Python3 implementation of the approach
import sys

# Function to return the minimum
# deletions to get minimum OR
def findMinDel(arr, n) :

    # To store the minimum element
    min_num = sys.maxsize;

    # Find the minimum element
    # from the array
    for i in range(n) :
        min_num = min(arr[i], min_num);

    # To store the frequency of
    # the minimum element
    cnt = 0;

    # Find the frequency of the
    # minimum element
    for i in range(n) :
        if (arr[i] == min_num) :
            cnt += 1;

    # Return the final answer
    return n - cnt;

# Driver code
if __name__ == "__main__" :

    arr = [ 3, 3, 2 ];
    n = len(arr);

    print(findMinDel(arr, n));

# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;

class GFG
{

// Function to return the minimum
// deletions to get minimum OR
static int findMinDel(int []arr, int n)
{

    // To store the minimum element
    int min_num = int.MaxValue;

    // Find the minimum element
    // from the array
    for (int i = 0; i < n; i++)
        min_num = Math.Min(arr[i],
                           min_num);

    // To store the frequency of
    // the minimum element
    int cnt = 0;

    // Find the frequency of the
    // minimum element
    for (int i = 0; i < n; i++)
        if (arr[i] == min_num)
            cnt++;

    // Return the readonly answer
    return n - cnt;
}

// Driver code
public static void Main(String[] args)
{
    int []arr = { 3, 3, 2 };
    int n = arr.Length;

    Console.Write(findMinDel(arr, n));
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// Javascript implementation of the approach

// Function to return the minimum
// deletions to get minimum OR
function findMinDel(arr, n)
{

    // To store the minimum element
    var min_num = 1000000000;

    // Find the minimum element
    // from the array
    for (var i = 0; i < n; i++)
        min_num = Math.min(arr[i], min_num);

    // To store the frequency of
    // the minimum element
    var cnt = 0;

    // Find the frequency of the
    // minimum element
    for (var i = 0; i < n; i++)
        if (arr[i] == min_num)
            cnt++;

    // Return the final answer
    return n - cnt;
}

// Driver code
var arr = [3, 3, 2];
var n = arr.length;
document.write( findMinDel(arr, n));

// This code is contributed by noob2000.
</script>

Output:

2

Time Complexity: O(N)

Auxiliary Space: O(1)