Minimum number of elements that should be removed to make the array good

Given an array arr[], the task is to find the minimum number of elements that must be removed to make the array good. A sequence a₁, a₂ … a_n is called good if for each element a_i, there exists an element a_j (i not equals to j) such that a_i + a_j is a power of two i.e. 2^d for some non-negative integer d.
Examples:

Input: arr[] = {4, 7, 1, 5, 4, 9} 
Output: 1 
Remove 5 from the array to make the array good.
Input: arr[] = {1, 3, 1, 1} 
Output: 0 

Approach: We should delete only such a_i for which there is no a_j (i not equals to j) such that a_i + a_j is a power of 2. 
For each value let’s find the number of its occurrences in the array. We can use the map data-structure.
Now we can easily check that a_i doesn’t have a pair a_j. Let’s iterate over all possible sums, S = 2⁰, 2¹, …, 2³⁰ and for each S calculate S – a[i] whether it exists in the map.
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 number
// of elements that must be removed
// to make the given array good
int minimumRemoval(int n, int a[])
{

    map<int, int> c;

    // Count frequency of each element
    for (int i = 0; i < n; i++)
        c[a[i]]++;

    int ans = 0;

    // For each element check if there
    // exists another element that makes
    // a valid pair
    for (int i = 0; i < n; i++) {
        bool ok = false;
        for (int j = 0; j < 31; j++) {
            int x = (1 << j) - a[i];
            if (c.count(x) && (c[x] > 1
                       || (c[x] == 1 && x != a[i]))) {
                ok = true;
                break;
            }
        }

        // If does not exist then
        // increment answer
        if (!ok)
            ans++;
    }

    return ans;
}

// Driver code
int main()
{
    int a[] = { 4, 7, 1, 5, 4, 9 };
    int n = sizeof(a) / sizeof(a[0]);
    cout << minimumRemoval(n, a);

    return 0;
}

Java

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

class GFG
{

// Function to return the minimum number
// of elements that must be removed
// to make the given array good
static int minimumRemoval(int n, int a[])
{

    Map<Integer,Integer> c = new HashMap<>();

    // Count frequency of each element
    for (int i = 0; i < n; i++)
        if(c.containsKey(a[i]))
        {
            c.put(a[i], c.get(a[i])+1);
        }
        else
        {
            c.put(a[i], 1);
        }

    int ans = 0;

    // For each element check if there
    // exists another element that makes
    // a valid pair
    for (int i = 0; i < n; i++)
    {
        boolean ok = false;
        for (int j = 0; j < 31; j++)
        {
            int x = (1 << j) - a[i];
            if ((c.get(x) != null && (c.get(x) > 1)) ||
                c.get(x) != null && (c.get(x) == 1 && x != a[i]))
            {
                ok = true;
                break;
            }
        }

        // If does not exist then
        // increment answer
        if (!ok)
            ans++;
    }

    return ans;
}

// Driver code
public static void main(String[] args)
{
    int a[] = { 4, 7, 1, 5, 4, 9 };
    int n = a.length;
    System.out.println(minimumRemoval(n, a));
}
}

/* This code contributed by PrinciRaj1992 */

Python3

# Python3 implementation of the approach

# Function to return the minimum number
# of elements that must be removed
# to make the given array good
def minimumRemoval(n, a) :

    c = dict.fromkeys(a, 0);

    # Count frequency of each element
    for i in range(n) :
        c[a[i]] += 1;

    ans = 0;

    # For each element check if there
    # exists another element that makes
    # a valid pair
    for i in range(n) :
        ok = False;
        for j in range(31) :

            x = (1 << j) - a[i];
            if (x in c and (c[x] > 1 or
               (c[x] == 1 and x != a[i]))) :

                ok = True;
                break;

        # If does not exist then
        # increment answer
        if (not ok) :
            ans += 1;

    return ans;

# Driver Code
if __name__ == "__main__" :

    a = [ 4, 7, 1, 5, 4, 9 ];
    n = len(a) ;

    print(minimumRemoval(n, a));

# This code is contributed by Ryuga

C#

// C# implementation of the approach
using System.Linq;
using System;

class GFG
{
// Function to return the minimum number
// of elements that must be removed
// to make the given array good
static int minimumRemoval(int n, int []a)
{

    int[] c = new int[1000];

    // Count frequency of each element
    for (int i = 0; i < n; i++)
        c[a[i]]++;

    int ans = 0;

    // For each element check if there
    // exists another element that makes
    // a valid pair
    for (int i = 0; i < n; i++)
    {
        bool ok = true;
        for (int j = 0; j < 31; j++)
        {
            int x = (1 << j) - a[i];
            if (c.Contains(x) && (c[x] > 1 ||
                    (c[x] == 1 && x != a[i])))
            {
                ok = false;
                break;
            }
        }

        // If does not exist then
        // increment answer
        if (!ok)
            ans++;
    }

    return ans;
}

// Driver code
static void Main()
{
    int []a = { 4, 7, 1, 5, 4, 9 };
    int n = a.Length;
    Console.WriteLine(minimumRemoval(n, a));
}
}

// This code is contributed by mits

Javascript

<script>

// JavaScript implementation of the approach

// Function to return the minimum number
// of elements that must be removed
// to make the given array good
function minimumRemoval( n, a)
{

    var c = {};

    // Count frequency of each element
    for (let i = 0; i < n; i++){
        if(a[i] in c)
          c[a[i]]++;
        else
          c[a[i]] = 1;
    }

    var ans = 0;

    // For each element check if there
    // exists another element that makes
    // a valid pair
    for (let i = 0; i < n; i++) {
        var ok = false;
        for (let j = 0; j < 31; j++) {
            let x = (1 << j) - a[i];

            if ((x in c && c[x] > 1)
                       || (c[x] == 1 && x != a[i])) {

                ok = true;
                break;
            }
        }

        // If does not exist then
        // increment answer
        if (!ok)
            ans++;
    }

    return ans;
}

// Driver code
var a = new Array( 4, 7, 1, 5, 4, 9 );
var n = a.length;
console.log(minimumRemoval(n, a));

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

Output:

1

Time Complexity: O(nlogn)

Auxiliary Space: O(n)