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

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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 … an is called good if for each element ai, there exists an element aj (i not equals to j) such that ai + aj is a power of two i.e. 2d for some non-negative integer d.
Examples: 

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

Approach: We should delete only such ai for which there is no aj (i not equals to j) such that ai + aj 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 ai doesn’t have a pair aj. 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)

Another Approach:

  1. Create a function “isPowerOfTwo” to check if a number is a power of two.
  2. Create a function “countRemovedElements” that takes an array as input and returns the minimum number of elements that must be removed to make the array good.
  3. Initialize a count variable to keep track of the number of removed elements.
  4. Loop through the array and for each element, check if there exists another element in the array that can form a power of two with it.
  5. If such an element is found, continue to the next element in the array.
  6. If no such element is found, increment the count variable.
  7. Return the count of removed elements.
  8. In the “main” function, create an example array and call the “countRemovedElements” function to get the minimum number of elements that must be removed to make the array good.
  9. Print the output.

Below is the implementation of the above approach:

C++




#include <iostream>
#include <vector>
#include <cmath>
 
using namespace std;
 
// function to check if a number is a power of two
bool isPowerOfTwo(int x) {
    return (x && !(x & (x - 1)));
}
 
// function to count the minimum number of elements that must be removed to make the array good
int countRemovedElements(vector<int>& arr) {
    int n = arr.size();
    int count = 0;
    for (int i = 0; i < n; i++) {
        bool found = false;
        // loop through the array to find another element that
       // can form a power of two with the current element
        for (int j = 0; j < n; j++) {
            // check if the current element is not the same as the other element,
           // and their sum is a power of two
            if (i != j && isPowerOfTwo(arr[i] + arr[j])) {
                found = true;
                break;
            }
        }
        // if no other element is found to form a power of two with
       // the current element, increment the count of removed elements
        if (!found) {
            count++;
        }
    }
    // return the count of removed elements
    return count;
}
 
int main() {
    // example array
    vector<int> arr = {4, 7, 1, 5, 4, 9};
    // call the function to count the minimum number of elements
   // that must be removed to make the array good
    cout << countRemovedElements(arr) << endl; // Output: 1
     
    return 0;
}
//This code is contributed by rudra1807raj


Java




import java.util.*;
 
public class Main {
    // function to check if a number is a power of two
    public static boolean isPowerOfTwo(int x)
    {
        return (x & (x - 1)) == 0;
    }
 
    // function to count the minimum number of elements that
    // must be removed to make the array good
    public static int
    countRemovedElements(ArrayList<Integer> arr)
    {
        int n = arr.size();
        int count = 0;
        for (int i = 0; i < n; i++) {
            boolean found = false;
            // loop through the array to find another
            // element that can form a power of two with the
            // current element
            for (int j = 0; j < n; j++) {
                // check if the current element is not the
                // same as the other element, and their sum
                // is a power of two
                if (i != j
                    && isPowerOfTwo(arr.get(i)
                                    + arr.get(j))) {
                    found = true;
                    break;
                }
            }
            // if no other element is found to form a power
            // of two with the current element, increment
            // the count of removed elements
            if (!found) {
                count++;
            }
        }
        // return the count of removed elements
        return count;
    }
 
    public static void main(String[] args)
    {
        // example array
        ArrayList<Integer> arr = new ArrayList<>(
            Arrays.asList(4, 7, 1, 5, 4, 9));
        // call the function to count the minimum number of
        // elements that must be removed to make the array
        // good
        System.out.println(
            countRemovedElements(arr)); // Output: 1
    }
}


Python3




import math
 
# function to check if a number is a power of two
def isPowerOfTwo(x):
    return (x and not (x & (x - 1)))
 
# function to count the minimum number of elements
# that must be removed to make the array good
def countRemovedElements(arr):
    n = len(arr)
    count = 0
    for i in range(n):
        found = False
        # loop through the array to find another element
        # that can form a power of two with the current element
        for j in range(n):
            # check if the current element is not
            # the same as the other element, and
            # their sum is a power of two
            if i != j and isPowerOfTwo(arr[i] + arr[j]):
                found = True
                break
        # if no other element is found to form
        # a power of two with the current element,
        # increment the count of removed elements
        if not found:
            count += 1
    # return the count of removed elements
    return count
 
 
# example array
arr = [4, 7, 1, 5, 4, 9]
 
# call the function to count the minimum number of elements
# that must be removed to make the array good
print(countRemovedElements(arr))  # Output: 1


C#




using System;
using System.Collections.Generic;
 
class Program
{
    // Function to check if a number is a power of two
    static bool IsPowerOfTwo(int x)
    {
        return (x != 0) && ((x & (x - 1)) == 0);
    }
 
    // Function to count the minimum number of elements that must be removed to make the array good
    static int CountRemovedElements(List<int> arr)
    {
        int n = arr.Count;
        int count = 0;
 
        for (int i = 0; i < n; i++)
        {
            bool found = false;
 
            // Loop through the array to find another element that can form a power of two with the current element
            for (int j = 0; j < n; j++)
            {
                // Check if the current element is not the same as the other element,
                // and their sum is a power of two
                if (i != j && IsPowerOfTwo(arr[i] + arr[j]))
                {
                    found = true;
                    break;
                }
            }
 
            // If no other element is found to form a power of two with the current element, increment the count of removed elements
            if (!found)
            {
                count++;
            }
        }
 
        // Return the count of removed elements
        return count;
    }
 
    static void Main()
    {
        // Example list
        List<int> arr = new List<int> { 4, 7, 1, 5, 4, 9 };
 
        // Call the function to count the minimum number of elements that must be removed to make the list good
        Console.WriteLine(CountRemovedElements(arr)); // Output: 1
    }
}


Javascript




// Function to check if a number is a power of two
function isPowerOfTwo(x) {
    return (x && !(x & (x - 1)));
}
 
// Function to count the minimum number of elements that must be removed to make the array good
function countRemovedElements(arr) {
    const n = arr.length;
    let count = 0;
     
    for (let i = 0; i < n; i++) {
        let found = false;
         
        // Loop through the array to find another element that can form a power of two with the current element
        for (let j = 0; j < n; j++) {
            // Check if the current element is not the same as the other element, and their sum is a power of two
            if (i !== j && isPowerOfTwo(arr[i] + arr[j])) {
                found = true;
                break;
            }
        }
         
        // If no other element is found to form a power of two with the current element, increment the count of removed elements
        if (!found) {
            count++;
        }
    }
     
    // Return the count of removed elements
    return count;
}
 
// Example array
const arr = [4, 7, 1, 5, 4, 9];
 
// Call the function to count the minimum number of elements that must be removed to make the array good
console.log(countRemovedElements(arr)); // Output: 1
 
// This code is contributed by shivamgupta0987654321


Output

1

Time Complexity: The time complexity of the given code is O(n^2), where n is the size of the input array. This is because the code contains a nested loop, where for each element in the array, it loops through the entire array to find another element that can form a power of two with it. Therefore, the time complexity is proportional to the square of the input size.

Auxiliary Space: The space complexity of the code is O(1) because the code uses only a constant amount of additional memory to perform its operations, regardless of the size of the input array. The code does not create any additional data structures or use any recursive calls, and therefore the amount of memory used is constant.



Last Updated : 31 Oct, 2023
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