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 … 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: 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 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++
#include <bits/stdc++.h>
using namespace std;
int minimumRemoval( int n, int a[])
{
map< int , int > c;
for ( int i = 0; i < n; i++)
c[a[i]]++;
int ans = 0;
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 (!ok)
ans++;
}
return ans;
}
int main()
{
int a[] = { 4, 7, 1, 5, 4, 9 };
int n = sizeof (a) / sizeof (a[0]);
cout << minimumRemoval(n, a);
return 0;
}
|
Java
import java.util.*;
class GFG
{
static int minimumRemoval( int n, int a[])
{
Map<Integer,Integer> c = new HashMap<>();
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 ( 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 (!ok)
ans++;
}
return ans;
}
public static void main(String[] args)
{
int a[] = { 4 , 7 , 1 , 5 , 4 , 9 };
int n = a.length;
System.out.println(minimumRemoval(n, a));
}
}
|
Python3
def minimumRemoval(n, a) :
c = dict .fromkeys(a, 0 );
for i in range (n) :
c[a[i]] + = 1 ;
ans = 0 ;
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 ( not ok) :
ans + = 1 ;
return ans;
if __name__ = = "__main__" :
a = [ 4 , 7 , 1 , 5 , 4 , 9 ];
n = len (a) ;
print (minimumRemoval(n, a));
|
C#
using System.Linq;
using System;
class GFG
{
static int minimumRemoval( int n, int []a)
{
int [] c = new int [1000];
for ( int i = 0; i < n; i++)
c[a[i]]++;
int ans = 0;
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 (!ok)
ans++;
}
return ans;
}
static void Main()
{
int []a = { 4, 7, 1, 5, 4, 9 };
int n = a.Length;
Console.WriteLine(minimumRemoval(n, a));
}
}
|
Javascript
<script>
function minimumRemoval( n, a)
{
var c = {};
for (let i = 0; i < n; i++){
if (a[i] in c)
c[a[i]]++;
else
c[a[i]] = 1;
}
var ans = 0;
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 (!ok)
ans++;
}
return ans;
}
var a = new Array( 4, 7, 1, 5, 4, 9 );
var n = a.length;
console.log(minimumRemoval(n, a));
</script>
|
Time Complexity: O(nlogn)
Auxiliary Space: O(n)
Another Approach:
- Create a function “isPowerOfTwo” to check if a number is a power of two.
- 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.
- Initialize a count variable to keep track of the number of removed elements.
- 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.
- If such an element is found, continue to the next element in the array.
- If no such element is found, increment the count variable.
- Return the count of removed elements.
- 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.
- Print the output.
Below is the implementation of the above approach:
C++
#include <iostream>
#include <vector>
#include <cmath>
using namespace std;
bool isPowerOfTwo( int x) {
return (x && !(x & (x - 1)));
}
int countRemovedElements(vector< int >& arr) {
int n = arr.size();
int count = 0;
for ( int i = 0; i < n; i++) {
bool found = false ;
for ( int j = 0; j < n; j++) {
if (i != j && isPowerOfTwo(arr[i] + arr[j])) {
found = true ;
break ;
}
}
if (!found) {
count++;
}
}
return count;
}
int main() {
vector< int > arr = {4, 7, 1, 5, 4, 9};
cout << countRemovedElements(arr) << endl;
return 0;
}
|
Java
import java.util.*;
public class Main {
public static boolean isPowerOfTwo( int x)
{
return (x & (x - 1 )) == 0 ;
}
public static int
countRemovedElements(ArrayList<Integer> arr)
{
int n = arr.size();
int count = 0 ;
for ( int i = 0 ; i < n; i++) {
boolean found = false ;
for ( int j = 0 ; j < n; j++) {
if (i != j
&& isPowerOfTwo(arr.get(i)
+ arr.get(j))) {
found = true ;
break ;
}
}
if (!found) {
count++;
}
}
return count;
}
public static void main(String[] args)
{
ArrayList<Integer> arr = new ArrayList<>(
Arrays.asList( 4 , 7 , 1 , 5 , 4 , 9 ));
System.out.println(
countRemovedElements(arr));
}
}
|
Python3
import math
def isPowerOfTwo(x):
return (x and not (x & (x - 1 )))
def countRemovedElements(arr):
n = len (arr)
count = 0
for i in range (n):
found = False
for j in range (n):
if i ! = j and isPowerOfTwo(arr[i] + arr[j]):
found = True
break
if not found:
count + = 1
return count
arr = [ 4 , 7 , 1 , 5 , 4 , 9 ]
print (countRemovedElements(arr))
|
C#
using System;
using System.Collections.Generic;
class Program
{
static bool IsPowerOfTwo( int x)
{
return (x != 0) && ((x & (x - 1)) == 0);
}
static int CountRemovedElements(List< int > arr)
{
int n = arr.Count;
int count = 0;
for ( int i = 0; i < n; i++)
{
bool found = false ;
for ( int j = 0; j < n; j++)
{
if (i != j && IsPowerOfTwo(arr[i] + arr[j]))
{
found = true ;
break ;
}
}
if (!found)
{
count++;
}
}
return count;
}
static void Main()
{
List< int > arr = new List< int > { 4, 7, 1, 5, 4, 9 };
Console.WriteLine(CountRemovedElements(arr));
}
}
|
Javascript
function isPowerOfTwo(x) {
return (x && !(x & (x - 1)));
}
function countRemovedElements(arr) {
const n = arr.length;
let count = 0;
for (let i = 0; i < n; i++) {
let found = false ;
for (let j = 0; j < n; j++) {
if (i !== j && isPowerOfTwo(arr[i] + arr[j])) {
found = true ;
break ;
}
}
if (!found) {
count++;
}
}
return count;
}
const arr = [4, 7, 1, 5, 4, 9];
console.log(countRemovedElements(arr));
|
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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