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++
// 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> |
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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; // 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 |
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.
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