Largest subset where absolute difference of any two element is a power of 2

Given an array arr[] of distinct elements -109 ≤ ai ≤ 109. The task is to find the largest sub-set from the given array such that the absolute difference between any two numbers in the sub-set is a positive power of two. If it is not possible to make such sub-set then print -1.

Examples:

Input: arr[] = {3, 4, 5, 6, 7}
Output: 3 5 7
|3 – 5| = 21, |5 – 7| = 21 and |3 – 7| = 22.

Input: arr[] = {2, 5, 8}
Output: -1

Approach: Let’s prove that the size of the sub-set will not be > 3. Suppose a, b, c and d are four elements from a sub-set and a < b < c < d.
Let abs(a – b) = 2k and abs(b – c) = 2l then abs(a – c) = abs(a – b) + abs(b – c) = 2k + 2l = 2m. It means that k = l. Conditions must hold for the triple (b, c, d) too. Now it is easy to see that if abs(a – b) = abs(b – c) = abs(c – d) = 2k then abs(a – d) = abs(a – b) * 3 which is not a power of two. So the size of the sub-set will never be greater than 3.



  • Let’s check if the answer is 3. Iterate over the given array for middle elements of the sub-set and for powers of two from 1 to 30 inclusively. Let xi be the middle element of the sub-set and j the current power of two. Then if there are elements xi-2j and xi+2j in the array then the answer is 3.
  • Else check if the answer is 2. repeat the previous step but here one can get either left point xi-2j or xi+2j.
  • If the answer is neither 2 nor 3 then print -1.

Below is the implementation of the above approach:

C++

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// CPP program to find sub-set with
// maximum possible size
#include <bits/stdc++.h>
using namespace std;
  
// Function to find sub-set with
// maximum possible size
void PowerOfTwo(vector<int> x, int n)
{
    // Sort the given array
    sort(x.begin(), x.end());
  
    // To store required sub-set
    vector<int> res;
  
    for (int i = 0; i < n; ++i) {
  
        // Iterate for all powers of two
        for (int j = 1; j < 31; ++j) {
  
            // Left number
            int lx = x[i] - (1 << j);
  
            // Right number
            int rx = x[i] + (1 << j);
  
            // Predefined binary search in c++
            bool isl = binary_search(x.begin(), x.end(), lx);
            bool isr = binary_search(x.begin(), x.end(), rx);
  
            // If possible to get sub-set of size 3
            if (isl && isr && int(res.size()) < 3)
                res = { lx, x[i], rx };
  
            // If possible to get sub-set of size 2
            if (isl && int(res.size()) < 2)
                res = { lx, x[i] };
  
            // If possible to get sub-set of size 2
            if (isr && int(res.size()) < 2)
                res = { x[i], rx };
        }
    }
  
    // If not possible to get sub-set
    if (!res.size()) {
        cout << -1;
        return;
    }
  
    // Print the sub-set
    for (auto it : res)
        cout << it << " ";
}
  
// Driver Code
int main()
{
  
    vector<int> a = { 3, 4, 5, 6, 7 };
  
    int n = a.size();
    PowerOfTwo(a, n);
  
    return 0;
}

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Java

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// Java program to find sub-set with
// maximum possible size
import java.util.*;
  
class GFG
{
      
// Function to find sub-set with
// maximum possible size
static void PowerOfTwo(int []x, int n)
{
    // Sort the given array
    Arrays.sort(x);
  
    // To store required sub-set
    ArrayList<Integer> res = new ArrayList<Integer>();
  
    for (int i = 0; i < n; ++i) 
    {
  
        // Iterate for all powers of two
        for (int j = 1; j < 31; ++j) 
        {
  
            // Left number
            int lx = x[i] - (1 << j);
  
            // Right number
            int rx = x[i] + (1 << j);
  
            // Predefined binary search in Java
            boolean isl = Arrays.binarySearch(x,lx) <
                                    0 ? false : true;
            boolean isr = Arrays.binarySearch(x,rx) < 
                                    0 ? false : true;
  
            // If possible to get sub-set of size 3
            if (isl && isr && res.size() < 3)
            {
                res.clear();
                res.add(lx);
                res.add(x[i]);
                res.add(rx);
            }
  
            // If possible to get sub-set of size 2
            if (isl && res.size() < 2)
            {
                res.clear();
                res.add(lx);
                res.add(x[i]);
            }
            // If possible to get sub-set of size 2
            if (isr && res.size() < 2)
            {
                res.clear();
                res.add(x[i]);
                res.add(rx);
            }
        }
    }
  
    // If not possible to get sub-set
    if (res.size() == 0)
    {
        System.out.println("-1");
        return;
    }
  
    // Print the sub-set
    for (int i = 0; i < res.size(); i++)
        System.out.print(res.get(i) + " ");
}
  
// Driver Code
public static void main (String[] args) 
{
    int[] a = {3, 4, 5, 6, 7};
    int n = a.length;
    PowerOfTwo(a, n);
}
}
  
// This code is Contributed by chandan_jnu

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Python3

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# Python3 program to find sub-set with 
# maximum possible size 
  
# Function to find sub-set with 
# maximum possible size 
def PowerOfTwo(x, n) : 
      
    # Sort the given array 
    x.sort()
  
    # To store required sub-set 
    res = [] 
  
    for i in range(n) : 
  
        # Iterate for all powers of two 
        for j in range(1, 31) :
              
            # Left number 
            lx = x[i] - (1 << j)
  
            # Right number 
            rx = x[i] + (1 << j) 
  
            if lx in x :
                isl = True
            else :
                isl = False
              
            if rx in x :
                isr = True
            else :
                isr = False
              
            # If possible to get sub-set of size 3 
            if (isl and isr and len(res) < 3) :
                res = [ lx, x[i], rx ] 
  
            # If possible to get sub-set of size 2 
            if (isl and len(res) < 2) :
                res = [ lx, x[i] ] 
  
            # If possible to get sub-set of size 2 
            if (isr and len(res) < 2) :
                res = [ x[i], rx ]
  
    # If not possible to get sub-set 
    if (not len(res)) :
        print(-1
        return
      
    # Print the sub-set 
    for it in res :
        print(it, end = " "
  
# Driver Code 
if __name__ == "__main__" :
  
    a = [ 3, 4, 5, 6, 7
  
    n = len(a)
    PowerOfTwo(a, n) 
  
# This code is contributed by Ryuga

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C#

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// C# program to find sub-set with
// maximum possible size
using System;
using System.Collections;
  
class GFG
{
      
// Function to find sub-set with
// maximum possible size
static void PowerOfTwo(int[] x, int n)
{
    // Sort the given array
    Array.Sort(x);
  
    // To store required sub-set
    ArrayList res = new ArrayList();
  
    for (int i = 0; i < n; ++i) 
    {
  
        // Iterate for all powers of two
        for (int j = 1; j < 31; ++j) 
        {
  
            // Left number
            int lx = x[i] - (1 << j);
  
            // Right number
            int rx = x[i] + (1 << j);
  
            // Predefined binary search in C#
            bool isl = Array.IndexOf(x, lx) < 0? false : true;
            bool isr = Array.IndexOf(x, rx) < 0? false : true;
  
            // If possible to get sub-set of size 3
            if (isl && isr && res.Count < 3)
            {
                res.Clear();
                res.Add(lx);
                res.Add(x[i]);
                res.Add(rx);
            }
  
            // If possible to get sub-set of size 2
            if (isl && res.Count < 2)
            {
                res.Clear();
                res.Add(lx);
                res.Add(x[i]);
            }
            // If possible to get sub-set of size 2
            if (isr && res.Count < 2)
            {
                res.Clear();
                res.Add(x[i]);
                res.Add(rx);
            }
        }
    }
  
    // If not possible to get sub-set
    if (res.Count == 0)
    {
        Console.Write("-1");
        return;
    }
  
    // Print the sub-set
    for (int i = 0; i < res.Count; i++)
        Console.Write(res[i] + " ");
}
  
// Driver Code
public static void Main() 
{
    int[] a = {3, 4, 5, 6, 7};
    int n = a.Length;
    PowerOfTwo(a, n);
}
}
  
// This code is Contributed by chandan_jnu

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PHP

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<?php
// PHP program to find sub-set with
// maximum possible size
  
// Function to find sub-set with
// maximum possible size
function PowerOfTwo($x, $n)
{
    // Sort the given array
    sort($x);
  
    // To store required sub-set
    $res = array();
  
    for ($i = 0; $i < $n; ++$i
    {
  
        // Iterate for all powers of two
        for ($j = 1; $j < 31; ++$j
        {
  
            // Left number
            $lx = $x[$i] - (1 << $j);
  
            // Right number
            $rx = $x[$i] + (1 << $j);
  
            // Predefined binary search in PHP
            $isl = in_array($lx, $x);
            $isr = in_array($rx, $x);
  
            // If possible to get sub-set of size 3
            if ($isl && $isr && count($res) < 3)
            {
                unset($res);
                $res = array();
                array_push($res, $lx);
                array_push($res, $x[$i]);
                array_push($res, $rx);
            }
  
            // If possible to get sub-set of size 2
            if ($isl && count($res) < 2)
            {
                unset($res);
                $res = array();
                array_push($res, $lx);
                array_push($res, $x[$i]);
            }
  
            // If possible to get sub-set of size 2
            if ($isr && count($res) < 2)
            {
                unset($res);
                $res = array();
                array_push($res, $x[$i]);
                array_push($res, $rx);
            }
        }
    }
  
    // If not possible to get sub-set
    if (!count($res))
    {
        echo "-1";
        return;
    }
  
    // Print the sub-set
    for ($i = 0; $i < count($res); $i++)
        echo $res[$i] . " ";
}
  
// Driver Code
$a = array( 3, 4, 5, 6, 7 );
  
$n = count($a);
PowerOfTwo($a, $n);
  
// This code is contributed by chandan_jnu
?>

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Output:

3 5 7

Time Complexity : O(N*logN)
Auxiliary Space : O(1)

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