给定不同的元件的阵列ARR [] -10 9≤A I≤10 9。任务是从给定的数组中找到最大的子集，以使子集中任意两个数字之间的绝对差为2的正幂。如果无法进行这样的设置，则打印-1 。
例子：
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
方法：让我们证明子集的大小不会> 3 。假设a ， b ， c和d是来自子集的四个元素，并且a 。
假设abs(a – b)= 2 k，而abs(b – c)= 2 l，则abs(a – c)= abs(a – b)+ abs(b – c)= 2 k + 2 l = 2 m 。这意味着k = l 。三元组(b，c，d)的条件也必须成立。现在可以很容易地看出，如果abs(a – b)= abs(b – c)= abs(c – d)= 2 k，则abs(a – d)= abs(a – b)* 3不是二的幂。因此，子集的大小永远不会大于3。
- 让我们检查答案是否为3 。在给定数组上迭代子集的中间元素以及2的幂(从1到30)(包括1和30)。令xi为子集的中间元素，令j为两者的当前幂。然后，如果数组中存在元素x i -2 j和x i +2 j ，则答案为3 。
- 否则检查答案是否为2 。重复上一步，但是在这里可以得到左点x i -2 j或x i +2 j 。
- 如果答案既不是2也不是3，则打印-1 。
下面是上述方法的实现：
C++
// CPP program to find sub-set with
// maximum possible size
#include
using namespace std;
// Function to find sub-set with
// maximum possible size
void PowerOfTwo(vector x, int n)
{
// Sort the given array
sort(x.begin(), x.end());
// To store required sub-set
vector 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 a = { 3, 4, 5, 6, 7 };
int n = a.size();
PowerOfTwo(a, n);
return 0;
} Java
// 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 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 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 Python3
# 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 RyugaC#
// 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_jnuPHP
输出：
3 5 7
时间复杂度：O(N * logN)
辅助空间：O(1)
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