给定一个由2 * N个整数组成的数组arr [] ，任务是检查是否有可能重新排列数组元素，使得每第i个arr [2 * i + 1] = 2 * arr [2 * i]指数。如果可能的话，请打印“是。否则，打印“否” 。
例子：
Input: arr[] = {4, -2, 2, -4}, N = 2
Output: Yes
Explanation: Rearrange the array as arr[] = {-2, -4, 2, 4}.
Input: arr[] = {3, 1, 3, 6}, N = 2
Output: No
方法：解决给定问题的想法是使用Map，并观察到一个人需要N个不同的对，以使一个元素是另一元素的两倍。请按照以下步骤解决问题：
- 初始化映射
- 遍历数组arr []并更新Map count中每个元素的计数。
- 遍历映射计数并执行以下操作：
- 如果X小于0 ，则初始化一个变量例如want与当前元素X组成一对，并分配want = X / 2 。否则，分配want = 2 * X。
- 检查X是否小于0且X为奇数，或者X的计数大于want的计数，然后打印“否” ，因为不可能与数组的任何其他元素形成一对剩余的X。
- 否则，想更新地图中的计数数作为计数(想) -数(X)。
- 完成上述步骤后，打印“是” ，因为存在满足给定属性的元素的任何组合。
下面是上述方法的实现：
C++
// cpp program of the above approach
#include
using namespace std;
// Function to check if it is possible
// to rearrange the array elements
// that satisfies the given conditions
string canReorderDoubled(vector A)
{
// Stores the count of elements
map count;
// Update the hash table
for (int a : A)
count[a]++;
// Traverse the hash table
for (auto x : count) {
// If the count of current
// element is zero
if (x.second == 0)
continue;
// Stores the element needed
// to form a pair with the
// current element
int xx = x.first;
int want = xx < 0 ? xx / 2 : xx * 2;
// If x is less than zero and odd
if (xx < 0 && xx % 2 != 0)
return "No";
// If count of x is greater
// than count of want
if (x.second
> count[want])
return "No";
// Update the count of want
// in the hash table
count[want] -= x.second;
}
// Return true if none of the
// above cases satisfies
return "Yes";
}
// Driver Code
int main()
{
vector arr = { 4, -2, 2, -4 };
int N = 2;
string res = canReorderDoubled(arr);
// Print the result obtained
cout<<(res);
}
// This code is contributed by mohit kumar 29. Java
// Java program of the above approach
import java.io.*;
import java.util.*;
class GFG {
// Function to check if it is possible
// to rearrange the array elements
// that satisfies the given conditions
public static String canReorderDoubled(int[] A)
{
// Stores the count of elements
Map count
= new TreeMap<>();
// Update the hash table
for (int a : A)
count.put(
a, count.getOrDefault(a, 0) + 1);
// Traverse the hash table
for (int x : count.keySet()) {
// If the count of current
// element is zero
if (count.get(x) == 0)
continue;
// Stores the element needed
// to form a pair with the
// current element
int want = x < 0 ? x / 2 : x * 2;
// If x is less than zero and odd
if (x < 0 && x % 2 != 0)
return "No";
// If count of x is greater
// than count of want
if (count.get(x)
> count.getOrDefault(want, 0))
return "No";
// Update the count of want
// in the hash table
count.put(want,
count.get(want)
- count.get(x));
}
// Return true if none of the
// above cases satisfies
return "Yes";
}
// Driver Code
public static void main(String[] args)
{
int[] arr = { 4, -2, 2, -4 };
int N = 2;
String res = canReorderDoubled(arr);
// Print the result obtained
System.out.println(res);
}
} Python3
# Python 3 program of the above approach
# Function to check if it is possible
# to rearrange the array elements
# that satisfies the given conditions
def canReorderDoubled(A):
# Stores the count of elements
count = {}
# Update the hash table
for a in A:
if a in count:
count[a] += 1
else:
count[a] = 1
# Traverse the hash table
for key,value in count.items():
# If the count of current
# element is zero
if (value == 0):
continue
# Stores the element needed
# to form a pair with the
# current element
xx = key
if xx < 0:
want = xx / 2
else:
want = xx * 2
# If x is less than zero and odd
if (xx < 0 and xx % 2 != 0):
return "No"
# If count of x is greater
# than count of want
if (want in count and value > count[want]):
return "No"
# Update the count of want
# in the hash table
if want in count:
count[want] -= value
# Return true if none of the
# above cases satisfies
return "Yes"
# Driver Code
if __name__ == '__main__':
arr = [4, -2, 2, -4]
N = 2
res = canReorderDoubled(arr)
# Print the result obtained
print(res)
# This code is contributed by bgangwar59.输出：
Yes时间复杂度： O(N * log N)
辅助空间： O(N)