给定一个大小为N的正整数数组arr[] ,任务是检查arr[] 中是否存在两个不相交的子数组,使得所有可能的2 (subarr[i])的总和和所有可能的2的总和(subarr2[j])是相等的。
例子:
Input: arr[] = {4, 3, 0, 1, 2, 0}
Output: YES
Explanation: Expressing every array element in the form of 2arr[i], the array is modified to { 16, 8, 1, 2, 4, 1 }.
Therefore, two valid subarrays are { 16 } and { 8, 1, 2, 4, 1 } whose sum are equal.
Input: arr[]={ 3, 4 }
Output: NO
方法:由于 2 的所有幂的二进制表示是唯一的,因此只有在该数组中存在任何重复元素时才能获得两个 sch 子数组。否则,这是不可能的。
请按照以下步骤解决问题:
- 按升序对给定数组进行排序。
- 遍历数组并检查任何一对相邻元素是否相等。
- 如果找到任何这样的对,打印“YES” 。否则,打印“NO” 。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
// Function to check if two non-intersecting
// subarrays with equal sum exists or not
void findSubarrays(int arr[], int N)
{
// Sort the given array
sort(arr, arr + N);
int i = 0;
// Traverse the array
for (i = 0; i < N - 1; i++) {
// Check for duplicate elements
if (arr[i] == arr[i + 1]) {
cout << "YES" << endl;
return;
}
}
// If no duplicate element is
// present in the array
cout << "NO" << endl;
}
// Driver Code
int main()
{
// Given array
int arr[] = { 4, 3, 0, 1, 2, 0 };
// Size of array
int N = sizeof(arr) / sizeof(arr[0]);
findSubarrays(arr, N);
return 0;
}
Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to check if two non-intersecting
// subarrays with equal sum exists or not
static void findSubarrays(int arr[], int N)
{
// Sort the given array
Arrays.sort(arr);
int i = 0;
// Traverse the array
for(i = 0; i < N - 1; i++)
{
// Check for duplicate elements
if (arr[i] == arr[i + 1])
{
System.out.println("YES");
return;
}
}
// If no duplicate element is
// present in the array
System.out.println("NO");
}
// Driver code
public static void main(String[] args)
{
// Given array
int[] arr = { 4, 3, 0, 1, 2, 0 };
// Size of array
int N = arr.length;
findSubarrays(arr, N);
}
}
// This code is contributed by susmitakundugoaldanga
Python3
# Python program for the above approach
# Function to check if two non-intersecting
# subarrays with equal sum exists or not
def findSubarrays(arr, N):
# Sort the given array
arr.sort();
i = 0;
# Traverse the array
for i in range(N - 1):
# Check for duplicate elements
if (arr[i] == arr[i + 1]):
print("YES");
return;
# If no duplicate element is
# present in the array
print("NO");
# Driver code
if __name__ == '__main__':
# Given array
arr = [4, 3, 0, 1, 2, 0];
# Size of array
N = len(arr);
findSubarrays(arr, N);
# This code is contributed by 29AjayKumar
C#
// C# program for the above approach
using System;
class GFG{
// Function to check if two non-intersecting
// subarrays with equal sum exists or not
static void findSubarrays(int[] arr, int N)
{
// Sort the given array
Array.Sort(arr);
int i = 0;
// Traverse the array
for(i = 0; i < N - 1; i++)
{
// Check for duplicate elements
if (arr[i] == arr[i + 1])
{
Console.WriteLine("YES");
return;
}
}
// If no duplicate element is
// present in the array
Console.WriteLine("NO");
}
// Driver code
public static void Main()
{
// Given array
int[] arr = { 4, 3, 0, 1, 2, 0 };
// Size of array
int N = arr.Length;
findSubarrays(arr, N);
}
}
// This code is contributed by sanjoy_62
Javascript
输出:
YES
时间复杂度: O(NLogN)
辅助空间: O(1)
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