给定一个偶数大小的整数数组。我们需要找到是否有可能将数组元素分为两组，从而满足以下条件。
- 两个子集的大小相同。
- 机器人集中的元素总数是相同的。
- 每个元素都是两个集合之一的一部分。
例子 ：
Input: arr[] = {1, 3, 2, 1, 2, 1}
Output : Yes
Explanation: We can get two subsets
as {1, 3, 1} and {2, 2, 1}
Input: {1, 2, 3, 4, 5, 6}
Output : No
这个想法是基于以下帖子的方法1。
在大小为n的给定数组中打印r元素的所有可能组合
我们生成大小为n / 2的所有子集。对于每个子集，我们检查其总和是否为2的总和。
C++
// Program to check if we can divide an array into two
// halves such that sum of the two is same.
#include
using namespace std;
bool combinationUtil(int arr[], int half[], int start, int end,
int index, int n, int sum);
// Returns true if it is possible to divide array into two halves.
// of same sum. This function mainly uses combinationUtil()
bool isPossible(int arr[], int n)
{
// If size of array is not even.
if (n % 2 != 0)
return false;
// If sum of array is not even.
int sum = accumulate(arr, arr + n, 0);
if (sum % 2 != 0)
return false;
// A temporary array to store all combination one by one
int half[n / 2];
// Print all combination using temporary array 'half[]'
return combinationUtil(arr, half, 0, n - 1, 0, n, sum);
}
/* arr[] ---> Input Array
half[] ---> Temporary array to store current combination
of size n/2
start & end ---> Staring and Ending indexes in arr[]
index ---> Current index in half[] */
bool combinationUtil(int arr[], int half[], int start, int end,
int index, int n, int sum)
{
// Current combination is ready to be printed, print it
if (index == n / 2) {
int curr_sum = accumulate(half, half + n / 2, 0);
return (curr_sum + curr_sum == sum);
}
// replace index with all possible elements. The condition
// "end-i+1 >= n/2-index" makes sure that including one element
// at index will make a combination with remaining elements
// at remaining positions
for (int i = start; i <= end && end - i + 1 >= n/2 - index; i++) {
half[index] = arr[i];
if (combinationUtil(arr, half, i + 1, end, index + 1, n, sum))
return true;
}
return false;
}
// Driver program to test above functions
int main()
{
int arr[] = { 1, 2, 4, 4, 5, 6 };
int n = sizeof(arr) / sizeof(arr[0]);
if (isPossible(arr, n))
cout << "Yes";
else
cout << "No";
return 0;
} Java
// Java program to check if we can
// divide an array into two halves
// such that sum of the two is same.
public class DivideArray{
static int accumulate(int arr[], int first,
int last)
{
int init = 0;
for (int i = first; i< last; i++) {
init = init + arr[i];
}
return init;
}
// Returns true if it is possible to divide
// array into two halves of same sum.
// This function mainly uses combinationUtil()
static Boolean isPossible(int arr[], int n)
{
// If size of array is not even.
if (n % 2 != 0)
return false;
// If sum of array is not even.
int sum = accumulate(arr, 0, n);
if (sum % 2 != 0)
return false;
// A temporary array to store all
// combination one by one int k=n/2;
int half[] = new int[n/2];
// Print all combination using temporary
// array 'half[]'
return combinationUtil(arr, half, 0, n - 1,
0, n, sum);
}
/* arr[] ---> Input Array
half[] ---> Temporary array to store current
combination of size n/2
start & end ---> Staring and Ending indexes in arr[]
index ---> Current index in half[] */
static Boolean combinationUtil(int arr[], int half[],
int start, int end,
int index, int n,
int sum)
{
// Current combination is ready to
// be printed, print it
if (index == n / 2) {
int curr_sum = accumulate(half, 0 , n/2);
return (curr_sum + curr_sum == sum);
}
// replace index with all possible elements.
// The condition "end-i+1 >= n/2-index" makes
// sure that including one element at index
// will make a combination with remaining
// elements at remaining positions
for (int i = start; i <= end && end - i + 1 >=
n/2 - index; i++) {
half[index] = arr[i];
if (combinationUtil(arr, half, i + 1, end,
index + 1, n, sum))
return true;
}
return false;
}
// Driver code
public static void main(String[] s)
{
int arr[] = {1, 2, 4, 4, 5, 6 };
if (isPossible(arr, arr.length))
System.out.println("Yes");
else
System.out.println("NO");
}
}
// This code is contributed by Prerna SainiPython3
# Python3 program to check if
# we can divide an array into two
# halves such that sum of the two is same.
# Returns true if it is possible to divide
# array into two halves of same sum.
# This function mainly uses combinationUtil()
def isPossible(arr, n):
# If size of array is not even
if n % 2 != 0:
return False
# If sum of array is not even.
s = sum(arr)
if s % 2 != 0:
return False
# A temporary array to store
# all combination one by one
half = [0] * (n // 2)
# Print all combination using
# temporary array 'half[]'
return combinationUtil(arr, half, 0,
n - 1, 0, n, s)
'''
arr[] ---> Input Array
half[] ---> Temporary array to store current
combination of size n/2
start & end ---> Staring and Ending indexes in arr[]
index ---> Current index in half[]
'''
def combinationUtil(arr, half, start, end, index, n, s):
# Current combination is
# ready to be printed, print it
if index == n // 2:
curr_sum = sum(half)
if (curr_sum + curr_sum == s):
return True
else:
return False
# replace index with all possible elements.
# The condition "end-i+1 >= n/2-index"
# makes sure that including one element
# at index will make a combination with
# remaining elements at remaining positions
i = start
while i <= end and (end - i + 1) >= (n // 2 - index):
half[index] = arr[i]
if combinationUtil(arr, half, i + 1, end,
index + 1, n, s):
return True
i += 1
return False
# Driver Code
if __name__ == "__main__":
arr = [1, 2, 4, 4, 5, 6]
n = len(arr)
if isPossible(arr, n):
print("Yes")
else:
print("No")
# This code is contributed by
# sanjeev2552C#
// C# program to check if we can
// divide an array into two halves
// such that sum of the two is same.
using System;
public class DivideArray{
static int accumulate(int []arr, int first,
int last)
{
int init = 0;
for (int i = first; i< last; i++) {
init = init + arr[i];
}
return init;
}
// Returns true if it is possible to divide
// array into two halves of same sum.
// This function mainly uses combinationUtil()
static Boolean isPossible(int []arr, int n)
{
// If size of array is not even.
if (n % 2 != 0)
return false;
// If sum of array is not even.
int sum = accumulate(arr, 0, n);
if (sum % 2 != 0)
return false;
// A temporary array to store all
// combination one by one int k=n/2;
int []half = new int[n/2];
// Print all combination using temporary
// array 'half[]'
return combinationUtil(arr, half, 0, n - 1,
0, n, sum);
}
/* arr[] ---> Input Array
half[] ---> Temporary array to store current
combination of size n/2
start & end ---> Staring and Ending indexes in arr[]
index ---> Current index in half[] */
static Boolean combinationUtil(int []arr, int []half,
int start, int end,
int index, int n,
int sum)
{
// Current combination is ready to
// be printed, print it
if (index == n / 2) {
int curr_sum = accumulate(half, 0 , n/2);
return (curr_sum + curr_sum == sum);
}
// replace index with all possible elements.
// The condition "end-i+1 >= n/2-index" makes
// sure that including one element at index
// will make a combination with remaining
// elements at remaining positions
for (int i = start; i <= end && end - i + 1 >=
n/2 - index; i++) {
half[index] = arr[i];
if (combinationUtil(arr, half, i + 1, end,
index + 1, n, sum))
return true;
}
return false;
}
// Driver code
public static void Main()
{
int []arr = {1, 2, 4, 4, 5, 6 };
if (isPossible(arr, arr.Length))
Console.WriteLine("Yes");
else
Console.WriteLine("NO");
}
}
// This code is contributed by PrinciRaj1992输出：
Yes