具有二进制等价的给定数组的子序列计数

给定一个由N 个整数组成的数组 arr[] ，任务是找到具有 Binary Equivalence的不同子序列的总数。

A subsequence has Binary Equivalence if the sum of the count of set and unset bits in the binary representations of all the decimal numbers across the subsequence are equal.

例子：

Input: arr[] = {2, 7, 10}
Output: 0011
Explanation:
2 → 0010→1’s = 1, 0’s = 3
7 → 0111→1’s = 3, 0’s = 1
10 → 1010→1’s = 2, 0’s = 2
The subsequence [2, 7, 10] has Binary Equivalence because the number of 0’s and 1’s across the subsequence is 6 each.
Similarly, [2, 7] also has Binary Equivalence of 4 each. 
But [7, 10] does not have Binary Equivalence. 
Likewise, [10] has Binary Equivalence of 2 each. 
The total number of unique subsequences where Binary Equivalence is possible is 3. 
Since 10 is the largest element in the given array and the number of bits required to represent 10 in binary is 4. Hence, the number of bits present in the output needs to be 4.

Input: arr[] = {5, 7, 9, 12}
Output: 0111

方法：思路是求表示数组最大元素所需的总位数。按照以下步骤解决这个问题：

1. 找到最大元素和最大元素的二进制表示的长度。
2. 在二进制表示的其他元素前面附加0 ，使每个元素的位数等于最大位数。
3. 查找给定数组的所有子序列。
4. 找出具有 Binary Equivalence的子序列的总数。
5. 如果总数的长度小于最大数的长度，则将总数转换为二进制并附加0，使两者的长度相等。

下面是上述方法的实现：

Python3

# Python program for the above approach
import itertools
 
# Function to find the number of
# subsequences having Binary Equivalence
def numberOfSubsequence(arr):
 
    # Find the maximum array element
    Max_element = max(arr)
 
    # Convert the maximum element
    # to its binary equivalent
    Max_Binary = "{0:b}".format(int(
        Max_element))
 
    # Dictionary to store the count of
    # set and unset bits of all array elements
    Dic = {}
 
    for i in arr:
        Str = "{0:b}".format(int(i))
 
        if len(Str) <= len(Max_Binary):
            diff = len(Max_Binary)-len(Str)
 
            # Add the extra zeros before all
            # the elements which have length
            # smaller than the maximum element
            Str = ('0'*diff)+Str
 
        zeros = Str.count('0')
        ones = Str.count('1')
 
        # Fill the dictionary with number
        # of 0's and 1's
        Dic[int(i)] = [zeros, ones]
 
    all_combinations = []
 
    # Find all the combination
    for r in range(len(arr)+1):
 
        comb = itertools.combinations(arr, r)
        comlist = list(comb)
        all_combinations += comlist
    count = 0
 
    # Find all the combinations where
    # sum_of_zeros == sum_of_ones
    for i in all_combinations[1:]:
        sum0 = 0
        sum1 = 0
        for j in i:
            sum0 += Dic[j][0]
            sum1 += Dic[j][1]
 
        # Count the total combinations
        # where sum_of_zeros = sum_of_ones
        if sum0 == sum1:
            count += 1
 
    # Convert the count number to its
    # binary equivalent
    Str = "{0:b}".format(int(count))
    if len(Str) <= len(Max_Binary):
        diff = len(Max_Binary)-len(Str)
 
        # Append leading zeroes to
        # the answer if its length is
        # smaller than the maximum element
        Str = ('0'*diff) + Str
 
    # Print the result
    print(Str)
 
 
# Driver Code
 
# Give array arr[]
arr = [5, 7, 9, 12]
 
# Function Call
numberOfSubsequence(arr)

输出：

0111

时间复杂度： O(2 ^N )
辅助空间： O(N ² )

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