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📜  XOR值为奇数的最长子序列的长度

📅  最后修改于: 2021-05-25 05:34:19             🧑  作者: Mango

给定N个正整数的数组arr [] ,任务是找到最长子序列的长度,以使子序列中所有整数的按位XOR奇数。

例子:

天真的方法:天真的想法是生成给定数组的所有可能的子序列,并检查任何子序列的按位XOR是否为奇数。如果存在按位异或为奇数的子序列,则在这些子序列中打印该子序列的最大长度。

时间复杂度: O(N * 2 N )
辅助空间: O(1)

高效的方法:为了优化上述朴素的方法,其思想是计算给定数组中奇数和偶数元素的数目,并找到最长子序列的长度,如下所示:

  1. 计算arr []中偶数和奇数元素的数量。
  2. 如果奇数值的数量等于数组大小,即N ,那么我们有两种可能的情况:
    1. 如果数组的大小为奇数,则最大长度将等于N
    2. 否则,最大长度将等于N – 1
  3. 如果偶数的计数等于数组大小,则最大长度将为零。
  4. 现在,如果给定数组中同时存在两种类型的元素,则最大长度将包括所有偶数元素,对于奇数元素,如果奇数值的数量为奇数,则我们将包括所有所有元素,否则,将包括奇数– 1个元素。
  5. 完成上述步骤后,打印最长子序列的最大长度。

下面是上述方法的实现:

C++
// C++ program for the above approach
#include 
using namespace std;
 
// Function for find max XOR subsequence
// having odd value
int maxXORSubsequence(int arr[], int n)
{
 
    // Initialize odd and even count
    int odd = 0, even = 0;
 
    // Count the number of odd and even
    // numbers in given array
    for (int i = 0; i < n; i++) {
        if (arr[i] & 1)
            odd++;
        else
            even++;
    }
 
    int maxlen;
 
    if (odd == n) {
 
        // if all values are odd
        // in given array
        if (odd % 2 == 0)
            maxlen = n - 1;
        else
            maxlen = n;
    }
    else if (even == n) {
 
        // if all values are even
        // in given array
        maxlen = 0;
    }
    else {
 
        // if both odd and even are
        // present in given array
        if (odd % 2 == 0)
            maxlen = even + odd - 1;
        else
            maxlen = even + odd;
    }
}
 
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 2, 3, 4, 5, 6, 7 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    cout << maxXORSubsequence(arr, n);
}


Java
// Java program for the above approach
class GFG{
     
// Function for find max XOR subsequence
// having odd value
static int maxXORSubsequence(int arr[], int n)
{
     
    // Initialize odd and even count
    int i, odd = 0, even = 0;
 
    // Count the number of odd and even
    // numbers in given array
    for(i = 0; i < n; i++)
    {
        if ((arr[i] & 1) != 0)
            odd++;
        else
            even++;
    }
 
    int maxlen;
 
    if (odd == n)
    {
         
        // If all values are odd
        // in given array
        if (odd % 2 == 0)
            maxlen = n - 1;
        else
            maxlen = n;
    }
    else if (even == n)
    {
 
        // If all values are even
        // in given array
        maxlen = 0;
    }
    else
    {
 
        // If both odd and even are
        // present in given array
        if (odd % 2 == 0)
            maxlen = even + odd - 1;
        else
            maxlen = even + odd;
    }
    return maxlen;
}
 
// Driver Code
public static void main (String []args)
{
     
    // Given array arr[]
    int arr[] = { 2, 3, 4, 5, 6, 7 };
    int n = arr.length;
 
    // Function Call
    System.out.print( maxXORSubsequence(arr, n));
}
}
 
// This code is contributed by chitranayal


Python3
# Python3 program for the above approach
 
# Function for find max XOR subsequence
# having odd value
def maxXorSubsequence(arr, n):
     
    # Initialize odd and even count
    odd = 0
    even = 0
     
    # Count the number of odd and even
    # numbers in given array
    for i in range(0, n):
        if arr[i] % 2 != 0:
            odd += 1
        else:
            even += 1
    if odd == n:
         
        # If all values are odd
        # in given array
        if odd % 2 == 0:
            maxlen = n - 1
        else:
            maxlen = n
             
    elif even == n:
         
        # If all values are even
        # in given array
        maxlen = 0
    else:
         
        # If both odd and even are
        # present in given array
        if odd % 2 == 0:
            maxlen = even + odd - 1
        else:
            maxlen = even + odd
             
    return maxlen
 
# Driver code
if __name__ == '__main__':
     
    # Given array arr[]
    arr = [ 2, 3, 4, 5, 6, 7 ]
    n = len(arr)
     
    # Function Call
    print(maxXorSubsequence(arr,n))
     
# This code is contributed by virusbuddah_


C#
// C# program for the above approach
using System;
class GFG{
      
// Function for find max XOR subsequence
// having odd value
static int maxXORSubsequence(int[] arr, int n)
{
      
    // Initialize odd and even count
    int i, odd = 0, even = 0;
  
    // Count the number of odd and even
    // numbers in given array
    for(i = 0; i < n; i++)
    {
        if ((arr[i] & 1) != 0)
            odd++;
        else
            even++;
    }
  
    int maxlen;
  
    if (odd == n)
    {
          
        // If all values are odd
        // in given array
        if (odd % 2 == 0)
            maxlen = n - 1;
        else
            maxlen = n;
    }
    else if (even == n)
    {
  
        // If all values are even
        // in given array
        maxlen = 0;
    }
    else
    {
  
        // If both odd and even are
        // present in given array
        if (odd % 2 == 0)
            maxlen = even + odd - 1;
        else
            maxlen = even + odd;
    }
    return maxlen;
}
  
// Driver Code
public static void Main (string []args)
{
      
    // Given array arr[]
    int []arr = { 2, 3, 4, 5, 6, 7 };
    int n = arr.Length;
  
    // Function Call
    Console.Write( maxXORSubsequence(arr, n));
}
}
  
// This code is contributed by rock_cool


Javascript


输出:
6

时间复杂度: O(N)
辅助空间: O(1)