长度为K的子序列的最大按位OR值

给定N个正整数和数字K的数组arr [] ，任务是找到大小K的子序列的按位OR的最大值。

例子：

Input: arr[] = {2, 5, 3, 6, 11, 13}, k = 3 
Output: 15 
Explanation: 
The sub-sequence wil maximum OR value is 2, 11, 13.

Input: arr[] = {5, 9, 7, 19}, k = 3 
Output: 31 
Explanation: 
The maximum value of bitwise OR of the subsequence of size K = 3 is 31. 
 

天真的方法：天真的方法是生成长度为K的所有子序列，并找到所有子序列的按位或值。所有这些中的最大数目将是答案。

时间复杂度： O(N ² )
辅助空间： O(K)

高效方法：要优化上述方法，请尝试实施贪婪方法。步骤如下：

1. 初始化大小为32的整数数组bit [] ，所有值均等于0。
2. 现在，对bit []数组的每个索引从31迭代到0，并检查bit数组的^第i^个值是否为0，然后在给定的数组中进行迭代，并找到一个元素，该元素对我们的bit数组贡献最大。
3. 取该元素并相应地更改位数组，如果k> 0，则每次也将k减1。否则将退出循环。
4. 现在将bit []数组转换为十进制数字以获得最终答案。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to convert bit array to
// decimal number
int build_num(int bit[])
{
    int ans = 0;
 
    for (int i = 0; i < 32; i++)
        if (bit[i])
            ans += (1 << i);
 
    // Return the final result
    return ans;
}
 
// Function to find the maximum Bitwise
// OR value of subsequence of length K
int maximumOR(int arr[], int n, int k)
{
    // Initialize bit array of
    // size 32 with all value as 0
    int bit[32] = { 0 };
 
    // Iterate for each index of bit[]
    // array from 31 to 0, and check if
    // the ith value of bit array is 0
    for (int i = 31; i >= 0; i--) {
 
        if (bit[i] == 0 && k > 0) {
            int temp = build_num(bit);
            int temp1 = temp;
            int val = -1;
 
            for (int j = 0; j < n; j++) {
 
                // Check for maximum
                // contributing element
                if (temp1 < (temp | arr[j])) {
                    temp1 = temp | arr[j];
                    val = arr[j];
                }
            }
 
            // Update the bit array
            // if max_contributing
            // element is found
            if (val != -1) {
 
                // Decrement the value of K
                k--;
                for (int j = 0; j < 32; j++) {
                    if (val & (1 << j))
                        bit[j]++;
                }
            }
        }
    }
 
    // Return the result
    return build_num(bit);
}
 
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 5, 9, 7, 19 };
 
    // Length of subsequence
    int k = 3;
    int n = sizeof arr / sizeof arr[0];
 
    // Function Call
    cout << maximumOR(arr, n, k);
    return 0;
}

Java

// Java program for the above approach
class GFG{
 
// Function to convert bit array to
// decimal number
static int build_num(int []bit)
{
    int ans = 0;
 
    for(int i = 0; i < 32; i++)
       if (bit[i] == 1)
           ans += (1 << i);
           ans += 32;
 
    // Return the final result
    return ans;
}
 
// Function to find the maximum Bitwise
// OR value of subsequence of length K
static int maximumOR(int []arr, int n, int k)
{
     
    // Initialize bit array of
    // size 32 with all value as 0
    int bit[] = new int[32];
 
    // Iterate for each index of bit[]
    // array from 31 to 0, and check if
    // the ith value of bit array is 0
    for(int i = 31; i >= 0; i--)
    {
       if (bit[i] == 0 && k > 0)
       {
           int temp = build_num(bit);
           int temp1 = temp;
           int val = -1;
            
           for(int j = 0; j < n; j++)
           {
                
              // Check for maximum
              // contributing element
              if (temp1 < (temp | arr[j]))
              {
                  temp1 = temp | arr[j];
                  val = arr[j];
              }
           }
            
           // Update the bit array
           // if max_contributing
           // element is found
           if (val != -1)
           {
                
               // Decrement the value of K
               k--;
               for(int j = 0; j < 32; j++)
               {
                  bit[j]++;
               }
           }
       }
    }
     
    // Return the result
    return build_num(bit);
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given array arr[]
    int arr[] = { 5, 9, 7, 19 };
 
    // Length of subsequence
    int k = 3;
    int n = arr.length;
 
    // Function call
    System.out.println(maximumOR(arr, n, k));
}
}
 
// This code is contributed by rock_cool

Python3

# Python3 program to implement
# above approach
 
# Function to convert bit array to
# decimal number
def build_num(bit):
 
    ans = 0
    for i in range(0, 32):
        if (bit[i]):
            ans += (1 << i)
 
    # Return the final result
    return ans;
 
# Function to find the maximum Bitwise
# OR value of subsequence of length K
def maximumOR(arr, n, k):
     
    # Initialize bit array of
    # size 32 with all value as 0
    bit = [0] * 32
 
    # Iterate for each index of bit[]
    # array from 31 to 0, and check if
    # the ith value of bit array is 0
    for i in range(31, -1, -1):
        if (bit[i] == 0 and k > 0):
            temp = build_num(bit)
            temp1 = temp
            val = -1
             
            for j in range(0, n):
                 
                # Check for maximum
                # contributing element
                if (temp1 < (temp | arr[j])):
                    temp1 = temp | arr[j]
                    val = arr[j]
 
            # Update the bit array
            # if max_contributing
            # element is found
            if (val != -1):
 
                # Decrement the value of K
                k -= 1
                for j in range(0, 32):
                    if (val & (1 << j)):
                        bit[j] += 1
 
    # Return the result
    return build_num(bit)
 
# Driver Code
 
# Given array arr[]
arr = [ 5, 9, 7, 19 ]
 
# Length of subsequence
k = 3;
n = len(arr)
 
# Function call
print(maximumOR(arr, n, k))
 
# This code is contributed by sanjoy_62

C#

// C# program for the above approach
using System;
class GFG{
 
// Function to convert bit array to
// decimal number
static int build_num(int []bit)
{
    int ans = 0;
 
    for(int i = 0; i < 32; i++)
       if (bit[i] == 1)
           ans += (1 << i);
           ans += 32;
 
    // Return the final result
    return ans;
}
 
// Function to find the maximum Bitwise
// OR value of subsequence of length K
static int maximumOR(int []arr, int n, int k)
{
     
    // Initialize bit array of
    // size 32 with all value as 0
    int []bit = new int[32];
 
    // Iterate for each index of bit[]
    // array from 31 to 0, and check if
    // the ith value of bit array is 0
    for(int i = 31; i >= 0; i--)
    {
       if (bit[i] == 0 && k > 0)
       {
           int temp = build_num(bit);
           int temp1 = temp;
           int val = -1;
            
           for(int j = 0; j < n; j++)
           {
                
              // Check for maximum
              // contributing element
              if (temp1 < (temp | arr[j]))
              {
                  temp1 = temp | arr[j];
                  val = arr[j];
              }
           }
            
           // Update the bit array
           // if max_contributing
           // element is found
           if (val != -1)
           {
                
               // Decrement the value of K
               k--;
               for(int j = 0; j < 32; j++)
               {
                  bit[j]++;
               }
           }
       }
    }
     
    // Return the result
    return build_num(bit);
}
 
// Driver Code
public static void Main()
{
     
    // Given array arr[]
    int []arr = { 5, 9, 7, 19 };
 
    // Length of subsequence
    int k = 3;
    int n = arr.Length;
 
    // Function call
    Console.Write(maximumOR(arr, n, k));
}
}
 
// This code is contributed by Code_Mech

Javascript

 
// Javascript program for the above approach
 
// Function to convert bit array to
// decimal number
function build_num(bit)
{
    let ans = 0;
 
    for(let i = 0; i < 32; i++)
        if (bit[i] > 0)
            ans += (1 << i);
 
    // Return the final result
    return ans;
}
 
// Function to find the maximum Bitwise
// OR value of subsequence of length K
function maximumOR(arr, n, k)
{
     
    // Initialize bit array of
    // size 32 with all value as 0
    let bit = new Array(32);
    bit.fill(0);
 
    // Iterate for each index of bit[]
    // array from 31 to 0, and check if
    // the ith value of bit array is 0
    for(let i = 31; i >= 0; i--)
    {
        if (bit[i] == 0 && k > 0)
        {
            let temp = build_num(bit);
            let temp1 = temp;
            let val = -1;
 
            for(let j = 0; j < n; j++)
            {
                 
                // Check for maximum
                // contributing element
                if (temp1 < (temp | arr[j]))
                {
                    temp1 = temp | arr[j];
                    val = arr[j];
                }
            }
 
            // Update the bit array
            // if max_contributing
            // element is found
            if (val != -1)
            {
                 
                // Decrement the value of K
                k--;
                for(let j = 0; j < 32; j++)
                {
                    if ((val & (1 << j)) > 0)
                        bit[j]++;
                }
            }
        }
    }
 
    // Return the result
    return build_num(bit);
}
     
// Driver code
 
// Given array arr[]
let arr = [ 5, 9, 7, 19 ];
 
// Length of subsequence
let k = 3;
let n = arr.length;
 
// Function Call
document.write(maximumOR(arr, n, k));
     
// This code is contributed by divyeshrabadiya07   
     

输出：

31

时间复杂度： O(N * log N)
辅助空间： O(1)
相似的文章：长度为K的子序列的最大按位与值

如果您希望与行业专家一起参加现场课程，请参阅《 Geeks现场课程》和《 Geeks现场课程美国》。