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📜  长度为 K 的子序列的最大按位 OR 值

📅  最后修改于: 2021-10-25 11:28:01             🧑  作者: Mango

给定一个由N 个正整数组成的数组arr[]和一个数字K ,任务是找到大小为 K 的子序列的按位 OR 的最大值。

例子:

朴素的方法:朴素的方法是生成所有长度为 K 的子序列,并找到所有子序列的按位或值。所有这些中的最大值将是答案。

时间复杂度: O(N 2 )
辅助空间: O(K)

高效方法:为了优化上述方法,尝试实现贪婪方法。以下是步骤:

  1. 初始化一个大小为 32 的整数数组bit[] ,所有值都等于 0。
  2. 现在迭代 bit[] 数组的每个索引从 31 到 0,并检查位数组的i值是否为 0,然后在给定的数组中迭代并找到一个元素,该元素在获取后对我们的位数组贡献最大为 1。
  3. 取该元素并相应地更改位数组,如果 k > 0,则每次将 k 减 1。否则跳出循环。
  4. 现在将 bit[] 数组转换为十进制数以获得最终答案。

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


输出:
31

时间复杂度: O(N*log N)
辅助空间: O(1)
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