给定长度为N的数组arr [] ,任务是从给定数组中找到所有可能的子序列之和的按位或。
例子:
Input: arr[] = {4, 2, 5}
Output: 15
Explanation: All subsequences from the given array and their corresponding sums:
{4} – 4
{2} – 2
{5} – 5
{4, 2} – 6
{4, 5} – 9
{2, 5} – 7
{4, 2, 5} -11
Therefore, the Bitwise OR of all sums = 4 | 2 | 5 | 6 | 9 | 7 | 11 = 15.
Input: arr[] = {1, 9, 8}
Output: 27
Explanation: All subsequences from the given array and their corresponding sums:
{1} – 1
{9} – 9
{8} – 8
{1, 9} – 10
{9, 8} – 17
{1, 8} – 9
{1, 9, 8} – 18
Therefore, Bitwise OR of all sums = 1 | 9 | 8 | 10 | 17 | 9 | 18 = 27.
天真的方法:最简单的方法是从给定数组生成所有可能的子序列,并找到它们各自的和。现在,在计算了它们的总和之后,打印所有获得的总和的按位或。
时间复杂度: O(2 N )
辅助空间: O(1)
高效的方法:为了优化上述方法,该想法是基于以下观察结果:
- 数组元素中的所有设置位也都设置在最终结果中。
- 给定数组的前缀和数组中设置的所有位也都设置在最终结果中。
请按照以下步骤解决以上问题:
- 用0初始化一个变量结果,该结果存储给定数组arr []的每个子序列之和的按位或。
- 用0初始化一个变量prefixSum ,该变量在任何时刻存储arr []的前缀和。
- 使用变量i遍历[0,N]范围内的数组元素。
- 更新prefixSum作为prefixSum + = ARR [i]中。
- 结果更新结果| = arr [i] | prefixSum。
- 完成上述步骤后,将结果值打印为答案。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include
using namespace std;
// Function to calculate Bitwise OR of
// sums of all subsequences
int findOR(int nums[], int N)
{
// Stores the prefix sum of nums[]
int prefix_sum = 0;
// Stores the bitwise OR of
// sum of each subsequence
int result = 0;
// Iterate through array nums[]
for (int i = 0; i < N; i++) {
// Bits set in nums[i] are
// also set in result
result |= nums[i];
// Calculate prefix_sum
prefix_sum += nums[i];
// Bits set in prefix_sum
// are also set in result
result |= prefix_sum;
}
// Return the result
return result;
}
// Driver Code
int main()
{
// Given array arr[]
int arr[] = { 4, 2, 5 };
int N = sizeof(arr) / sizeof(arr[0]);
// Function Call
cout << findOR(arr, N);
return 0;
}
Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to calculate Bitwise OR of
// sums of all subsequences
static int findOR(int nums[], int N)
{
// Stores the prefix sum of nums[]
int prefix_sum = 0;
// Stores the bitwise OR of
// sum of each subsequence
int result = 0;
// Iterate through array nums[]
for (int i = 0; i < N; i++) {
// Bits set in nums[i] are
// also set in result
result |= nums[i];
// Calculate prefix_sum
prefix_sum += nums[i];
// Bits set in prefix_sum
// are also set in result
result |= prefix_sum;
}
// Return the result
return result;
}
// Driver Code
public static void main(String[] args)
{
// Given array arr[]
int arr[] = { 4, 2, 5 };
int N = arr.length;
System.out.print(findOR(arr, N));
}
}
Python3
# Python3 program for the
# above approach
# Function to calculate
# Bitwise OR of sums of
# all subsequences
def findOR(nums, N):
# Stores the prefix
# sum of nums[]
prefix_sum = 0
# Stores the bitwise OR of
# sum of each subsequence
result = 0
# Iterate through array nums[]
for i in range(N):
# Bits set in nums[i] are
# also set in result
result |= nums[i]
# Calculate prefix_sum
prefix_sum += nums[i]
# Bits set in prefix_sum
# are also set in result
result |= prefix_sum
# Return the result
return result
# Driver Code
if __name__ == "__main__":
# Given array arr[]
arr = [4, 2, 5]
N = len(arr)
# Function Call
print(findOR(arr, N))
# This code is contributed by Chitranayal
C#
// C# program for the above approach
using System;
class GFG{
// Function to calculate Bitwise OR of
// sums of all subsequences
static int findOR(int[] nums, int N)
{
// Stores the prefix sum of nums[]
int prefix_sum = 0;
// Stores the bitwise OR of
// sum of each subsequence
int result = 0;
// Iterate through array nums[]
for(int i = 0; i < N; i++)
{
// Bits set in nums[i] are
// also set in result
result |= nums[i];
// Calculate prefix_sum
prefix_sum += nums[i];
// Bits set in prefix_sum
// are also set in result
result |= prefix_sum;
}
// Return the result
return result;
}
// Driver Code
public static void Main()
{
// Given array arr[]
int[] arr = { 4, 2, 5 };
// Size of array
int N = arr.Length;
// Function call
Console.Write(findOR(arr, N));
}
}
// This code is contributed by code_hunt
Javascript
15
时间复杂度: O(N)
辅助空间: O(1)