从 1 到 N 的倍数翻转位后 1 的计数

给定最初包含所有 0 的N大小的二进制数组A[] 。任务是在从 1 到N的倍数翻转位后找到 1 的最终计数。

例子：

Input: A[] = [0, 0, 0, 0]
Output: 2
Explanation: 
Flipping bits at multiples of 1 – [1, 1, 1, 1]
Flipping bits at multiples of 2 – [1, 0, 1, 0]
Flipping bits at multiples of 3 – [1, 0, 0, 0]
Flipping bits at multiples of 4 – [1, 0, 0, 1] 
Therefore count of 1’s after final flipping is 2.        

Input: A[] = [0, 0, 0, 0, 0, 0, 0]
Output: 2
Explanation:
Flipping bits at multiples of 1 – [1, 1, 1, 1, 1, 1, 1]
Flipping bits at multiples of 2 – [1, 0, 1, 0, 1, 0, 1]
Flipping bits at multiples of 3 – [1, 0, 0, 0, 1, 1, 1]
Flipping bits at multiples of 4 – [1, 0, 0, 1, 1, 1, 1]
Flipping bits at multiples of 5 – [1, 0, 0, 1, 0, 1, 1]
Flipping bits at multiples of 6 – [1, 0, 0, 1, 0, 0, 1]
Flipping bits at multiples of 7 – [1, 0, 0, 1, 0, 0, 0]
Therefore count of 1’s after final flipping is 2

朴素方法：解决方案的基本思想是基于贪婪方法。

For each element from 1 to N, flip all the elements at its multiples. The final count of 1 in the array is the answer.

请按照以下步骤实施该方法：

• 创建一个大小为N的数组并用 0 填充它。
• 运行一个循环i = 1 到 i = N，对于每个i，
  • 运行循环j = i – 1 到 N ，增量 = i。
    • 翻转每个j的位。
• 遍历数组并计算 1 的个数。

下面给出上述方法的实现。

C++

// C++ code to implement the approach
#include 
using namespace std;
 
// Function to calculate number of 1
// in the final array
int findOnes(int N, int arr[])
{
    int count = 0;
 
    // Loop to flip the elements
    // at multiples of i
    for (int i = 1; i <= N; i++) {
        for (int j = i - 1; j < N; j += i) {
            if (arr[j] == 0)
                arr[j] = 1;
            else
                arr[j] = 0;
        }
    }
 
    // Loop to determine 1s at final array
    for (int i = 0; i < N; i++) {
        if (arr[i] == 1)
            count++;
    }
 
    return count;
}
 
// Driver Code
int main()
{
    int N = 4;
    int arr[N];
 
    for (int i = 0; i < N; i++)
        arr[i] = 0;
 
    int count = findOnes(N, arr);
    cout << count;
 
    return 0;
}

Java

// Java code to implement the approach
import java.io.*;
 
class GFG {
 
  // Function to calculate number of 1
  // in the final array
  static int findOnes(int N, int arr[])
  {
    int count = 0;
 
    // Loop to flip the elements
    // at multiples of i
    for (int i = 1; i <= N; i++) {
      for (int j = i - 1; j < N; j += i) {
        if (arr[j] == 0)
          arr[j] = 1;
        else
          arr[j] = 0;
      }
    }
 
    // Loop to determine 1s at final array
    for (int i = 0; i < N; i++) {
      if (arr[i] == 1)
        count++;
    }
 
    return count;
  }
 
  // Driver Code
  public static void main (String[] args) {
    int N = 4;
    int[] arr = new int[N];
 
    for (int i = 0; i < N; i++)
      arr[i] = 0;
 
    int count = findOnes(N, arr);
    System.out.println(count);
  }
}
 
// This code is contributed by hrithikgarg03188.

Python3

# Python code to implement the approach
 
# Function to calculate number of 1
# in the final array
def findOnes(N, arr):
    count = 0
 
    # Loop to flip the elements
    # at multiples of i
    for i in range(1, N + 1):
        for j in range(i - 1, N, i):
            if (arr[j] == 0):
                arr[j] = 1
            else:
                arr[j] = 0
 
    # Loop to determine 1s at final array
    for i in range(N):
        if (arr[i] == 1):
            count += 1
 
    return count
 
# Driver Code
N = 4
arr = [0]*N
 
for i in range(N):
    arr[i] = 0
 
count = findOnes(N, arr)
print(count)
 
# This code is contributed by shinjanpatra

C#

// C# code to implement the approach
using System;
 
class GFG {
 
  // Function to calculate number of 1
  // in the final array
  static int findOnes(int N, int[] arr)
  {
    int count = 0;
 
    // Loop to flip the elements
    // at multiples of i
    for (int i = 1; i <= N; i++) {
      for (int j = i - 1; j < N; j += i) {
        if (arr[j] == 0)
          arr[j] = 1;
        else
          arr[j] = 0;
      }
    }
 
    // Loop to determine 1s at final array
    for (int i = 0; i < N; i++) {
      if (arr[i] == 1)
        count++;
    }
 
    return count;
  }
 
  // Driver Code
  public static void Main()
  {
    int N = 4;
    int[] arr = new int[N];
 
    for (int i = 0; i < N; i++)
      arr[i] = 0;
 
    int count = findOnes(N, arr);
    Console.WriteLine(count);
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

 
// JavaScript code to implement the approach
 
// Function to calculate number of 1
// in the final array
function findOnes(N, arr)
{
    let count = 0;
 
    // Loop to flip the elements
    // at multiples of i
    for (let i = 1; i <= N; i++) {
        for (let j = i - 1; j < N; j += i) {
            if (arr[j] == 0)
                arr[j] = 1;
            else
                arr[j] = 0;
        }
    }
 
    // Loop to determine 1s at final array
    for (let i = 0; i < N; i++) {
        if (arr[i] == 1)
            count++;
    }
 
    return count;
}
 
// Driver Code
 
let N = 4;
let arr = new Array(N);
 
for (let i = 0; i < N; i++)
    arr[i] = 0;
 
let count = findOnes(N, arr);
document.write(count);
 
// This code is contributed by shinjanpatra
 

输出

2

时间复杂度： O(N*log N)
辅助空间： O(N)

有效方法：解决此问题的有效方法基于以下数学观察：

After flipping all the bits at multiples of numbers in range 1 to N, there will be only floor(√N) 1s left. 

这个关系可以证明如下：

Initially, all the elements are 0. 

Lemma 1: At any index i, the final element will be 1 if it is flipped odd number of times. This is trivial.

Lemma 2: For any index i, the number of times element at that index will be flipped equals number of factors of that coin.
Since we flip elements at multiples of 1, 2, 3, 4, … N. For each factor of any index i, it will be flipped. 

Lemma 3: From Lemma 1 and Lemma 2, all the indices having odd number of factors will have 1 as its final element .
For any natural number N, we can write it in its prime factorization form:

• N = α^a x β^b x γ^c x δ^d ….
  where α < β < γ < δ …. are prime numbers and a, b, c, d are whole numbers.
• Then, total number of factors of N = (a+1) x (b+1) x (c+1) x (d+1) x …
• Hence, we want ((a+1) x (b+1) x (c+1) x (d+1) x …) to be odd.
• => ((a+1) x (b+1) x (c+1) x (d+1) x …) is odd if (a+1) x (b+1) x (c+1) x (d+1) x … is odd
  => (a+1) x (b+1) x (c+1) x (d+1) x …        is odd if   (a+1), (b+1), (c+1), (d+1) … is odd
  => (a+1), (b+1), (c+1), (d+1)                   is odd if    a, b, c, d, …. is even
• Hence, N = α^a x β^b x γ^c x δ^d …. should has a, b, c, d, … as even whole numbers. This is only possible if N is a perfect square.

因此，所有完美平方的索引都将 1 作为其最终元素。

Number of perfect squares below N = ⌊√N⌋ 

下面给出上述观察的实施。

C++

// C++ code to implement the approach
#include 
#include 
using namespace std;
 
// Function to count number of 1's
// in final array
int findOnes(int N, int arr[])
{
    return floor(sqrt(N));
}
 
// Driver Code
int main()
{
    int N = 4;
    int arr[] = { 0, 0, 0, 0 };
 
    int count = findOnes(N, arr);
    cout << count;
 
    return 0;
}

Java

// JAVA code to implement the approach
import java.util.*;
class GFG
{
 
  // Function to count number of 1's
  // in final array
  public static int findOnes(int N, int arr[])
  {
    return (int)Math.floor((int)Math.sqrt(N));
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    int N = 4;
    int arr[] = new int[] { 0, 0, 0, 0 };
 
    int count = findOnes(N, arr);
    System.out.print(count);
  }
}
 
// This code is contributed by Taranpreet

C＃

// C# code to implement the approach
using System;
class GFG {
 
    // Function to count number of 1's
    // in final array
    static int findOnes(int N, int[] arr)
    {
        return (int)(Math.Floor(Math.Sqrt(N)));
    }
 
    // Driver Code
    public static void Main()
    {
        int N = 4;
        int[] arr = { 0, 0, 0, 0 };
 
        int count = findOnes(N, arr);
        Console.Write(count);
    }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

// Javascript code to implement the approach
 
// Function to count number of 1's
// in final array
function findOnes(N, arr)
{
    return Math.floor(Math.sqrt(N));
}
 
// Driver Code
let N = 4;
let arr = [ 0, 0, 0, 0 ];
 
let count = findOnes(N, arr);
document.write(count);
 
// This code is contributed by Samim Hossain Mondal.

Python3

# Python code to implement the approach
 
# Function to count number of 1's
# in final array
import math
 
 
def findOnes(N, arr):
 
    return math.floor(math.sqrt(N))
 
 
# Driver Code
N = 4
arr = [ 0, 0, 0, 0 ]
 
count = findOnes(N, arr)
print(count)
 
# This code is contributed by shinjanpatra

输出

2

时间复杂度： O(1)
辅助空间： O(1)