Count of 1’s after flipping the bits at multiples from 1 to N

Given N sized binary array A[] containing all 0’s initially. The task is to find the final count of 1’s after flipping the bits at multiples from 1 to N.

Examples:

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

Naive Approach: The basic idea of the solution is based on greedy approach. 

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.

Follow the steps below to implement the approach:

• Create an array of size N and fill it with 0.
• Run a loop i = 1 to i = N, and for each i,
  • run a loop j = i – 1 to N with increment = i.
    • Flip the bits at each j.
• Iterate over the array and calculate number of 1.

Implementation of the above discussed approach is given below.

C++

// C++ code to implement the approach
#include <iostream>
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

<script>

// 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

</script>

Output

2

Time Complexity: O(N*log N)
Auxiliary Space: O(N)

Efficient Approach: The efficient approach to solve this problem is based on following mathematical observation:

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

This relation can be proved as per shown below:

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.

Therefore, all indices which are perfect squares will have 1 as their final element.

Number of perfect squares below N = ⌊√N⌋ 

Implementation of the above discussed observation is given below.

C++

// C++ code to implement the approach
#include <cmath>
#include <iostream>
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

<script>
// 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.
</script>

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

Output

2

Time Complexity: O(logN) as it is using inbuilt sqrt function
Auxiliary Space: O(1)