Find the number of integers x in range (1,N) for which x and x+1 have same number of divisors
Given an integer N. The task is to find the number of integers 1 < x < N, for which x and x + 1 have the same number of positive divisors.
Examples:
Input: N = 3
Output: 1
Divisors(1) = 1
Divisors(2) = 1 and 2
Divisors(3) = 1 and 3
Only valid x is 2.Input: N = 15
Output: 2
Approach: Find the number of divisors of all numbers below N and store them in an array. And count the number of integers x such that x and x + 1 have the same number of positive divisors by running a loop.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; #define N 100005 // To store number of divisors and // Prefix sum of such numbers int d[N], pre[N]; // Function to find the number of integers // 1 < x < N for which x and x + 1 have // the same number of positive divisors void Positive_Divisors() { // Count the number of divisors for ( int i = 1; i < N; i++) { // Run a loop upto sqrt(i) for ( int j = 1; j * j <= i; j++) { // If j is divisor of i if (i % j == 0) { // If it is perfect square if (j * j == i) d[i]++; else d[i] += 2; } } } int ans = 0; // x and x+1 have same number of // positive divisors for ( int i = 2; i < N; i++) { if (d[i] == d[i - 1]) ans++; pre[i] = ans; } } // Driver code int main() { // Function call Positive_Divisors(); int n = 15; // Required answer cout << pre[n] << endl; return 0; } |
Java
// Java implementation of the approach class GFG { static int N = 100005 ; // To store number of divisors and // Prefix sum of such numbers static int d[] = new int [N], pre[] = new int [N]; // Function to find the number of integers // 1 < x < N for which x and x + 1 have // the same number of positive divisors static void Positive_Divisors() { // Count the number of divisors for ( int i = 1 ; i < N; i++) { // Run a loop upto sqrt(i) for ( int j = 1 ; j * j <= i; j++) { // If j is divisor of i if (i % j == 0 ) { // If it is perfect square if (j * j == i) d[i]++; else d[i] += 2 ; } } } int ans = 0 ; // x and x+1 have same number of // positive divisors for ( int i = 2 ; i < N; i++) { if (d[i] == d[i - 1 ]) ans++; pre[i] = ans; } } // Driver code public static void main(String[] args) { // Function call Positive_Divisors(); int n = 15 ; // Required answer System.out.println(pre[n]); } } /* This code contributed by PrinciRaj1992 */ |
Python3
# Python3 implementation of the above approach from math import sqrt; N = 100005 # To store number of divisors and # Prefix sum of such numbers d = [ 0 ] * N pre = [ 0 ] * N # Function to find the number of integers # 1 < x < N for which x and x + 1 have # the same number of positive divisors def Positive_Divisors() : # Count the number of divisors for i in range (N) : # Run a loop upto sqrt(i) for j in range ( 1 , int (sqrt(i)) + 1 ) : # If j is divisor of i if (i % j = = 0 ) : # If it is perfect square if (j * j = = i) : d[i] + = 1 else : d[i] + = 2 ans = 0 # x and x+1 have same number of # positive divisors for i in range ( 2 , N) : if (d[i] = = d[i - 1 ]) : ans + = 1 pre[i] = ans # Driver code if __name__ = = "__main__" : # Function call Positive_Divisors() n = 15 # Required answer print (pre[n]) # This code is contributed by Ryuga |
C#
// C# implementation of the approach using System; class GFG { static int N =100005; // To store number of divisors and // Prefix sum of such numbers static int []d = new int [N]; static int []pre = new int [N]; // Function to find the number of integers // 1 < x < N for which x and x + 1 have // the same number of positive divisors static void Positive_Divisors() { // Count the number of divisors for ( int i = 1; i < N; i++) { // Run a loop upto sqrt(i) for ( int j = 1; j * j <= i; j++) { // If j is divisor of i if (i % j == 0) { // If it is perfect square if (j * j == i) d[i]++; else d[i] += 2; } } } int ans = 0; // x and x+1 have same number of // positive divisors for ( int i = 2; i < N; i++) { if (d[i] == d[i - 1]) ans++; pre[i] = ans; } } // Driver code public static void Main(String[] args) { // Function call Positive_Divisors(); int n = 15; // Required answer Console.WriteLine(pre[n]); } } // This code has been contributed by 29AjayKumar |
PHP
<?php // PHP implementation of the approach $N = 100005; // To store number of divisors and // Prefix sum of such numbers $d = array_fill (0, $N ,NULL); $pre = array_fill (0, $N ,NULL); // Function to find the number of integers // 1 < x < N for which x and x + 1 have // the same number of positive divisors function Positive_Divisors() { global $N , $d , $pre ; // Count the number of divisors for ( $i = 1; $i < $N ; $i ++) { // Run a loop upto sqrt(i) for ( $j = 1; $j * $j <= $i ; $j ++) { // If j is divisor of i if ( $i % $j == 0) { // If it is perfect square if ( $j * $j == $i ) $d [ $i ]++; else $d [ $i ] += 2; } } } $ans = 0; // x and x+1 have same number of // positive divisors for ( $i = 2; $i < $N ; $i ++) { if ( $d [ $i ] == $d [ $i - 1]) $ans ++; $pre [ $i ] = $ans ; } } // Driver code // Function call Positive_Divisors(); $n = 15; // Required answer echo $pre [ $n ] ; return 0; // This code is contributed by ChitraNayal ?> |
Javascript
<script> // Javascript implementation of the approach const N = 100005; // To store number of divisors and // Prefix sum of such numbers let d = new Array(N).fill(0); let pre = new Array(N).fill(0); // Function to find the number of integers // 1 < x < N for which x and x + 1 have // the same number of positive divisors function Positive_Divisors() { // Count the number of divisors for (let i = 1; i < N; i++) { // Run a loop upto sqrt(i) for (let j = 1; j * j <= i; j++) { // If j is divisor of i if (i % j == 0) { // If it is perfect square if (j * j == i) d[i]++; else d[i] += 2; } } } let ans = 0; // x and x+1 have same number of // positive divisors for (let i = 2; i < N; i++) { if (d[i] == d[i - 1]) ans++; pre[i] = ans; } } // Driver code // Function call Positive_Divisors(); let n = 15; // Required answer document.write(pre[n]); // This code is contributed by souravmahato348 </script> |
2
Time complexity: O(N3/2)
Auxiliary Space: O(N)
Approach 2:
One approach to finding the number of integers 1 < x < N for which x and x + 1 have the same number of positive divisors is to use a prime factorization method. This involves finding the prime factors of each number and using the formula for the number of divisors of a number based on its prime factorization.
Here is the step-by-step algorithm for implementing the approach:
- Define a function Positive_Divisors() to calculate the number of positive divisors for each integer in the range [2, N).
- In the function Positive_Divisors(), initialize an array d of size N to store the number of divisors for each integer in the range [2, N).
- Iterate over each integer i in the range [2, N) using a loop.
- For each integer i, calculate its prime factorization and compute the number of divisors using the formula for the number of divisors of a positive integer. Store the result in d[i].
- Initialize a variable ans to 0 to store the final answer.
- Iterate over each integer i in the range [2, N) using another loop.
- For each integer i, check if d[i] == d[i – 1], If this condition is true, then increment ans by 1.
- Initialize an array pre of size N to store the prefix sum of the answer.
- For each integer i in the range [2, N), set pre[i] = ans.
- In the main() function, call the function Positive_Divisors().
- Print the value of pre[n], where n is a given integer.
C++
#include <bits/stdc++.h> using namespace std; #define N 100005 // To store number of divisors and // Prefix sum of such numbers int d[N], pre[N]; // Function to find the number of integers // 1 < x < N for which x and x + 1 have // the same number of positive divisors void Positive_Divisors() { // Count the number of divisors for ( int i = 2; i < N; i++) { int num = i; int divisors = 1; for ( int j = 2; j * j <= num; j++) { int count = 0; while (num % j == 0) { count++; num /= j; } divisors *= (count + 1); } if (num > 1) divisors *= 2; d[i] = divisors; } int ans = 0; // x and x+1 have same number of // positive divisors for ( int i = 2; i < N; i++) { if (d[i] == d[i - 1]) ans++; pre[i] = ans; } } // Driver code int main() { // Function call Positive_Divisors(); int n = 15; // Required answer cout << pre[n] << endl; return 0; } |
2
Time Complexity: O(N log N)
Auxiliary Space: O(N)
Explaination :
The time complexity of this code is O(N log N) because calculating the prime factorization of each integer in the range [2, N) takes O(log N) time and this operation is performed for each integer in the range [2, N).
The auxiliary space complexity is O(N) because two arrays of size N are used.
Please Login to comment...