Find politeness of a number

Difficulty Level : Medium
Last Updated : 20 Jan, 2021

Given an integer n. Find politeness of number n. The politeness of a number is defined as the number of ways it can be expressed as the sum of consecutive integers.

Examples:

Input: n = 15
Output: 3
Explanation:
There are only three ways to express
15 as sum of consecutive integers i.e.,
15 = 1 + 2 + 3 + 4 + 5
15 = 4 + 5 + 6
15 = 7 + 8
Hence answer is 3

Input: n = 9;
Output:  2
There are two ways of representation:
9 = 2 + 3 + 4
9 = 4 + 5

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Naive approach:

Run a loop one inside another and find the sum of every consecutive integer up to n. The time complexity of this approach will be O(n²) which will not be sufficient for large value of n.

Efficient Approach:

Use factorization. We factorize the number n and count the number of odd factors. A total number of odd factors (except 1) is equal to the politeness of the number. Refer this for proof of this fact. In general, if a number can be represented as a^p * b^q * c^r … where a, b, c, … are prime factors of n. If a = 2 (even) then discard it and count total number of odd factors which can be written as [(q + 1) * (r + 1) * …] – 1 (Here 1 is subtracted because single term in representation is not allowed).
How does the above formula work? The fact is if a number is expressed as a^p * b^q * c^r … where a, b, c, … are prime factors of n, then a number of divisors is (p+1)*(q+1)*(r+1) ……
To simplify, let there be one factor, and the number is expressed as a^p. Divisors are 1, a, a², …. a^p. The count of divisors is p+1. Now let us take a slightly more complicated case a^pb^p. The divisors are :
1, a, a², …. a^p
b, ba, ba², …. ba^p
b², b²a, b²a², …. b²a^p
…………….
…………….
b^q, b^qa, b^qa², …. b^qa^p
The count of the above terms is (p+1)*(q+1). Similarly, we can prove for more prime factors.
Illustration : For n = 90, decomposition of prime factors will be as follows:-
=> 90 = 2 * 3² * 5¹. The power of odd prime factors 3, 5 are 2 and 1 respectively. Apply above formula as: (2 + 1) * (1 + 1) -1 = 5. Hence 5 will be the answer. We can crosscheck it. All odd factors are 3, 5, 9, 15 and 45.
Below is the program of the above steps:

C++

// C+ program to find politeness of number
#include <iostream>
using namespace std;
 
// A function to count all odd prime factors
// of a given number n
int countOddPrimeFactors(int n)
{
    int result = 1;
 
    // Eliminate all even prime 
    // factor of number of n
    while (n % 2 == 0)
        n /= 2;
 
    // n must be odd at this point, 
    // so iterate for only
    // odd numbers till sqrt(n)
    for (int i = 3; i * i <= n; i += 2) {
        int divCount = 0;
 
        // if i divides n, then start counting of
        // Odd divisors
        while (n % i == 0) {
            n /= i;
            ++divCount;
        }
 
        result *= divCount + 1;
    }
 
    // If n odd prime still remains then count it
    if (n > 2)
        result *= 2;
 
    return result;
}
 
int politness(int n)
{
    return countOddPrimeFactors(n) - 1;
}
 
// Driver program to test above function
int main()
{
    int n = 90;
    cout << "Politness of " << n << " = "
         << politness(n) << "\n";
 
    n = 15;
    cout << "Politness of " << n << " = "
         << politness(n) << "\n";
    return 0;
}

Java

// Java program to find politeness of a number
 
public class Politeness {
    // A function to count all odd prime factors
    // of a given number n
    static int countOddPrimeFactors(int n)
    {
        int result = 1;
 
        // Eliminate all even prime 
        // factor of number of n
        while (n % 2 == 0)
            n /= 2;
 
        // n must be odd at this point, so iterate
        // for only odd numbers till sqrt(n)
        for (int i = 3; i * i <= n; i += 2) {
            int divCount = 0;
 
            // if i divides n, then start counting of
            // Odd divisors
            while (n % i == 0) {
                n /= i;
                ++divCount;
            }
 
            result *= divCount + 1;
        }
        // If n odd prime still remains then count it
        if (n > 2)
            result *= 2;
 
        return result;
    }
 
    static int politness(int n)
    {
        return countOddPrimeFactors(n) - 1;
    }
 
    public static void main(String[] args)
    {
        int n = 90;
        System.out.println("Politness of " + n + " = "
                           + politness(n));
 
        n = 15;
        System.out.println("Politness of " + n + " = "
                           + politness(n));
    }
}
 
// This code is contributed by Saket Kumar

Python

# Python program to find politeness of number
 
# A function to count all odd prime factors
# of a given number n
def countOddPrimeFactors(n) :
    result = 1;
  
    # Eliminate all even prime factor of 
    # number of n
    while (n % 2 == 0) :
        n /= 2
  
    # n must be odd at this point, so iterate
    # for only odd numbers till sqrt(n)
    i = 3
    while i * i <= n :
        divCount = 0
  
        # if i divides n, then start counting 
        # of Odd divisors
        while (n % i == 0) :
            n /= i
            divCount = divCount + 1
        
        result = result * divCount + 1
        i = i + 2
  
    # If n odd prime still remains then count it
    if (n > 2) :
        result = result * 2
         
    return result
     
  
def politness( n) :
    return countOddPrimeFactors(n) - 1;
 
# Driver program to test above function
n = 90
print "Politness of ", n, " = ", politness(n)
n = 15
print "Politness of ", n, " = ", politness(n)
 
# This code is contributed by Nikita Tiwari.

C#

// C# program to find politeness of a number.
using System;
 
public class GFG {
     
    // A function to count all odd prime 
    // factors of a given number n
    static int countOddPrimeFactors(int n)
    {
         
        int result = 1;
 
        // Eliminate all even prime factor 
        // of number of n
        while (n % 2 == 0)
            n /= 2;
 
        // n must be odd at this point, so
        // iterate for only odd numbers
        // till sqrt(n)
        for (int i = 3; i * i <= n; i += 2)
        {
            int divCount = 0;
 
            // if i divides n, then start
            // counting of Odd divisors
            while (n % i == 0) {
                n /= i;
                ++divCount;
            }
 
            result *= divCount + 1;
        }
         
        // If n odd prime still remains 
        // then count it
        if (n > 2)
            result *= 2;
 
        return result;
    }
 
    static int politness(int n)
    {
        return countOddPrimeFactors(n) - 1;
    }
 
    // Driver code
    public static void Main()
    {
        int n = 90;
        Console.WriteLine("Politness of "
               + n + " = " + politness(n));
 
        n = 15;
        Console.WriteLine("Politness of "
               + n + " = " + politness(n));
    }
}
 
// This code is contributed by nitin mittal.

PHP

<?php
// PHP program to find 
// politeness of number
 
// A function to count all
// odd prime factors of a 
// given number n
function countOddPrimeFactors($n)
{
    $result = 1;
 
    // Eliminate all even prime 
    // factor of number of n
    while($n % 2 == 0)
        $n /= 2;
 
    // n must be odd at this 
    // point, so iterate for only
    // odd numbers till sqrt(n)
    for ($i = 3; $i * $i <= $n; $i += 2) 
    {
        $divCount = 0;
 
        // if i divides n, then
        // start counting of
        // Odd divisors
        while($n % $i == 0)
        {
            $n /= $i;
            ++$divCount;
        }
 
        $result *= $divCount + 1;
    }
 
    // If n odd prime still 
    // remains then count it
    if ($n > 2)
        $result *= 2;
 
    return $result;
}
 
function politness($n)
{
    return countOddPrimeFactors($n) - 1;
}
 
    // Driver Code
    $n = 90;
    echo "Politness of " , $n , " = "
               , politness($n), "\n";
 
    $n = 15;
    echo "Politness of " , $n , " = "
               , politness($n) ,"\n";
 
// This code is contributed by nitin mittal.
?>

Output:
Politness of 90 = 5
Politness of 15 = 3

Time complexity: O(sqrt(n))
Auxiliary space: O(1)
Reference: Wikipedia

Another Efficient approach:

Calculate if an AP can be generated for the given length domain [2, sqrt(2*n)]. The reason to calculate for length till sqrt(2*n) is-

max length wil be for the AP 1, 2, 3…

Length for this AP is -

n= ( len * (len+1) ) / 2

len2 + len - (2*n) =0

so len≈sqrt(2*n)

so we can check for each len from 1 to sqrt(2*n) ,if AP can be generated with this len. The formula to get the first term of the AP is –

n= ( len/2) * ( (2*A1) + len-1 )

Below is the implementation of the above approach:

C++

// CPP program for the above approach
#include <iostream>
#include <math.h>
using namespace std;
 
// Function to find politeness
int politness(int n)
{
    int count = 0;
   
    // sqrt(2*n) as max length 
    // will be when the sum starts
    // from 1
    // which follows the equation n^2 - n - (2*sum) = 0
    for (int i = 2; i <= sqrt(2 * n); i++) {
        int a;
        if ((2 * n) % i != 0)
            continue;
        a = 2 * n;
        a /= i;
        a -= (i - 1);
        if (a % 2 != 0)
            continue;
        a /= 2;
        if (a > 0) {
            count++;
        }
    }
    return count;
}
 
// Driver program to test above function
int main()
{
    int n = 90;
    cout << "Politness of " << n << " = " << politness(n)
         << "\n";
 
    n = 15;
    cout << "Politness of " << n << " = " << politness(n)
         << "\n";
    return 0;
}
 
// This code is contributed by Prajjwal Chittori

Java

// Java program for the above approach
import java.lang.Math;
public class Main {
   
  // Function to find politeness
  static int politness(int n)
  {
    int count = 0;
 
    // sqrt(2*n) as max length 
    // will be when the sum
    // starts from 1
    // which follows the 
    // equation n^2 - n - (2*sum) = 0
    for (int i = 2; i <= Math.sqrt(2 * n); i++) {
      int a;
      if ((2 * n) % i != 0)
        continue;
      a = 2 * n;
      a /= i;
      a -= (i - 1);
      if (a % 2 != 0)
        continue;
      a /= 2;
      if (a > 0) {
        count++;
      }
    }
    return count;
  }
 
  // Driver Code
  public static void main(String[] args)
  {
    int n = 90;
    System.out.println("Politness of " + n + " = "
                       + politness(n));
 
    n = 15;
    System.out.println("Politness of " + n + " = "
                       + politness(n));
  }
}
 
// This code is contributed by Prajjwal Chittori

Python

# python program for the above approach
import math
 
# Function to find politeness
def politness(n):
    count = 0
     
    # sqrt(2*n) as max length will be 
    # when the sum starts from 1
    # which follows the equation 
    # n^2 - n - (2*sum) = 0
    for i in range(2, int(math.sqrt(2 * n)) + 1):
        if ((2 * n) % i != 0):
            continue
        a = 2 * n
        a = a / i
        a = a - (i - 1)
        if (a % 2 != 0):
            continue
        a /= 2
        if (a > 0):
            count = count + 1
    return count
 
 
# Driver program to test above function
n = 90
print "Politness of ", n, " = ", politness(n)
n = 15
print "Politness of ", n, " = ", politness(n)
 
# This code is contributed by Prajjwal Chittori

Output

Politness of 90 = 5
Politness of 15 = 3

Time complexity: O(sqrt(2*n)) ≈ O(sqrt(n)) Auxiliary space: O(1)

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