Find politeness of a number

20 Apr, 2022
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
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 ap * bq * cr ... 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 ap * bq * cr ... 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 ap. Divisors are 1, a, a2, .... ap. The count of divisors is p+1. Now let us take a slightly more complicated case apbp. The divisors are : 1, a, a2, .... ap b, ba, ba2, .... bap b2, b2a, b2a2, .... b2ap ................ ................ bq, bqa, bqa2, .... bqapThe 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++
Java
Python3
C#
PHP
Javascript
// C+ program to find politeness of number#include <iostream>using namespace std; // A function to count all odd prime factors// of a given number nint 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 politeness(int n){    return countOddPrimeFactors(n) - 1;} // Driver program to test above functionint main(){    int n = 90;    cout << "Politeness of " << n << " = "         << politeness(n) << "\n";     n = 15;    cout << "Politeness of " << n << " = "         << politeness(n) << "\n";    return 0;}
// 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 politeness(int n)    {        return countOddPrimeFactors(n) - 1;    }     public static void main(String[] args)    {        int n = 90;        System.out.println("Politeness of " + n + " = "                           + politeness(n));         n = 15;        System.out.println("Politeness of " + n + " = "                           + politeness(n));    }} // This code is contributed by Saket Kumar
# Python program to find politeness of number # A function to count all odd prime factors# of a given number ndef 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 politeness( n) :    return countOddPrimeFactors(n) - 1; # Driver program to test above functionn = 90print ("Politeness of ", n, " = ", politeness(n))n = 15print ("Politeness of ", n, " = ", politeness(n)) # This code is contributed by Nikita Tiwari.
// 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 politeness(int n)    {        return countOddPrimeFactors(n) - 1;    }     // Driver code    public static void Main()    {        int n = 90;        Console.WriteLine("Politeness of "               + n + " = " + politeness(n));         n = 15;        Console.WriteLine("Politeness of "               + n + " = " + politeness(n));    }} // This code is contributed by nitin mittal.
<?php// PHP program to find// politeness of number // A function to count all// odd prime factors of a// given number nfunction 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 $result;} function politeness($n){    return countOddPrimeFactors($n) - 1;}     // Driver Code    $n = 90;    echo "Politeness of " , $n , " = "               , politeness($n), "\n";     $n = 15;    echo "Politeness of " , $n , " = "               , politeness($n) ,"\n"; // This code is contributed by nitin mittal.?>
<script>// JavaScript program for the above approach      // A function to count all odd prime    // factors of a given number n    function countOddPrimeFactors(n)    {                  let 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 (let i = 3; i * i <= n; i += 2)        {            let 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 politeness(n)    {        return countOddPrimeFactors(n) - 1;    } // Driver Code     let n = 90;        document.write("Politeness of "               + n + " = " + politeness(n) + "<br />");          n = 15;        document.write("Politeness of "               + n + " = " + politeness(n)); // This code is contributed by splevel62.</script>
Output:
Politeness of 90 = 5
Politeness 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 will 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++
Java
Python3
C#
Javascript
// CPP program for the above approach#include <iostream>#include <math.h>using namespace std; // Function to find politenessint 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 functionint 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 program for the above approachimport 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 program for the above approachimport math # Function to find politenessdef 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 functionn = 90print ("Politness of ", n, " = ", politness(n))n = 15print ("Politness of ", n, " = ", politness(n)) # This code is contributed by Prajjwal Chittori
// C# program for the above approachusing System;public class GFG {     // 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;    Console.WriteLine("Politness of " + n + " = "                       + politness(n));     n = 15;    Console.WriteLine("Politness of " + n + " = "                       + politness(n));  }} // This code is contributed by gauravrajput1
<script> // JavaScript program for the above approach   // Function to find politeness function politness(n)  {    let 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 (let i = 2; i <= Math.sqrt(2 * n); i++) {      let a;      if ((2 * n) % i != 0)        continue;      a = 2 * n;      a = Math.floor(a / i);      a -= (i - 1);      if (a % 2 != 0)        continue;      a = Math.floor(a / 2);      if (a > 0) {        count++;      }    }    return count;  } // Driver Code      let n = 90;    document.write("Politness of " + n + " = "                       + politness(n) + "<br/>");     n = 15;    document.write("Politness of " + n + " = "                       + politness(n)); </script>
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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