Given a number “n”, find its total number of divisors is even or odd.

Examples:

Input: n = 10 Output: Even Input: n = 100 Output: Odd Input: n = 125 Output: Even

A **naive approach** would be to find all the divisors and then see if the total number of divisors is even or odd.

The time complexity for such a solution would be O(sqrt(n))

`// Naive Solution to ` `// find if count of ` `// divisors is even ` `// or odd ` `import` `java.io.*; ` `import` `java.math.*; ` ` ` `class` `GFG { ` ` ` ` ` `// Function to count ` ` ` `// the divisors ` ` ` `static` `void` `countDivisors(` `int` `n) ` ` ` `{ ` ` ` `// Initialize count ` ` ` `// of divisors ` ` ` `int` `count = ` `0` `; ` ` ` ` ` `// Note that this ` ` ` `// loop runs till ` ` ` `// square root ` ` ` `for` `(` `int` `i = ` `1` `; i <= Math.sqrt(n) + ` `1` `; i++) { ` ` ` `if` `(n % i == ` `0` `) ` ` ` ` ` `// If divisors are ` ` ` `// equal increment ` ` ` `// count by one ` ` ` `// Otherwise increment ` ` ` `// count by 2 ` ` ` `count += (n / i == i) ? ` `1` `: ` `2` `; ` ` ` `} ` ` ` ` ` `if` `(count % ` `2` `== ` `0` `) ` ` ` `System.out.println(` `"Even"` `); ` ` ` ` ` `else` ` ` `System.out.println(` `"Odd"` `); ` ` ` `} ` ` ` ` ` `// Driver program ` ` ` `public` `static` `void` `main(String args[]) ` ` ` `{ ` ` ` `System.out.println(` `"The count of divisor:"` `); ` ` ` `countDivisors(` `10` `); ` ` ` `} ` `} ` `/* This code is contributed by Nikita Tiwari.*/` |

*chevron_right*

*filter_none*

**Output:**

The count of divisor: Even

**Efficient Solution: **

We can observe that the number of divisors is odd only in case of perfect squares. Hence the best solution would be to check if the given number is perfect square or not. If it’s a perfect square, then the number of divisors would be odd, else it’d be even.

`// Java program for ` `// Efficient Solution to find ` `// if count of divisors is ` `// even or odd ` `import` `java.io.*; ` `import` `java.math.*; ` ` ` `class` `GFG { ` ` ` ` ` `// Function to find if count ` ` ` `// of divisors is even or ` ` ` `// odd ` ` ` `static` `void` `countDivisors(` `int` `n) ` ` ` `{ ` ` ` `int` `root_n = (` `int` `)(Math.sqrt(n)); ` ` ` ` ` `// If n is a perfect square, ` ` ` `// then, it has odd divisors ` ` ` `if` `(root_n * root_n == n) ` ` ` `System.out.println(` `"Odd"` `); ` ` ` ` ` `else` ` ` `System.out.println(` `"Even"` `); ` ` ` `} ` ` ` ` ` `/* Driver program to test above function */` ` ` `public` `static` `void` `main(String args[]) ` ` ` `throws` `IOException ` ` ` `{ ` ` ` `System.out.println(` `"The count of "` ` ` `+ ` `"divisors of 10 is: "` `); ` ` ` ` ` `countDivisors(` `10` `); ` ` ` `} ` `} ` ` ` `/* This code is contributed by Nikita Tiwari. */` |

*chevron_right*

*filter_none*

**Output:**

The count of divisors of 10 is: Even

Please refer complete article on Check if count of divisors is even or odd for more details!

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready.