Last Updated: 31 Dec, 2020

Perfect Number

Easy
Asked in companies
IntuitDefine Next

Problem statement

You are given a positive integer K. Your task is to find out whether K is a perfect number or not.

Note :
Perfect numbers are those numbers that are equal to the sum of all its proper divisors. 

Proper divisors are all the divisors of a number excluding the number itself. 
For example :
For K = 6, the proper divisors of 6 are 1, 2, 3 and its sum is 6 so 6 is a perfect number. For K = 8, the proper divisors of 8 are 1, 2,4 and its sum is 7 so 8 is not a perfect number.
Input format :
The first line of input contains a single integer T, representing the number of test cases. 

The first and the only line of each test case contains a single positive integer K, as described in the problem statement.
Output Format :
For each test case, return true if K is a perfect number or else return false.
Note :
You do not need to print anything. It has already been taken care of. Just implement the given function.
Constraint :
1 <= T <= 100
2 <= K <= 10^8

Time Limit: 1 sec
Follow Up :
Can you do this in O(sqrt(N)) time and constant space?

Approaches

01 Approach

  • Iterate through all the numbers from 1 to n-1 and if the number divides k then add it to the sum.
  • If the sum is equal to k then it is a perfect number.

02 Approach

  • One key observation is the divisor occurs in pairs. Let say A and B are two number and A * B = K, let’s say that A < sqrt(K) then B should be greater than sqrt(K). So it is enough to iterate through numbers less than or equal to sqrt(k) to get all of its divisors.
  • Iterate through all the numbers from 1 to sqrt(K), let's say we are currently at ith number and k is divisible by i then add both i and k/i to the sum.
  • If the sum of all proper divisors is equal to k then return true else return false.
  • In case of i == 1 or i == sqrt(K) just add i to the sum and not k/i.