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Sum of all Perfect numbers lying in the range [L, R]
  • Last Updated : 12 Apr, 2021

Given two numbers L, R which signifies the range [L, R], the task is to find the sum of all perfect numbers lying in the range [L, R].
Examples: 
 

Input: L = 6, R = 10 
Output:
Explanation: 
From 6 to 10, the only perfect number is 6.
Input: L = 6, R = 28 
Output: 34 
Explanation: 
There are two perfect numbers in the range [6, 28]. They are, {6, 28} 
6 + 28 = 34. 
 

 

Naive Approach: The naive approach for this problem is to check if a number is a perfect number or not by iterating through every number in the range [L, R]. If there are N numbers in the range, the time complexity for this approach is O(N * sqrt(K)) where K is the maximum number(R) in the range [L, R].
Efficient Approach: The idea is to use the concept of prefix sum array to perform precomputations and store the sum of all the numbers in an array. 
 

  • Initialize an array up to the max size such that every index ‘i’ of the array represents the sum of all the perfect numbers from [0, i].
  • Iterate over the array and for every index, check if it is a perfect number or not.
  • If it is a perfect number, then add the sum stored at the previous index (i – 1) and the current index ‘i’.
  • If it is not a perfect number, then add the sum stored at the previous index (i – 1) and the value 0.
  • Finally, for every query [L, R], return the value: 
     
sum = arr[R] - arr[L - 1]

Below is the implementation of the above approach:
 



C++




// C++ implementation to find the
// sum of all perfect numbers
// lying in the range [L, R]
 
#include <bits/stdc++.h>
#define ll int
using namespace std;
 
// Array to store the sum
long long pref[100010];
 
// Function to check if a number is
// a perfect number or not
int isPerfect(int n)
{
    int sum = 1;
 
    // Iterating till the square root
    // of the number and checking if
    // the sum of divisors is equal
    // to the number or not
    for (int i = 2; i * i <= n; i++) {
        if (n % i == 0) {
            if (i * i != n)
                sum = sum + i + n / i;
            else
                sum = sum + i;
        }
    }
 
    // If it is a perfect number, then
    // return the number
    if (sum == n && n != 1)
        return n;
 
    // Else, return 0
    return 0;
}
 
// Function to precompute the sum
// of perfect squares and store
// then in an array
void precomputation()
{
    for (int i = 1; i <= 100000; ++i) {
        pref[i] = pref[i - 1] + isPerfect(i);
    }
}
 
int main()
{
 
    int L = 6, R = 28;
 
    precomputation();
 
    cout << pref[R] - pref[L - 1];
 
    return 0;
}

Java




// Java implementation to find the
// sum of all perfect numbers
// lying in the range [L, R]
class GFG {
     
    // Array to store the sum
    static int pref [] = new int[10000];
     
    // Function to check if a number is
    // a perfect number or not
    static int isPerfect(int n)
    {
        int sum = 1;
     
        // Iterating till the square root
        // of the number and checking if
        // the sum of divisors is equal
        // to the number or not
        for (int i = 2; i * i <= n; i++) {
            if (n % i == 0) {
                if (i * i != n)
                    sum = sum + i + n / i;
                else
                    sum = sum + i;
            }
        }
     
        // If it is a perfect number, then
        // return the number
        if (sum == n && n != 1)
            return n;
     
        // Else, return 0
        return 0;
    }
     
    // Function to precompute the sum
    // of perfect squares and store
    // then in an array
    static void precomputation()
    {
        for (int i = 1; i < 10000; ++i) {
            pref[i] = pref[i - 1] + isPerfect(i);
        }
    }
     
    public static void main (String[] args)
    {
     
        int L = 6, R = 28;
     
        precomputation();
     
        System.out.println(pref[R] - pref[L - 1]);
     
    }
 
}
 
// This code is contributed by AnkitRai01

Python3




# Python3 implementation to find the
# sum of all perfect numbers
# lying in the range [L, R]
 
from math import sqrt
 
# Array to store the sum
pref = [0]*10000;
 
# Function to check if a number is
# a perfect number or not
def isPerfect(n)  :
 
    sum = 1;
 
    # Iterating till the square root
    # of the number and checking if
    # the sum of divisors is equal
    # to the number or not
    for i in range(2, int(sqrt(n)) + 1) :
        if (n % i == 0) :
            if (i * i != n) :
                sum = sum + i + n // i;
            else :
                sum = sum + i;
                 
    # If it is a perfect number, then
    # return the number
    if (sum == n and n != 1) :
        return n;
 
    # Else, return 0
    return 0;
 
# Function to precompute the sum
# of perfect squares and store
# then in an array
def precomputation() :
 
    for i in range(1, 10000) :
        pref[i] = pref[i - 1] + isPerfect(i);
 
if __name__ == "__main__" :
 
 
    L = 6; R = 28;
 
    precomputation();
 
    print(pref[R] - pref[L - 1]);
 
# This code is contributed by AnkitRai01

C#




// C# implementation to find the
// sum of all perfect numbers
// lying in the range [L, R]
using System;
 
public class GFG {
      
    // Array to store the sum
    static int []pref = new int[10000];
      
    // Function to check if a number is
    // a perfect number or not
    static int isPerfect(int n)
    {
        int sum = 1;
      
        // Iterating till the square root
        // of the number and checking if
        // the sum of divisors is equal
        // to the number or not
        for (int i = 2; i * i <= n; i++) {
            if (n % i == 0) {
                if (i * i != n)
                    sum = sum + i + n / i;
                else
                    sum = sum + i;
            }
        }
      
        // If it is a perfect number, then
        // return the number
        if (sum == n && n != 1)
            return n;
      
        // Else, return 0
        return 0;
    }
      
    // Function to precompute the sum
    // of perfect squares and store
    // then in an array
    static void precomputation()
    {
        for (int i = 1; i < 10000; ++i) {
            pref[i] = pref[i - 1] + isPerfect(i);
        }
    }
      
    public static void Main(String[] args)
    {
      
        int L = 6, R = 28;
      
        precomputation();
      
        Console.WriteLine(pref[R] - pref[L - 1]);
      
    }
  
}
// This code contributed by Rajput-Ji

Javascript




<script>
 
// Javascript implementation to find the
// sum of all perfect numbers
// lying in the range [L, R]
 
// Array to store the sum
var pref = Array(100010).fill(0);
 
// Function to check if a number is
// a perfect number or not
function isPerfect(n)
{
    var sum = 1;
 
    // Iterating till the square root
    // of the number and checking if
    // the sum of divisors is equal
    // to the number or not
    for (var i = 2; i * i <= n; i++) {
        if (n % i == 0) {
            if (i * i != n)
                sum = sum + i + n / i;
            else
                sum = sum + i;
        }
    }
 
    // If it is a perfect number, then
    // return the number
    if (sum == n && n != 1)
        return n;
 
    // Else, return 0
    return 0;
}
 
// Function to precompute the sum
// of perfect squares and store
// then in an array
function precomputation()
{
    for (var i = 1; i <= 100000; ++i) {
        pref[i] = pref[i - 1] + isPerfect(i);
    }
}
 
var L = 6, R = 28;
precomputation();
document.write( pref[R] - pref[L - 1]);
 
// This code is contributed by noob2000.
</script>
Output: 
34

 

Time Complexity: 
 

  • The time taken for the precomputation is O(K * sqrt(K)) where K is the number upto which we are performing the precomputation
  • After precomputation, each query is answered in O(1).

 

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