Sum of all Perfect numbers lying in the range [L, R]

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: 6
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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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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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