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++
// C++ implementation to find the// sum of all perfect numbers// lying in the range [L, R]#include <bits/stdc++.h>#define ll intusing namespace std;// Array to store the sumlong long pref[100010];// Function to check if a number is// a perfect number or notint 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 arrayvoid 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 sumvar pref = Array(100010).fill(0);// Function to check if a number is// a perfect number or notfunction 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 arrayfunction 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).
Auxiliary Space: O(105)
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