all divisors of a number gfg practice

GFG Weekly Coding Contest. Examples: It is returning duplicates, like for 6 it returns. Practice Given a natural number n, print all distinct divisors of it. According to this post, we can get all divisors of a number through the following codes. #mathematics One of the divisors in every pair is smaller than or equal to sqrt(n) Assume Y = p1 * p2 * p3 where p1,p2,p3 are prime and p1,p2,p3 > N [Explained above]. Find sum of inverse of the divisors when sum of divisors and the number is given. By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Find sum of divisors of all the divisors of a natural number, Kth element in permutation of first N natural numbers having all even numbers placed before odd numbers in increasing order, Sum of all proper divisors of natural numbers in an array, Maximum possible prime divisors that can exist in numbers having exactly N divisors, Find sum of inverse of the divisors when sum of divisors and the number is given, Sum of all proper divisors of a natural number, Divisors of n-square that are not divisors of n, Check if count of even divisors of N is equal to count of odd divisors, Check if a number has an odd count of odd divisors and even count of even divisors, Minimum value exceeding X whose count of divisors has different parity with count of divisors of X, Mathematical and Geometric Algorithms - Data Structure and Algorithm Tutorials, Learn Data Structures with Javascript | DSA Tutorial, Introduction to Max-Heap Data Structure and Algorithm Tutorials, Introduction to Set Data Structure and Algorithm Tutorials, Introduction to Map Data Structure and Algorithm Tutorials, A-143, 9th Floor, Sovereign Corporate Tower, Sector-136, Noida, Uttar Pradesh - 201305, We use cookies to ensure you have the best browsing experience on our website. Here the task is simpler, we need to count divisors.First of all store all primes from 2 to max_size in an array so that we should only check for the prime divisors. Generating all divisors of a number using its prime factorization This article is contributed by Shivam Pradhan (anuj_charm). (Ep. slow) problem. Find divisors of a number - Math from scratch Check if a number has an odd count of odd divisors and even count of even divisors. However, it doesn't seem to be a good one. For example, the divisors of number 24 are 1 2 3 4 6 8 12 24. #xml, #android 3 Divisors | Practice | GeeksforGeeks Examples: Below is the implementation of the above approach: Time Complexity: O((A * B * C) + N3/2)Auxiliary Space: O(N). Is iMac FusionDrive->dual SSD migration any different from HDD->SDD upgrade from Time Machine perspective? Practice Given a positive integer n, we have to find the total number of divisors for n. Examples: Input : n = 25 Output : 3 Divisors are 1, 5 and 25. You will be notified via email once the article is available for improvement. acknowledge that you have read and understood our. Actually this doesn't work on large numbers. Example 1: Input: a = 12, b = 24 Output: 6 Explanation: all . xn) / b ) mod (m), Legendres formula (Given p and n, find the largest x such that p^x divides n! Thank you for your valuable feedback! A Naive Solution would be to iterate all the numbers from 1 to n, checking if that number divides n and printing it. First, find the prime factorization of n using this approach and for every prime factor, store it with the count of its occurrence. For example, primeFactors(12) will populate factors[] in this way: The third part setDivisors() recursively calls itself to calculate all the divisors of x, using the vector factors[] and puts them in vector divisors[]. 2.) Input: N = 16Output: 4 9Explanation: 4 and 9 have exactly three divisors. Generate the prime numbers from 1 to N using any sieve method efficiently, Print all the prime numbers(X) between 1 to N, such as X, If the above condition is satisfied, the number will be printed and the loop will continue till. Job-a-Thon. ----------------- Now if we find the factors of N, it will always have following combinations: 1*N. X*X. This article is being improved by another user right now. GCD of all the numbers is 1. Below is the implementation of the above approach: Time Complexity: O(N2)Auxiliary Space: O(1), Time Complexity: O(N*sqrt(N))Auxiliary Space: O(1). As we know the divisors of a number will definitely be lesser or equal to the number, all the numbers between 1 and the number, are the possible candidates for the divisors. 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Then find d(i.j.k) using pre computed number of divisors. Expected Time Complexity: O ( (B-A) sqrt (B)) Expected Space Complexity: O (1) Constraints: 1<=A<=B<=104 This article is being improved by another user right now. There's no reason to do that. To understand how to do this, let's look at some examples. Therefore, the auxiliary space used by the function is O(sqrt(x)). 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List all the divisors of number 6. . #interviewquestions You just iterate over the set of numbers corresponding to the possible combinations of exponents. #mobile, #java The input array has a space complexity of O(n). Given a positive integer n, we have to find the total number of divisors for n. We have discussed different approaches for printing all divisors (here and here). A Nave Solution would be to iterate all the numbers from 1 to sqrt(n), checking if that number divides n and incrementing number of divisors. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. Why not? 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Plenty of good solutions exist for finding all the prime factors of not too large numbers. If x<=y This article is being improved by another user right now. 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But Y is a factor of N and cannot be greater than N. Therefore, there is a contradiction, which implies that one of p1, p2, p3 must be less than N. Can you suggest me ways to optimize this code? Recommended: Please try your approach on {IDE} first, before moving on to the solution. Given a natural number n, print all distinct divisors of it. Find sum of divisors of all the divisors of a natural number. Connect and share knowledge within a single location that is structured and easy to search. #java Efficient Solution: The first part, sieve() is used to find the prime numbers and put them in primes[] array. Examples: Input : 18 Output : 6 Divisors of 18 are 1, 2, 3, 6, 9 and 18. Java Python3 C# Javascript #include <bits/stdc++.h> using namespace std; void factors (int n) { int i, j; cout << "1 -->1\n"; for (i = 2; i <= n; i++) { cout << i << " -->"; Please refer below for an O(sqrt(n)) time solution that prints divisors in sorted order.Find all divisors of a natural number | Set 2This article is contributed by Ashutosh Kumar. Given a positive integer N., The task is to find the value of where function F(i) for the number i be defined as the sum of all divisors of 'i'. Output: 1 5 25 125 Practice Given a number n, count all distinct divisors of it. Here is the Java Implementation of this approach: We can use modified sieve for getting all the factors for all numbers in range [1, N-1]. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 4. Now all the possible divisors can be generated recursively if the count of occurrence of every prime factor of n is known. Finally, we can check if all the integers in the array are divisors of X by iterating through the array and checking if X is divisible by each integer. Can something be logically necessary now but not in the future? Efficiently getting all divisors of a given number - Stack Overflow #java Maximum possible prime divisors that can exist in numbers having exactly N divisors, Find sum of divisors of all the divisors of a natural number, Find sum of inverse of the divisors when sum of divisors and the number is given, Count numbers up to N having exactly 5 divisors, Find Nth smallest number having exactly 4 divisors, Divisors of n-square that are not divisors of n, Check if count of even divisors of N is equal to count of odd divisors, Check if a number has an odd count of odd divisors and even count of even divisors, Minimum value exceeding X whose count of divisors has different parity with count of divisors of X, Count number of integers less than or equal to N which has exactly 9 divisors, Mathematical and Geometric Algorithms - Data Structure and Algorithm Tutorials, Learn Data Structures with Javascript | DSA Tutorial, Introduction to Max-Heap Data Structure and Algorithm Tutorials, Introduction to Set Data Structure and Algorithm Tutorials, Introduction to Map Data Structure and Algorithm Tutorials, A-143, 9th Floor, Sovereign Corporate Tower, Sector-136, Noida, Uttar Pradesh - 201305, We use cookies to ensure you have the best browsing experience on our website. For every prime factor pi, it can be included x times where 0 x ai. Not getting accurate results. Meanwhile push the divisor d for every multiples. Approach: As every number greater than 1 can be represented in its prime factorization as p1a1*p2a2**pkak, where pi is a prime number, k 1 and ai is a positive integer. This article is being improved by another user right now. So you have far fewer divisions. After searching some related posts, I did not find any good solutions. acknowledge that you have read and understood our. Hack-a-thon. #naivemethod Starting the Prompt Design Site: A New Home in our Stack Exchange Neighborhood, Temporary policy: Generative AI (e.g., ChatGPT) is banned. What's the significance of a C function declaration in parentheses apparently forever calling itself? Below is a program for the same: Time Complexity : O(n)Auxiliary Space : O(1). Note that this problem is different from finding all prime factors. How is the pion related to spontaneous symmetry breaking in QCD? The Overflow #186: Do large language models know what theyre talking about? Common divisors of N numbers - GeeksforGeeks Divisors of n-square that are not divisors of n. 5. Input: 6 Output: 1 2 3 6 Explanation Divisors of 6 are: 1, 2, 3, 6 Input: 10 Output: 1 2 5 10 Recommended: Please try your approach on {IDE} first, before moving on to the solution. I originally thought that you just wanted the number of distinct factors. I will look to adjust algorithm to incorporate square root. Examples: Input : n = 10 Output: 1 2 5 10 Input: n = 100 Output: 1 2 4 5 10 20 25 50 100 Input: n = 125 Output: 1 5 25 125 Note that this problem is different from finding all prime factors. Naive solution: By using our site, you Please include more context in your reply, see. X^3 N such that it divides the number, and another number Y such that N = X * Y. X consists of all the prime factor of N, which are less than N and Y contains all the prime factors that are greater. Let x be the smaller, i.e; x <= y Since we know that each integer in the array is a divisor of X, we can take the square root of X^2 to get X. 4. Therefore, we can multiply all the integers in the array to get X^2, take the square root of X^2 to get X, and then check if all the integers in the array are divisors of X. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. #factors Browse other questions tagged, Where developers & technologists share private knowledge with coworkers, Reach developers & technologists worldwide, The future of collective knowledge sharing. POTD. How can I write a fast function to calculate total divisors of a number? #equala, #java Menu. Not the answer you're looking for? Thank you for your valuable feedback! Example 1. What does "rooting for my alt" mean in Stranger Things? I would be more worried about the unmatched paren on the first line. That doesn't help if the original number is prime, but if it isn't then you'll do far fewer divisions. of digits in any base, Find element using minimum segments in Seven Segment Display, Find next greater number with same set of digits, Numbers having difference with digit sum more than s, Total numbers with no repeated digits in a range, Find number of solutions of a linear equation of n variables, Program for dot product and cross product of two vectors, Number of non-negative integral solutions of a + b + c = n, Check if a number is power of k using base changing method, Convert a binary number to hexadecimal number, Program for decimal to hexadecimal conversion, Converting a Real Number (between 0 and 1) to Binary String, Convert from any base to decimal and vice versa, Decimal to binary conversion without using arithmetic operators, Introduction to Primality Test and School Method, Efficient program to print all prime factors of a given number, Pollards Rho Algorithm for Prime Factorization, Find numbers with n-divisors in a given range, Modular Exponentiation (Power in Modular Arithmetic), Eulers criterion (Check if square root under modulo p exists), Find sum of modulo K of first N natural number, Exponential Squaring (Fast Modulo Multiplication), Trick for modular division ( (x1 * x2 . This article is contributed by Karun Anantharaman. 2. You will be notified via email once the article is available for improvement. Input: n = 100 1.) GFG Weekly Coding Contest. To solve this problem, we can use the fact that the product of all the integers in the array will give us X^2. 100000, Input: N = 5Output:1 >12 >1, 23 >1, 34 >1, 2, 45 >1, 5. The function uses a vector to store the divisors of x, which has a maximum size of sqrt(x). 1. Practice Given a positive integer 'n' and query 'q'. Approach: This is the basic approach. 3. #gfg Thanks for contributing an answer to Stack Overflow! By using our site, you According to this post, we can get all divisors of a number through the following codes. Given three integers A, B, C, the task is to findAi=1 Bj=1Ck=1 d(i.j.k), where d(x) is the number of divisors of x. The task is to find all the common divisors of all N integers. rev2023.7.14.43533. GCD of all the numbers is 6. #frequencycount The time complexity is O(N * log(N)) as the sum of harmonic series 1 + 1/2 + 1/3 + + 1/N can be approximated to log(N). Find centralized, trusted content and collaborate around the technologies you use most. acknowledge that you have read and understood our. How to insert ArrayList into While-condition? Figure all combinations of two integers of a product, Efficiently finding all divisors of a number. #lecture The remaining number can only have a maximum of 2 prime factors. The set of a number's divisors consists of all natural numbers that divide it evenly leaving no remainder. Find sum of inverse of the divisors when sum of divisors and the number is given. Finally, the divisor of number 6 is the number itself. Saved by Job-a-Thon. Sum of divisors | Practice | GeeksforGeeks While your answer may help to achieve what the original poster wants, your post offers no guidance or explanation that may help others understand/learn what exactly you are doing and why. How Does Military Budgeting Work? See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. Can we improve the above solution? A proper divisor of a natural number is the divisor that is strictly less than the number. Follow the below steps to solve the problem: Below is the implementation of the above approach: Time Complexity: O(N*log(log(N)))Auxiliary Space: O(N), Run a loop from 2 to sqrt(N) and check if the current element is prime or not, if it is so then print that number, but this method will increase the time complexity of the solution, Time Complexity: O(sqrt N2)Auxiliary Space: O(1). This article is being improved by another user right now. the output is not in a sorted fashion which we had got using the brute-force technique. (C++). If X is divisible by all integers in the array, we return X. As we know the possible candidates, we iterate upon all the candidates and check whether they divide the actual number. Code-only answers are discouraged. #howto But since all primes less than N have been absorbed by X, this is not possible.