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Find all the possible remainders when N is divided by all positive integers from 1 to N+1
• Difficulty Level : Hard
• Last Updated : 02 Jun, 2021

Given a large integer N, the task is to find all the possible remainders when N is divided by all the positive integers from 1 to N + 1.

Examples:

Input: N = 5
Output: 0 1 2 5
5 % 1 = 0
5 % 2 = 1
5 % 3 = 2
5 % 4 = 1
5 % 5 = 0
5 % 6 = 5

Input: N = 11
Output: 0 1 2 3 5 11

Naive approach: Run a loop from 1 to N + 1 and return all the unique remainders found when dividing N by any integer from the range. But this approach is not efficient for larger values of N.

Efficient approach: It can be observed that one part of the answer will always contain numbers between 0 to ceil(sqrt(n)). It can be proven by running the naive algorithm on smaller values of N and checking the remainders obtained or by solving the equation ceil(N / k) = x or x ≤ (N / k) < x + 1 where x is one of the remainders for all integers k when N is divided by k for k from 1 to N + 1
The solution to the above inequality is nothing but integers k from (N / (x + 1), N / x] of length N / x – N / (x + 1) = N / (x2 + x). Therefore, iterate from k = 1 to ceil(sqrt(N)) and store all the unique N % k. What if the above k is greater than ceil(sqrt(N))? They will always correspond to values 0 ≤ x < ceil(sqrt(N)). So, again start storing remainders from N / (ceil(sqrt(N)) – 1 to 0 and return the final answer with all the possible remainders.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `typedef` `long` `long` `int` `ll;` `// Function to find all the distinct``// remainders when n is divided by``// all the elements from``// the range [1, n + 1]``void` `findRemainders(ll n)``{` `    ``// Set will be used to store``    ``// the remainders in order``    ``// to eliminate duplicates``    ``set vc;` `    ``// Find the remainders``    ``for` `(ll i = 1; i <= ``ceil``(``sqrt``(n)); i++)``        ``vc.insert(n / i);``    ``for` `(ll i = n / ``ceil``(``sqrt``(n)) - 1; i >= 0; i--)``        ``vc.insert(i);` `    ``// Print the contents of the set``    ``for` `(``auto` `it : vc)``        ``cout << it << ``" "``;``}` `// Driver code``int` `main()``{``    ``ll n = 5;` `    ``findRemainders(n);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.util.*;` `class` `GFG``{` `// Function to find all the distinct``// remainders when n is divided by``// all the elements from``// the range [1, n + 1]``static` `void` `findRemainders(``long` `n)``{` `    ``// Set will be used to store``    ``// the remainders in order``    ``// to eliminate duplicates``    ``HashSet vc = ``new` `HashSet();` `    ``// Find the remainders``    ``for` `(``long` `i = ``1``; i <= Math.ceil(Math.sqrt(n)); i++)``        ``vc.add(n / i);``    ``for` `(``long` `i = (``long``) (n / Math.ceil(Math.sqrt(n)) - ``1``);``                                                ``i >= ``0``; i--)``        ``vc.add(i);` `    ``// Print the contents of the set``    ``for` `(``long` `it : vc)``        ``System.out.print(it+ ``" "``);``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``long` `n = ``5``;` `    ``findRemainders(n);``}``}` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 implementation of the approach``from` `math ``import` `ceil, floor, sqrt` `# Function to find all the distinct``# remainders when n is divided by``# all the elements from``# the range [1, n + 1]``def` `findRemainders(n):` `    ``# Set will be used to store``    ``# the remainders in order``    ``# to eliminate duplicates``    ``vc ``=` `dict``()` `    ``# Find the remainders``    ``for` `i ``in` `range``(``1``, ceil(sqrt(n)) ``+` `1``):``        ``vc[n ``/``/` `i] ``=` `1``    ``for` `i ``in` `range``(n ``/``/` `ceil(sqrt(n)) ``-` `1``, ``-``1``, ``-``1``):``        ``vc[i] ``=` `1` `    ``# Print the contents of the set``    ``for` `it ``in` `sorted``(vc):``        ``print``(it, end ``=` `" "``)` `# Driver code``n ``=` `5` `findRemainders(n)` `# This code is contributed by Mohit Kumar`

## C#

 `// C# implementation of the approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{` `// Function to find all the distinct``// remainders when n is divided by``// all the elements from``// the range [1, n + 1]``static` `void` `findRemainders(``long` `n)``{` `    ``// Set will be used to store``    ``// the remainders in order``    ``// to eliminate duplicates``    ``List<``long``> vc = ``new` `List<``long``>();` `    ``// Find the remainders` `    ``for` `(``long` `i = 1; i <= Math.Ceiling(Math.Sqrt(n)); i++)``        ``vc.Add(n / i);``    ``for` `(``long` `i = (``long``) (n / Math.Ceiling(Math.Sqrt(n)) - 1);``                                                 ``i >= 0; i--)``        ``vc.Add(i);``    ``vc.Reverse();``    ` `    ``// Print the contents of the set``    ``foreach` `(``long` `it ``in` `vc)``        ``Console.Write(it + ``" "``);``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``long` `n = 5;` `    ``findRemainders(n);``}``}` `// This code is contributed by PrinciRaj1992`

## Javascript

 ``
Output:
`0 1 2 5`

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