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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 <bits/stdc++.h>
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<ll> 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<Long> vc = new HashSet<Long>();
 
    // 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




<script>
 
// Javascript implementation of the approach
 
// Function to find all the distinct
// remainders when n is divided by
// all the elements from
// the range [1, n + 1]
function findRemainders(n)
{
     
    // Set will be used to store
    // the remainders in order
    // to eliminate duplicates
    var vc = new Set();
 
    // Find the remainders
    for(var i = 1; i <= Math.ceil(Math.sqrt(n)); i++)
        vc.add(parseInt(n / i));
    for(var i = parseInt(n / Math.ceil(Math.sqrt(n))) - 1;
            i >= 0; i--)
        vc.add(i);
 
    // Print the contents of the set
    [...vc].sort((a, b) => a - b).forEach(it => {
        document.write(it + " ");
    });
}
 
// Driver code
var n = 5;
 
findRemainders(n);
 
// This code is contributed by famously
 
</script>
Output: 
0 1 2 5

 




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