Given a function f(K) = floor(N/K) (N>0 and K>0), the task is to find all possible values of f(K) for a given N where K takes all values in the range [1, Inf].
Examples:
Input: N = 5
Output: 0 1 2 5
Explanation:
5 divide 1 = 5
5 divide 2 = 2
5 divide 3 = 1
5 divide 4 = 1
5 divide 5 = 1
5 divide 6 = 0
5 divide 7 = 0
So all possible distinct values of f(k) are {0, 1, 2, 5}.
Input: N = 11
Output: 0 1 2 3 5 11
Explanation:
11 divide 1 = 11
11 divide 2 = 5
11 divide 3 = 3
11 divide 4 = 2
11 divide 5 = 2
11 divide 6 = 1
11 divide 7 = 1
…
…
11 divided 11 = 1
11 divides 12 = 0
So all possible distinct values of f(k) are {0, 1, 2, 3, 5, 11}.
Naive Approach:
The simplest approach to the iterate over [1, N+1] and store in a set, all values of (N/i) ( 1 ? i ? N + 1) to avoid duplication.
Below is the implementation of the above approach:
C++
// C++ Program for the // above approach #include <bits/stdc++.h> using namespace std; // Function to print all // possible values of // floor(N/K) void allQuotients(int N) { set<int> s; // loop from 1 to N+1 for (int k = 1; k <= N + 1; k++) { s.insert(N / k); } for (auto it : s) cout << it << " "; } int main() { int N = 5; allQuotients(N); return 0; } |
Java
// Java program for the above approach import java.util.*; class GFG{ // Function to print all // possible values of // Math.floor(N/K) static void allQuotients(int N) { HashSet<Integer> s = new HashSet<Integer>(); // loop from 1 to N+1 for(int k = 1; k <= N + 1; k++) { s.add(N / k); } for(int it : s) System.out.print(it + " "); } // Driver code public static void main(String[] args) { int N = 5; allQuotients(N); } } // This code is contributed by Rajput-Ji |
Python3
# Python3 program for the above approach # Function to print all possible # values of floor(N/K) def allQuotients(N): s = set() # Iterate from 1 to N+1 for k in range(1, N + 2): s.add(N // k) for it in s: print(it, end = ' ') # Driver code if __name__ == '__main__': N = 5 allQuotients(N) # This code is contributed by himanshu77 |
C#
// C# program for the above approach using System; using System.Collections.Generic; class GFG{ // Function to print all possible // values of Math.floor(N/K) static void allQuotients(int N) { SortedSet<int> s = new SortedSet<int>(); // Loop from 1 to N+1 for(int k = 1; k <= N + 1; k++) { s.Add(N / k); } foreach(int it in s) { Console.Write(it + " "); } } // Driver code static void Main() { int N = 5; allQuotients(N); } } // This code is contributed by divyeshrabadiya07 |
0 1 2 5
Time Complexity: O(N)
Auxiliary Space: O(N)
Efficient Approach:
An optimized solution is to iterate over [1, ?N] and insert values K and (N/K) into the set.
C++
// C++ Program for the // above approach #include <bits/stdc++.h> using namespace std; // Function to print all // possible values of // floor(N/K) void allQuotients(int N) { set<int> s; s.insert(0); for (int k = 1; k <= sqrt(N); k++) { s.insert(k); s.insert(N / k); } for (auto it : s) cout << it << " "; } int main() { int N = 5; allQuotients(N); return 0; } |
Java
// Java program for the above approach import java.util.*; class GFG{ // Function to print all // possible values of // Math.floor(N/K) static void allQuotients(int N) { HashSet<Integer> s = new HashSet<Integer>(); s.add(0); // loop from 1 to N+1 for(int k = 1; k <= Math.sqrt(N); k++) { s.add(k); s.add(N / k); } for(int it : s) System.out.print(it + " "); } // Driver code public static void main(String[] args) { int N = 5; allQuotients(N); } } // This code is contributed by rock_cool |
Python3
# Python3 program for the above approach from math import * # Function to print all possible # values of floor(N/K) def allQuotients(N): s = set() s.add(0) for k in range(1, int(sqrt(N)) + 1): s.add(k) s.add(N // k) for it in s: print(it, end = ' ') # Driver code if __name__ == '__main__': N = 5 allQuotients(N) # This code is contributed by himanshu77 |
C#
// C# program for the above approach using System; using System.Collections.Generic; class GFG{ // Function to print all possible // values of Math.floor(N/K) static void allQuotients(int N) { SortedSet<int> s = new SortedSet<int>(); s.Add(0); // loop from 1 to N+1 for(int k = 1; k <= Math.Sqrt(N); k++) { s.Add(k); s.Add(N / k); } foreach(int it in s) { Console.Write(it + " "); } } // Driver code static void Main() { int N = 5; allQuotients(N); } } // This code is contributed by divyeshrabadiya07 |
0 1 2 5
Time Complexity: O(?N)
Auxiliary Space: O(N)
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