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Count of pairs from first N natural numbers with remainder at least K

  • Difficulty Level : Hard
  • Last Updated : 06 Aug, 2021

Given two positive integers N and K, the task is to find the number of pairs (a, b) over the range [1, N] such that a%b is at least K.

Examples:

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Input: N = 5, K = 2
Output: 7
Explanation: 
Following are the all possible pairs satisfying the given criteria:



  1. (2, 3): The value of 2%3 = 2(>= K).
  2. (5, 3): The value of 5%3 = 2(>= K).
  3. (2, 4): The value of 2%4 = 2(>= K).
  4. (3, 4): The value of 3%4 = 3(>= K).
  5. (2, 5): The value of 2%5 = 2(>= K).
  6. (3, 5): The value of 3%5 = 3(>= K).
  7. (4, 5): The value of 4%5 = 4(>= K).

Therefore, the total count of pairs is 7.

Input: N = 6, K = 0
Output: 36

Naive Approach: The simplest approach to solve the given problem is to generate all possible pairs (a, b) over the range [1, N] and if the value of a%b is at least K, then count this pair. After checking for all the pairs, print the total pairs obtained.

Time Complexity: O(N2)
Auxiliary Space: O(1)

Efficient Approach: The above approach can also be optimized by iterating over the range [1, N] and fix the second number in the pair, i.e., b. For each fixed b there will be a period of N/b and every period can be combiner with (b – K) elements. So, a total of (N/b)*(b – K) elements will be there. Now for the remaining elements that are N%b there will be max(0, n%b – k + 1) pairs. Follow the steps below to solve the problem:

  • If the value of K is 0, then print N2 as the resultant number of valid pairs.
  • Initialize the variable, say ans as 0 that stores the resultant count of pairs.
  • Iterate over the range [K + 1, N] using the variable b and perform the following steps:
    • Add the value of (N/b)*(b – K) to the variable ans.
    • Add the value of the maximum of (N % b – K + 1) or 0 to the variable ans.
  • After performing the above steps, print the value of ans as the resultant count of pairs.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to count the number of pairs
// (a, b) such that a%b is at least K
int countTotalPairs(int N, int K)
{
    // Base Case
    if (K == 0) {
        return N * N;
    }
 
    // Stores resultant count of pairs
    int ans = 0;
 
    // Iterate over the range [K + 1, N]
    for (int b = K + 1; b <= N; b++) {
 
        // Find the cycled elements
        ans += (N / b) * (b - K);
 
        // Find the remaining elements
        ans += max(N % b - K + 1, 0);
    }
 
    // Return the resultant possible
    // count of pairs
    return ans;
}
 
// Driver Code
int main()
{
    int N = 5, K = 2;
    cout << countTotalPairs(N, K);
 
    return 0;
}

Java




/*package whatever //do not write package name here */
import java.io.*;
 
class GFG {
 
  // Function to count the number of pairs
  // (a, b) such that a%b is at least K
  public static int countTotalPairs(int N, int K)
  {
 
    // Base case
    if (K == 0) {
      return N * N;
    }
 
    // Stores resultant count of pairs
    int ans = 0;
 
    // Iterate over the range [K + 1, N]
    for (int i = K + 1; i <= N; i++)
    {
 
      // Find the cycled element
      ans += (N / i) * (i - K);
      if ((N % i) - K + 1 > 0)
      {
 
        // Find the remaining element
        ans += (N % i) - K + 1;
      }
    }
 
    // Return the resultant possible
    // count of pairs
    return ans;
  }
 
  // Driver code
  public static void main(String[] args)
  {
    int N = 5, K = 2;
    System.out.println(countTotalPairs(N, K));
  }
}
 
// This code is contributed by maddler.

Python3




# Python program for the above approach;
 
# Function to count the number of pairs
# (a, b) such that a%b is at least K
def countTotalPairs(N, K):
    # Base Case
    if (K == 0) :
        return N * N
     
 
    # Stores resultant count of pairs
    ans = 0
 
    # Iterate over the range [K + 1, N]
    for b in range(K + 1, N + 1) :
 
        # Find the cycled elements
        ans += (N // b) * (b - K)
 
        # Find the remaining elements
        ans += max(N % b - K + 1, 0)
     
 
    # Return the resultant possible
    # count of pairs
    return ans
 
 
# Driver Code
 
N = 5
K = 2
print(countTotalPairs(N, K))
 
# This code is contributed by _saurabh_jaiswal

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to count the number of pairs
// (a, b) such that a%b is at least K
static int countTotalPairs(int N, int K)
{
    // Base Case
    if (K == 0) {
        return N * N;
    }
 
    // Stores resultant count of pairs
    int ans = 0;
 
    // Iterate over the range [K + 1, N]
    for (int b = K + 1; b <= N; b++) {
 
        // Find the cycled elements
        ans += (N / b) * (b - K);
 
        // Find the remaining elements
        ans += Math.Max(N % b - K + 1, 0);
    }
 
    // Return the resultant possible
    // count of pairs
    return ans;
}
 
// Driver Code
public static void Main()
{
    int N = 5, K = 2;
    Console.Write(countTotalPairs(N, K));
}
}
 
// This code is contributed by ipg2016107.

Javascript




<script>
 
        // JavaScript program for the above approach;
 
        // Function to count the number of pairs
        // (a, b) such that a%b is at least K
        function countTotalPairs(N, K) {
            // Base Case
            if (K == 0) {
                return N * N;
            }
 
            // Stores resultant count of pairs
            let ans = 0;
 
            // Iterate over the range [K + 1, N]
            for (let b = K + 1; b <= N; b++) {
 
                // Find the cycled elements
                ans += Math.floor((N / b) * (b - K));
 
                // Find the remaining elements
                ans += Math.max(N % b - K + 1, 0);
            }
 
            // Return the resultant possible
            // count of pairs
            return ans;
        }
 
        // Driver Code
 
        let N = 5, K = 2;
        document.write(countTotalPairs(N, K));
 
   // This code is contributed by Potta Lokesh
    </script>
Output: 
7

 

Time Complexity: O(N)
Auxiliary Space: O(1)




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