Given two integers N and K where N, K > 0, the task is to find the total number of pairs (a, b) where 1 ? a, b ? N such that a % b = K.
Examples:
Input: N = 4, K = 2
Output: 2
Only valid pairs are (2, 3) and (2, 4).Input: N = 11, K = 5
Output: 7
Naive approach: Run two loops from 1 to n and count all the pairs (i, j) where i % j = K. The time complexity of this approach will be O(n2).
Efficient approach: Initially total count = N – K because all the numbers from the range which are > K will give K as the remainder after dividing it. After that, for all i > K there are exactly (N – K) / i numbers which will give the remainder as K after getting divided by i.
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to return the count // of required pairs int CountAllPairs( int N, int K)
{ int count = 0;
if (N > K) {
// Initial count
count = N - K;
for ( int i = K + 1; i <= N; i++)
count = count + ((N - K) / i);
}
return count;
} // Driver code int main()
{ int N = 11, K = 5;
cout << CountAllPairs(N, K);
return 0;
} |
// Java implementation of the approach import java.io.*;
class GFG {
// Function to return the count
// of required pairs
static int CountAllPairs( int N, int K)
{
int count = 0 ;
if (N > K) {
// Initial count
count = N - K;
for ( int i = K + 1 ; i <= N; i++)
count = count + ((N - K) / i);
}
return count;
}
// Driver code
public static void main(String[] args)
{
int N = 11 , K = 5 ;
System.out.println(CountAllPairs(N, K));
}
} |
# Python3 implementation of the approach import math
# Function to return the count # of required pairs def CountAllPairs(N, K):
count = 0
if ( N > K):
# Initial count
count = N - K
for i in range (K + 1 , N + 1 ):
count = count + ((N - K) / / i)
return count
# Driver code N = 11
K = 5
print (CountAllPairs(N, K))
|
// C# implementation of the approach using System;
class GFG {
// Function to return the count
// of required pairs
static int CountAllPairs( int N, int K)
{
int count = 0;
if (N > K) {
// Initial count
count = N - K;
for ( int i = K + 1; i <= N; i++)
count = count + ((N - K) / i);
}
return count;
}
// Driver code
public static void Main()
{
int N = 11, K = 5;
Console.WriteLine(CountAllPairs(N, K));
}
} |
<?php // PHP implementation of the approach // Function to return the count // of required pairs function CountAllPairs( $N , $K )
{ $count = 0;
if ( $N > $K ){
// Initial count
$count = $N - $K ;
for ( $i = $K +1; $i <= $N ; $i ++)
{
$x = ((( $N - $K ) / $i ));
$count = $count + (int)( $x );
}
}
return $count ;
} // Driver code
$N = 11;
$K = 5;
echo (CountAllPairs( $N , $K ));
?> |
<script> // JavaScript implementation of the approach // Function to return the count
// of required pairs
function CountAllPairs( N, K)
{
let count = 0;
if (N > K) {
// Initial count
count = N - K;
for (let i = K + 1; i <= N; i++)
count = count + parseInt((N - K) / i);
}
return count;
}
// Driver code
let N = 11, K = 5;
document.write(CountAllPairs(N, K));
//contributed by sravan (171fa07058) </script> |
7
Time Complexity: O(N – K), where N and K are the given inputs.
Auxiliary Space: O(1), no extra space is required, so it is a constant.