Find the number of pairs (a, b) such that a % b = K
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++
#include <bits/stdc++.h>
using namespace std;
int CountAllPairs( int N, int K)
{
int count = 0;
if (N > K) {
count = N - K;
for ( int i = K + 1; i <= N; i++)
count = count + ((N - K) / i);
}
return count;
}
int main()
{
int N = 11, K = 5;
cout << CountAllPairs(N, K);
return 0;
}
|
Java
import java.io.*;
class GFG {
static int CountAllPairs( int N, int K)
{
int count = 0 ;
if (N > K) {
count = N - K;
for ( int i = K + 1 ; i <= N; i++)
count = count + ((N - K) / i);
}
return count;
}
public static void main(String[] args)
{
int N = 11 , K = 5 ;
System.out.println(CountAllPairs(N, K));
}
}
|
Python3
import math
def CountAllPairs(N, K):
count = 0
if ( N > K):
count = N - K
for i in range (K + 1 , N + 1 ):
count = count + ((N - K) / / i)
return count
N = 11
K = 5
print (CountAllPairs(N, K))
|
C#
using System;
class GFG {
static int CountAllPairs( int N, int K)
{
int count = 0;
if (N > K) {
count = N - K;
for ( int i = K + 1; i <= N; i++)
count = count + ((N - K) / i);
}
return count;
}
public static void Main()
{
int N = 11, K = 5;
Console.WriteLine(CountAllPairs(N, K));
}
}
|
PHP
<?php
function CountAllPairs( $N , $K )
{
$count = 0;
if ( $N > $K ){
$count = $N - $K ;
for ( $i = $K +1; $i <= $N ; $i ++)
{
$x = ((( $N - $K ) / $i ));
$count = $count + (int)( $x );
}
}
return $count ;
}
$N = 11;
$K = 5;
echo (CountAllPairs( $N , $K ));
?>
|
Javascript
<script>
function CountAllPairs( N, K)
{
let count = 0;
if (N > K) {
count = N - K;
for (let i = K + 1; i <= N; i++)
count = count + parseInt((N - K) / i);
}
return count;
}
let N = 11, K = 5;
document.write(CountAllPairs(N, K));
</script>
|
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.
Last Updated :
07 Dec, 2022
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