Count integers in the range [A, B] that are not divisible by C and D

0 0 4 minutes read

Given four integers A, B, C and D. The task is to find the count of integers in the range [A, B] that are not divisible by C and D .
Examples:

Input: A = 4, B = 9, C = 2, D = 3
Output: 2
5 and 7 are such integers.

Input: A = 10, B = 50, C = 4, D = 6
Output: 28

Approach: First include all the integers in the range in the required answer i.e. B – A + 1. Then remove all the numbers which are divisible by C and D and finally add all the numbers which are divisible by both C and D.
Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the count of
// integers from the range [a, b] that
// are not divisible by c and d
int countNums(int a, int b, int c, int d)
{
    // Numbers which are divisible by c
    int x = b / c - (a - 1) / c;
 
    // Numbers which are divisible by d
    int y = b / d - (a - 1) / d;
 
    // Find lowest common factor of c and d
    int k = (c * d) / __gcd(c, d);
 
    // Numbers which are divisible by both c and d
    int z = b / k - (a - 1) / k;
 
    // Return the required answer
    return b - a + 1 - x - y + z;
}
 
// Driver code
int main()
{
    int a = 10, b = 50, c = 4, d = 6;
 
    cout << countNums(a, b, c, d);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
 
// Function to return the count of
// integers from the range [a, b] that
// are not divisible by c and d
static int countNums(int a, int b, int c, int d)
{
    // Numbers which are divisible by c
    int x = b / c - (a - 1) / c;
 
    // Numbers which are divisible by d
    int y = b / d - (a - 1) / d;
 
    // Find lowest common factor of c and d
    int k = (c * d) / __gcd(c, d);
 
    // Numbers which are divisible by both c and d
    int z = b / k - (a - 1) / k;
 
    // Return the required answer
    return b - a + 1 - x - y + z;
}
static int __gcd(int a, int b) 
{ 
    if (b == 0) 
        return a; 
    return __gcd(b, a % b);     
}
 
// Driver code
public static void main(String []args) 
{
    int a = 10, b = 50, c = 4, d = 6;
 
    System.out.println(countNums(a, b, c, d));
}
}
 
// This code is contributed by PrinciRaj1992

Python3

# Python3 implementation of the approach
from math import gcd
 
# Function to return the count of 
# integers from the range [a, b] that 
# are not divisible by c and d 
def countNums(a, b, c, d) :
 
    # Numbers which are divisible by c 
    x = b // c - (a - 1) // c; 
 
    # Numbers which are divisible by d 
    y = b // d - (a - 1) // d; 
 
    # Find lowest common factor of c and d 
    k = (c * d) // gcd(c, d); 
 
    # Numbers which are divisible 
    # by both c and d 
    z = b // k - (a - 1) // k; 
 
    # Return the required answer 
    return (b - a + 1 - x - y + z); 
 
# Driver code 
if __name__ == "__main__" : 
 
    a = 10; b = 50; c = 4; d = 6; 
 
    print(countNums(a, b, c, d)); 
 
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
     
class GFG
{
 
// Function to return the count of
// integers from the range [a, b] that
// are not divisible by c and d
static int countNums(int a, int b, int c, int d)
{
    // Numbers which are divisible by c
    int x = b / c - (a - 1) / c;
 
    // Numbers which are divisible by d
    int y = b / d - (a - 1) / d;
 
    // Find lowest common factor of c and d
    int k = (c * d) / __gcd(c, d);
 
    // Numbers which are divisible by both c and d
    int z = b / k - (a - 1) / k;
 
    // Return the required answer
    return b - a + 1 - x - y + z;
}
 
static int __gcd(int a, int b) 
{ 
    if (b == 0) 
        return a; 
    return __gcd(b, a % b);     
}
 
// Driver code
public static void Main(String []args) 
{
    int a = 10, b = 50, c = 4, d = 6;
 
    Console.WriteLine(countNums(a, b, c, d));
}
}
 
// This code is contributed by Rajput-Ji

Javascript

<script>
 
// Javascript implementation of the approach
 
// Function to return the count of
// integers from the range [a, b] that
// are not divisible by c and d
function countNums(a, b, c, d)
{
    // Numbers which are divisible by c
    let x = parseInt(b / c) - parseInt((a - 1) / c);
 
    // Numbers which are divisible by d
    let y = parseInt(b / d) - parseInt((a - 1) / d);
 
    // Find lowest common factor of c and d
    let k = parseInt((c * d) / __gcd(c, d));
 
    // Numbers which are divisible by both c and d
    let z = parseInt(b / k) - parseInt((a - 1) / k);
 
    // Return the required answer
    return b - a + 1 - x - y + z;
}
 
function __gcd(a, b) 
{ 
    if (b == 0) 
        return a; 
    return __gcd(b, a % b);     
}
 
// Driver code
    let a = 10, b = 50, c = 4, d = 6;
 
    document.write(countNums(a, b, c, d));
 
</script>

Output:

Time Complexity: O(log(min(c, d)), where c and d are the given inputs.
Auxiliary Space: O(1), no extra space is required, so it is a constant.

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!