Given four integers a, b, c, and N. The task is to find the Nth term which is divisible by either of a, b or c.
Examples:
Input: a = 2, b = 3, c = 5, N = 10
Output: 14
Sequence is 2, 3, 4, 5, 6, 8, 9, 10, 12, 14, 15, 16…
Input: a = 3, b = 5, c = 7, N = 10
Output: 18
Naive Approach: A simple approach is to traverse over all the terms starting from 1 until we find the desired Nth term which is divisible by either a, b or c. This solution has time complexity of O(N).
Efficient Approach: The idea is to use Binary search. Here we can calculate how many numbers from 1 to num are divisible by either a, b or c by using the formula: (num / a) + (num / b) + (num / c) – (num / lcm(a, b)) – (num / lcm(b, c)) – (num / lcm(a, c)) + (num / lcm(a, b, c))
The above formula is derived using set theory
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return// gcd of a and bint gcd(int a, int b){ if (a == 0) return b; return gcd(b % a, a);}// Function to return the lcm of a and bint lcm(int a, int b){ return (a * b) / gcd(a, b);}// Function to return the count of numbers// from 1 to num which are divisible by a, b or cint divTermCount(int a, int b, int c, int num){ // Calculate number of terms divisible by a and // by b and by c then, remove the terms which is are // divisible by both a and b, both b and c, both // c and a and then add which are divisible by a and // b and c return ((num / a) + (num / b) + (num / c) - (num / lcm(a, b)) - (num / lcm(b, c)) - (num / lcm(a, c)) + (num / lcm(a, lcm(b, c))));}// Function to find the nth term// divisible by a, b or c// by using binary searchint findNthTerm(int a, int b, int c, int n){ // Set low to 1 and high to max (a, b, c) * n int low = 1, high = INT_MAX, mid; while (low < high) { mid = low + (high - low) / 2; // If the current term is less than // n then we need to increase low // to mid + 1 if (divTermCount(a, b, c, mid) < n) low = mid + 1; // If current term is greater than equal to // n then high = mid else high = mid; } return low;}// Driver codeint main(){ int a = 2, b = 3, c = 5, n = 10; cout << findNthTerm(a, b, c, n); return 0;} |
Java
// Java implementation of the approach class GFG{ // Function to return // gcd of a and b static int gcd(int a, int b) { if (a == 0) return b; return gcd(b % a, a); } // Function to return the lcm of a and b static int lcm(int a, int b) { return (a * b) / gcd(a, b); } // Function to return the count of numbers // from 1 to num which are divisible by a, b or c static int divTermCount(int a, int b, int c, int num) { // Calculate number of terms divisible by a and // by b and by c then, remove the terms which is are // divisible by both a and b, both b and c, both // c and a and then add which are divisible by a and // b and c return ((num / a) + (num / b) + (num / c) - (num / lcm(a, b)) - (num / lcm(b, c)) - (num / lcm(a, c)) + (num / lcm(a, lcm(b, c)))); } // Function to find the nth term // divisible by a, b or c // by using binary search static int findNthTerm(int a, int b, int c, int n) { // Set low to 1 and high to max (a, b, c) * n int low = 1, high = Integer.MAX_VALUE, mid; while (low < high) { mid = low + (high - low) / 2; // If the current term is less than // n then we need to increase low // to mid + 1 if (divTermCount(a, b, c, mid) < n) low = mid + 1; // If current term is greater than equal to // n then high = mid else high = mid; } return low; } // Driver code public static void main(String[] args) { int a = 2, b = 3, c = 5, n = 10; System.out.println(findNthTerm(a, b, c, n)); } }// This code is contributed by// Rajnis09 |
Python
# Python3 implementation of the approach# Function to return# gcd of a and bdef gcd(a, b): if (a == 0): return b return gcd(b % a, a)# Function to return the lcm of a and bdef lcm(a, b): return ((a * b) // gcd(a, b))# Function to return the count of numbers# from 1 to num which are divisible by a, b or cdef divTermCount(a, b, c, num): # Calculate number of terms divisible by a and # by b and by c then, remove the terms which is are # divisible by both a and b, both b and c, both # c and a and then add which are divisible by a and # b and c return ((num // a) + (num // b) + (num // c) - (num // lcm(a, b)) - (num // lcm(b, c)) - (num // lcm(a, c)) + (num // lcm(a, lcm(b, c))))# Function to find the nth term# divisible by a, b or c# by using binary searchdef findNthTerm(a, b, c, n): # Set low to 1 and high to max (a, b, c) * n low = 1 high = 10**9 mid=0 while (low < high): mid = low + (high - low) // 2 # If the current term is less than # n then we need to increase low # to mid + 1 if (divTermCount(a, b, c, mid) < n): low = mid + 1 # If current term is greater than equal to # n then high = mid else: high = mid return low# Driver codea = 2b = 3c = 5n = 10print(findNthTerm(a, b, c, n))# This code is contributed by mohit kumar 29 |
C#
// C# implementation of the approach using System; class GFG{ // Function to return // gcd of a and b static int gcd(int a, int b) { if (a == 0) return b; return gcd(b % a, a); } // Function to return the lcm of a and b static int lcm(int a, int b) { return (a * b) / gcd(a, b); } // Function to return the count of numbers // from 1 to num which are divisible by a, b or c static int divTermCount(int a, int b, int c, int num) { // Calculate number of terms divisible by a and // by b and by c then, remove the terms which is are // divisible by both a and b, both b and c, both // c and a and then add which are divisible by a and // b and c return ((num / a) + (num / b) + (num / c) - (num / lcm(a, b)) - (num / lcm(b, c)) - (num / lcm(a, c)) + (num / lcm(a, lcm(b, c)))); } // Function to find the nth term // divisible by a, b or c // by using binary search static int findNthTerm(int a, int b, int c, int n) { // Set low to 1 and high to max (a, b, c) * n int low = 1, high = int.MaxValue, mid; while (low < high) { mid = low + (high - low) / 2; // If the current term is less than // n then we need to increase low // to mid + 1 if (divTermCount(a, b, c, mid) < n) low = mid + 1; // If current term is greater than equal to // n then high = mid else high = mid; } return low; } // Driver code public static void Main(String[] args) { int a = 2, b = 3, c = 5, n = 10; Console.WriteLine(findNthTerm(a, b, c, n)); } }/* This code is contributed by PrinciRaj1992 */ |
Javascript
<script>// Javascript implementation of the approach// Function to return// gcd of a and bfunction gcd(a, b){ if (a == 0) return b; return gcd(b % a, a);}// Function to return the lcm of a and bfunction lcm(a, b){ return parseInt((a * b) / gcd(a, b));}// Function to return the count of numbers// from 1 to num which are divisible by a, b or cfunction divTermCount(a, b, c, num){ // Calculate number of terms divisible by a and // by b and by c then, remove the terms which is are // divisible by both a and b, both b and c, both // c and a and then add which are divisible by a and // b and c return (parseInt(num / a) + parseInt(num / b) + parseInt(num / c) - parseInt(num / lcm(a, b)) - parseInt(num / lcm(b, c)) - parseInt(num / lcm(a, c)) + parseInt(num / lcm(a, lcm(b, c))));}// Function to find the nth term// divisible by a, b or c// by using binary searchfunction findNthTerm(a, b, c, n){ // Set low to 1 and high to max (a, b, c) * n let low = 1, high = Number.MAX_VALUE, mid; while (low < high) { mid = low + parseInt((high - low) / 2); // If the current term is less than // n then we need to increase low // to mid + 1 if (divTermCount(a, b, c, mid) < n) low = mid + 1; // If current term is greater than equal to // n then high = mid else high = mid; } return low;}// Driver code let a = 2, b = 3, c = 5, n = 10; document.write(findNthTerm(a, b, c, n));</script> |
Output:
14
Time Complexity: O(log n * log(min(a, b))), where a and b are two parameters for GCD.
Auxiliary Space: O(log(min(a, b)))
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