Count possible values of K such that X – 1 and Y – 1 modulo K is same
Given two numbers X and Y, the task is to find the count of integers K which satisfies the equation (X – 1) % K = (Y – 1) % K.
Examples:
Input: X = 2, Y = 6
Output: 3
Explanation:
K = {1, 2, 4} satisfies the given equation. Therefore, the count is 3.Input: X = 4, Y = 9
Output: 2
Approach: Follow the steps below to solve the problem:
- The idea is to find the absolute difference of X and Y.
- Find the count of divisors of the calculated absolute difference and print the count as the required answer.
Below is the implementation of the above approach:
C++14
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to count integers K// satisfying given equationint condition(int a, int b){ // Calculate the absoluter // difference between a and b int d = abs(a - b), count = 0; // Iterate till sqrt of the diffeence for (int i = 1; i <= sqrt(d); i++) { if (d % i == 0) { if (d / i == i) count += 1; else count += 2; } } // Return the count return count;}// Driver Codeint main(){ int x = 2, y = 6; cout << condition(x, y) << endl; return 0;} |
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Java
// Java program for the above approachimport java.io.*;class GFG{ // Function to count integers K// satisfying given equationstatic int condition(int a, int b){ // Calculate the absoluter // difference between a and b int d = Math.abs(a - b), count = 0; // Iterate till sqrt of the diffeence for(int i = 1; i <= Math.sqrt(d); i++) { if (d % i == 0) { if (d / i == i) count += 1; else count += 2; } } // Return the count return count;} // Driver Codepublic static void main (String[] args){ int x = 2, y = 6; System.out.println(condition(x, y));}}// This code is contributed by sanjoy_62 |
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Python3
# Python3 program for the above approach# Function to count integers K# satisfying given equationdef condition(a, b): # Calculate the absoluter # difference between a and b d = abs(a - b) count = 0 # Iterate till sqrt of the diffeence for i in range(1, d + 1): if i * i > d: break if (d % i == 0): if (d // i == i): count += 1 else: count += 2 # Return the count return count# Driver Codeif __name__ == '__main__': x = 2 y = 6 print(condition(x, y))# This code is contributed by mohit kumar 29 |
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C#
// C# program for the // above approachusing System;class GFG{ // Function to count // integers K satisfying // given equationstatic int condition(int a, int b){ // Calculate the absoluter // difference between a and b int d = Math.Abs(a - b), count = 0; // Iterate till sqrt of // the diffeence for(int i = 1; i <= Math.Sqrt(d); i++) { if (d % i == 0) { if (d / i == i) count += 1; else count += 2; } } // Return the count return count;} // Driver Codepublic static void Main(String[] args){ int x = 2, y = 6; Console.WriteLine(condition(x, y));}}// This code is contributed by shikhasingrajput |
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Output:
3
Time Complexity: O(√abs(X – Y))
Auxiliary Space: O(1)
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