Given two integers X and Y, the task is to find and print the numbers that divide X and Y to produce the same remainder.
Examples:
Input: X = 1, Y = 5
Output: 1, 2, 4
Explanation:
Let the number be M. It can be any value in the range [1, 5]:
If M = 1, 1 % 1 = 0 and 5 % 1 = 0
If M = 2, 1 % 2 = 1 and 5 % 2 = 1
If M = 3, 1 % 3 = 1 and 5 % 3 = 2
If M = 4, 1 % 4 = 1 and 5 % 4 = 1
If M = 5, 1 % 5 = 1 and 5 % 5 = 0
Therefore, the possible M values are 1, 2, 4
Input: X = 8, Y = 10
Output: 1, 2
Naive Approach: The naive approach for this problem is to check the modulo value for all the possible values of M in the range [1, max(X, Y)] and print the value of M if the condition satisfies.
Below is the implementation of the above approach:
C++
// C++ program to find numbers// that divide X and Y// to produce the same remainder#include <iostream>using namespace std;// Function to find// the required number as Mvoid printModulus(int X, int Y){ // Finding the maximum // value among X and Y int n = max(X, Y); // Loop to iterate through // maximum value among X and Y. for (int i = 1; i <= n; i++) { // If the condition satisfies, then // print the value of M if (X % i == Y % i) cout << i << " "; }}// Driver codeint main(){ int X, Y; X = 10; Y = 20; printModulus(X, Y); return 0;} |
Java
// Java program to find numbers// that divide X and Y// to produce the same remainderimport java.util.*;import java.io.*;class GFG{ // Function to find// the required number as Mstatic void printModulus(int X, int Y){ // Finding the maximum // value among X and Y int n = Math.max(X, Y); // Loop to iterate through // maximum value among X and Y. for (int i = 1; i <= n; i++) { // If the condition satisfies, then // print the value of M if (X % i == Y % i) System.out.print(i + " "); }} // Driver codepublic static void main(String[] args){ int X, Y; X = 10; Y = 20; printModulus(X, Y);}}// This code is contributed by Princi Singh |
Python3
# Python program to find numbers# that divide X and Y# to produce the same remainder# Function to find# the required number as Mdef printModulus( X, Y): # Finding the maximum # value among X and Y n = max(X, Y) # Loop to iterate through # maximum value among X and Y. for i in range(1, n + 1): # If the condition satisfies, then # print the value of M if (X % i == Y % i): print(i,end=" ")# Driver codeX = 10Y = 20printModulus(X, Y)# This code is contributed by Atul_kumar_Shrivastava |
C#
// C# program to find numbers// that divide X and Y// to produce the same remainderusing System;class GFG{// Function to find// the required number as Mstatic void printModulus(int X, int Y){ // Finding the maximum // value among X and Y int n = Math.Max(X, Y); // Loop to iterate through // maximum value among X and Y. for (int i = 1; i <= n; i++) { // If the condition satisfies, then // print the value of M if (X % i == Y % i) Console.Write(i + " "); }}// Driver codepublic static void Main(){ int X, Y; X = 10; Y = 20; printModulus(X, Y);}}// This code is contributed by AbhiThakur |
Javascript
<script>// Javascript program to find numbers // that divide X and Y // to produce the same remainder // Function to find // the required number as M function printModulus(X, Y) { // Finding the maximum // value among X and Y var n = Math.max(X, Y); // Loop to iterate through // maximum value among X and Y. for (var i = 1; i <= n; i++) { // If the condition satisfies, then // print the value of M if (X % i == Y % i) document.write(i+" "); } } // Driver code X = 10; Y = 20; printModulus(X, Y); // This code is contributed by noob2000.</script> |
Output:
1 2 5 10
Time Complexity: O(max(X, Y))
Auxiliary Space: O(1)
Efficient Approach: Let’s assume that Y is greater than X by a difference of D.
- Then Y can be expressed as
Y = X + D
and
Y % M = (X + D) % M
= (X % M) + (D % M)
- Now, the condition becomes whether X % M and X % M + D % M are equal or not.
- Here, since X % M is common on both the sides, the value of M is true if for some M, D % M = 0.
- Therefore, the required values of M will be the factors of D.
Below is the implementation of the above approach:
CPP
// C++ program to find numbers// that divide X and Y to// produce the same remainder#include <iostream>using namespace std;// Function to print all the possible values// of M such that X % M = Y % Mvoid printModulus(int X, int Y){ // Finding the absolute difference // of X and Y int d = abs(X - Y); // Iterating from 1 int i = 1; // Loop to print all the factors of D while (i * i <= d) { // If i is a factor of d, then print i if (d % i == 0) { cout << i << " "; // If d / i is a factor of d, // then print d / i if (d / i != i) cout << d / i << " "; } i++; }}// Driver codeint main(){ int X = 10; int Y = 26; printModulus(X, Y); return 0;} |
Java
// Java program to find numbers// that divide X and Y to// produce the same remainderimport java.util.*;import java.io.*;class GFG{ // Function to print all the possible values// of M such that X % M = Y % Mstatic void printModulus(int X, int Y){ // Finding the absolute difference // of X and Y int d = Math.abs(X - Y); // Iterating from 1 int i = 1; // Loop to print all the factors of D while (i * i <= d) { // If i is a factor of d, then print i if (d % i == 0) { System.out.print(i+ " "); // If d / i is a factor of d, // then print d / i if (d / i != i) System.out.print(d / i+ " "); } i++; }} // Driver codepublic static void main(String[] args){ int X = 10; int Y = 26; printModulus(X, Y);}}// This code is contributed by Princi Singh |
Python3
# Python program to find numbers# that divide X and Y to# produce the same remainder# Function to print all the possible values# of M such that X % M = Y % Mdef printModulus(X, Y): # Finding the absolute difference # of X and Y d = abs(X - Y); # Iterating from 1 i = 1; # Loop to print all the factors of D while (i * i <= d): # If i is a factor of d, then pri if (d % i == 0): print(i, end=""); # If d / i is a factor of d, # then prd / i if (d // i != i): print(d // i, end=" "); i+=1; # Driver codeif __name__ == '__main__': X = 10; Y = 26; printModulus(X, Y);# This code contributed by Princi Singh |
C#
// C# program to find numbers// that divide X and Y to// produce the same remainderusing System;public class GFG{ // Function to print all the possible values// of M such that X % M = Y % Mstatic void printModulus(int X, int Y){ // Finding the absolute difference // of X and Y int d = Math.Abs(X - Y); // Iterating from 1 int i = 1; // Loop to print all the factors of D while (i * i <= d) { // If i is a factor of d, then print i if (d % i == 0) { Console.Write(i+ " "); // If d / i is a factor of d, // then print d / i if (d / i != i) Console.Write(d / i+ " "); } i++; }} // Driver codepublic static void Main(String[] args){ int X = 10; int Y = 26; printModulus(X, Y);}} // This code contributed by Princi Singh |
Javascript
<script> // Javascript program to find numbers // that divide X and Y to // produce the same remainder // Function to print all the possible values // of M such that X % M = Y % M function printModulus(X, Y) { // Finding the absolute difference // of X and Y var d = Math.abs(X - Y); // Iterating from 1 var i = 1; // Loop to print all the factors of D while (i * i <= d) { // If i is a factor of d, then print i if (d % i == 0) { document.write(i + " "); // If d / i is a factor of d, // then print d / i if (d / i != i) document.write(parseInt(d / i) + " "); } i++; } } // Driver code var X = 10; var Y = 26; printModulus(X, Y);// This code is contributed by rrrtnx. </script> |
Output:
1 16 2 8 4
Time Complexity Analysis O(sqrt(D)), where D is the difference between the values X and Y.
Auxiliary Space: O(1)
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