Given two numbers ‘a’ and ‘b’. Find the minimum difference between any terms in shifted infinite tables of ‘a’ and ‘b’, given shifts ‘x’ and ‘y’, where x, y >= 0.
Let us consider a = 6 and b = 16. Infinite tables of these numbers are.
Table of a : 6, 12, 18, 24, 30, 36, 42, 48.….
Table of b : 16, 32, 48, 64, 80, 96, 112, 128…..
Let given shifts be x = 5 and y = 2
Shifted Table of a : 11, 17, 23, 29, 35, 41, 47 …
Shifted Table of b : 18, 34, 50, 66, 82, 98, 114 …
The minimum difference between any two terms of above two shifted tables is 1 (See colored terms). So the output should be 1.
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Algorithm:
- Compute the Greatest Common Divisor (GCD) of the numbers ‘a’ and ‘b’. Let the GCD of two numbers be ‘g’.
- Compute the absolute difference “diff” of the shifts ‘x’ and ‘y’ under modulo ‘g’, i.e., diff = abs(a – b) % g
- The required shifted difference is minimum of ‘diff’ and ‘g – diff’.
Below is implementation of above algorithm.
C++
// C++ program to find the minimum difference // between any two terms of two tables #include <bits/stdc++.h> using namespace std; // Utility function to find GCD of a and b int gcd(int a, int b) { while (b != 0) { int t = b; b = a % b; a = t; } return a; } // Returns minimum difference between // any two terms of shifted tables of 'a' // and 'b'. 'x' is shift in table of 'a' // and 'y' is shift in table of 'b'. int findMinDiff(int a, int b, int x, int y) { // Calculate gcd of a nd b int g = gcd(a,b); // Calculate difference between x and y int diff = abs(x-y) % g; return min(diff, g - diff); } // Driver Code int main() { int a = 20, b = 52, x = 5, y = 7; cout << findMinDiff(a, b, x, y) << endl; return 0; } |
Java
// Java program to find the // minimum difference between // any two terms of two tables import java.io.*; class GFG { // Utility function to // find GCD of a and b static int gcd(int a, int b) { while (b != 0) { int t = b; b = a % b; a = t; } return a; } // Returns minimum difference // between any two terms of // shifted tables of 'a' and // 'b'. 'x' is shift in table // of 'a' and 'y' is shift in // table of 'b'. static int findMinDiff(int a, int b, int x, int y) { // Calculate gcd // of a nd b int g = gcd(a,b); // Calculate difference // between x and y int diff = Math.abs(x - y) % g; return Math.min(diff, g - diff); } // Driver Code public static void main (String[] args) { int a = 20, b = 52, x = 5, y = 7; System.out.println( findMinDiff(a, b, x, y)); } } // This code is contributed by ajit |
Python3
# python3 program to find the minimum difference # between any two terms of two tables import math as mt # Utility function to find GCD of a and b def gcd(a,b): while (b != 0): t = b b = a % b a = t return a # Returns minimum difference between # any two terms of shifted tables of 'a' # and 'b'. 'x' is shift in table of 'a' # and 'y' is shift in table of 'b'. def findMinDiff (a, b, x, y): # Calculate gcd of a nd b g = gcd(a,b) # Calculate difference between x and y diff = abs(x-y) % g return min(diff, g - diff) # Driver Code a,b,x,y = 20,52,5,7 print(findMinDiff(a, b, x, y)) #This code is contributed by Mohit kumar 29 |
C#
// C# program to find the minimum difference // between any two terms of two tables using System; class GFG { // Utility function to find GCD of a and b static int gcd(int a, int b) { while (b != 0) { int t = b; b = a % b; a = t; } return a; } // Returns minimum difference between any // two terms of shifted tables of 'a' and // 'b'. 'x' is shift in table of 'a' and // 'y' is shift in table of 'b'. static int findMinDiff(int a, int b, int x, int y) { // Calculate gcd // of a nd b int g = gcd(a, b); // Calculate difference // between x and y int diff = Math.Abs(x - y) % g; return Math.Min(diff, g - diff); } // Driver Code static void Main() { int a = 20, b = 52, x = 5, y = 7; Console.WriteLine(findMinDiff(a, b, x, y)); } } // This code is contributed by mits |
PHP
<?php // PHP program to find the minimum // difference between any two terms // of two tables // Utility function to // find GCD of a and b function gcd($a, $b) { while ($b != 0) { $t = $b; $b = $a % $b; $a = $t; } return $a; } // Returns minimum difference // between any two terms of // shifted tables of 'a' and // 'b'. 'x' is shift in table // of 'a' and 'y' is shift in // table of 'b'. function findMinDiff($a, $b, $x, $y) { // Calculate gcd of a nd b $g = gcd($a, $b); // Calculate difference // between x and y $diff = abs($x - $y) % $g; return min($diff, $g - $diff); } // Driver Code $a = 20; $b = 52; $x = 5; $y = 7; echo findMinDiff($a, $b, $x, $y), "\n"; // This code is contributed by aj_36 ?> |
Output :
2
Complexity: O(log n), to compute the GCD.
How does this work?
After shifting tables become (Assuming y > x and diff = y-x)
a+x, 2a+x, 3a+x, .. ab+x, …. and b+y, 2b+y, 3b+y, …., ab+y, ….
In general, difference is, abs[(am + x) – (bn + y)] where m >= 1 and n >= 1
An upper bound on minimum difference is “abs(x – y)”. We can always get this difference by putting m = b and n = a.
How can we reduce the absolute difference below “abs(x – y)”?
Let us rewrite “abs[(am + x) – (bn + y)]” as abs[(x – y) – (bn – am)]
Let the GCD of ‘a’ and ‘b’ be ‘g’. Let us consider the table of ‘g’. The table has all terms like a, 2a, 3a, … b, 2b, 3b, … and also the terms (bn – am) and (am – bn).
This article is contributed by Vaibhav Agarwal. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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