Given three integers A, B and C. In an infinite sequence, A is the first number, C is the common difference (Si – Si – 1 = C). The task is to check if the number B will appear in the sequence or not.
Examples:
Input: A = 1, B = 7, C = 3
Output: Yes
The sequence will be 1, 4, 7, 10, …
Input: A = 1, B = -4, C = 5
Output: No
Approach: There are two cases:
- When C = 0, print Yes if A = B else No as the sequence will consist only the number A
- When C > 0, for any non-negative integer k the equation B = A + k * C must be satisfied i.e. (B – A) / C must be a non-negative integer.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function that returns true if// the sequence will contain Bbool doesContainB(int a, int b, int c){ if (a == b) return true; if ((b - a) * c > 0 && (b - a) % c == 0) return true; return false;}// Driver codeint main(){ int a = 1, b = 7, c = 3; if (doesContainB(a, b, c)) cout << "Yes"; else cout << "No"; return 0;} |
Java
// Java implementation of the approachclass GFG { // Function that returns true if // the sequence will contain B static boolean doesContainB(int a, int b, int c) { if (a == b) { return true; } if ((b - a) * c > 0 && (b - a) % c == 0) { return true; } return false; } // Driver code public static void main(String[] args) { int a = 1, b = 7, c = 3; if (doesContainB(a, b, c)) { System.out.println("Yes"); } else { System.out.println("No"); } }}// This code contributed by Rajput-Ji |
Python3
# Python3 implementation of the approach# Function that returns true if# the sequence will contain Bdef doesContainB(a, b, c): if (a == b): return True if ((b - a) * c > 0 and (b - a) % c == 0): return True return False# Driver codeif __name__ == '__main__': a, b, c = 1, 7, 3 if (doesContainB(a, b, c)): print("Yes") else: print("No")# This code is contributed by 29AjayKumar |
C#
// C# implementation of the approachusing System;class GFG { // Function that returns true if // the sequence will contain B static bool doesContainB(int a, int b, int c) { if (a == b) { return true; } if ((b - a) * c > 0 && (b - a) % c == 0) { return true; } return false; } // Driver code public static void Main() { int a = 1, b = 7, c = 3; if (doesContainB(a, b, c)) { Console.WriteLine("Yes"); } else { Console.WriteLine("No"); } }}/* This code contributed by PrinciRaj1992 */ |
PHP
<?php// PHP implementation of the approach// Function that returns true if// the sequence will contain Bfunction doesContainB($a, $b, $c){ if ($a == $b) return true; if (($b - $a) * $c > 0 && ($b - $a) % $c == 0) return true; return false;}// Driver code$a = 1; $b = 7; $c = 3;if (doesContainB($a, $b, $c)) echo "Yes";else echo "No";// This code is contributed// by Akanksha Rai?> |
Javascript
<script>// javascript program for the above approach // Function that returns true if // the sequence will contain B function doesContainB(a, b, c) { if (a == b) { return true; } if ((b - a) * c > 0 && (b - a) % c == 0) { return true; } return false; }// Driver Code let a = 1, b = 7, c = 3; if (doesContainB(a, b, c)) { document.write("Yes"); } else { document.write("No"); } </script> |
Output:
Yes
Time Complexity: O(1), since there is only basic arithmetic which takes constant time.
Auxiliary Space: O(1), since no extra space has been taken.
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