Given 3 positive integers A, B, and C. Choose a positive integer M and multiply any one of A, B, or C with M. The task is to decide whether A, B and C would comply with Arithmetic Progression Mean (A.P.) Mean after performing the above-given operation once. The order of A, B, C cannot be changed.
Examples:
Input: A = 30, B = 5, C = 10
Output: YES
Explanation: B = 5 can be multiplied by M = 4.
Then 2 * B = A + C (2 * 20 = 30 + 10).Input: A = 2, B = 1, C = 1
Output: NO
Explanation: A + C = 3 which is odd. B = 1 which is odd.
Both these values cannot be made even (because 2*B is always even so need for making both of them even) and equal simultaneously.
Approach:
- Three integers are said to be in A.P. if
2 * B = A + C -(1)
This equation can also be written as
2 * B – C = A -(2)
and
2 * B – A = C -(3)
Below are the steps that use this relationship to determine whether the given three numbers can form an A.P. or not-
- Declare a variable new_b and initiate it equal to 2 * B.
- Return true if A, B, C are already in AP. Any of them can be multiplied by M=1 and the answer will always be “YES“.
- Return true if B can be multiplied by any number to make it equal to A+C i.e. (A + C) % new_b = 0. The remainder will always be 0 irrespective of the value of M.
- Now the only options remaining are to multiply either A or C to make their sum equal to new_b.
- If (A + C) > new_b, return false. In this case, multiplying A or C with M will only increase its value. The answer will always be “NO“.
- Else if (new_b – C) % A = 0, return true. In this case, A can be multiplied by some integer M. In this case, (new_b – C) can be seen as a multiple of A. Hence, on dividing (new_b – C) by A, the remainder will always be 0 irrespective of the value of M. The answer will be “YES” (a per equation (2)).
- Else if (new_b – A) % B = 0, return true. In this case, C can be multiplied by some integer M. In this case, (new_b – A) can be seen as a multiple of C. Hence, on dividing (new_b – A) by C, the remainder will always be 0 irrespective of the value of M. The answer will be “YES” (as per equation (3)).
- Else return false.
Illustration:
Input: A = 30, B = 5, C = 10
Output: YES
Explanation:
new_b = 2 * B
= 2 * 5
= 10
(A + C) % new_b
(30 + 10) % 10
40 % 10 = 0
This proves that the numbers 30 5 10 are in A.P. after multiplying with some number M.
Below is the implementation for the above approach-
C++
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to check if A, B, C// can be in A.P. after multiplying// by any integer Mbool isAP(int A, int B, int C){ int new_b = B * 2; // Check if A, B, C are already // in A.P. if (new_b == (A + C)) return true; // Check if multiplying B will // give A.P. if ((A + C) % new_b == 0) return true; if ((A + C) > new_b) return false; // Check if multiplying A will // give A.P. if ((new_b - C) % A == 0) return true; // Check if multiplying C will // give A.P. if ((new_b - A) % C == 0) return true; return false;}// Driver codeint main(){ int A, B, C; A = 30; B = 5; C = 10; // Function call bool ans = isAP(A, B, C); // Displaying the answer if (ans) cout << "YES"; else cout << "NO"; return 0;} |
Java
// Java program for the above approachimport java.io.*;import java.lang.*;import java.util.*;class GFG { // Function to check if A, B, C // can be in A.P. after multiplying // by any integer M static Boolean isAP(int A, int B, int C) { int new_b = B * 2; // Check if A, B, C are already // in A.P. if (new_b == (A + C)) return true; // Check if multiplying B will // give A.P. if ((A + C) % new_b == 0) return true; if ((A + C) > new_b) return false; // Check if multiplying A will // give A.P. if ((new_b - C) % A == 0) return true; // Check if multiplying C will // give A.P. if ((new_b - A) % C == 0) return true; return false; } // Driver code public static void main (String[] args) { int A, B, C; A = 30; B = 5; C = 10; // Function call Boolean ans = isAP(A, B, C); // Displaying the answer if (ans) System.out.print("YES"); else System.out.print("NO"); }}// This code is contributed by hrithikgarg03188 |
Python3
# Python code for the above approach# Function to check if A, B, C# can be in A.P. after multiplying# by any integer Mdef isAP(A, B, C): new_b = B * 2 # Check if A, B, C are already # in A.P. if (new_b == (A + C)): return True # Check if multiplying B will # give A.P. if ((A + C) % new_b == 0): return True if ((A + C) > new_b): return False # Check if multiplying A will # give A.P. if ((new_b - C) % A == 0): return True # Check if multiplying C will # give A.P. if ((new_b - A) % C == 0): return True return False# Driver codeA = 30B = 5C = 10# Function callans = isAP(A, B, C)# Displaying the answerif (ans): print("YES")else: print("NO")# This code is contributed by Saurabh Jaiswal |
C#
// C# program to implement// the above approachusing System;class GFG{ // Function to check if A, B, C // can be in A.P. after multiplying // by any integer M static bool isAP(int A, int B, int C) { int new_b = B * 2; // Check if A, B, C are already // in A.P. if (new_b == (A + C)) return true; // Check if multiplying B will // give A.P. if ((A + C) % new_b == 0) return true; if ((A + C) > new_b) return false; // Check if multiplying A will // give A.P. if ((new_b - C) % A == 0) return true; // Check if multiplying C will // give A.P. if ((new_b - A) % C == 0) return true; return false; } // Driver code public static int Main() { int A, B, C; A = 30; B = 5; C = 10; // Function call bool ans = isAP(A, B, C); // Displaying the answer if (ans) Console.Write("YES"); else Console.Write("NO"); return 0; }}// This code is contributed by Taranpreet |
Javascript
<script> // JavaScript code for the above approach // Function to check if A, B, C // can be in A.P. after multiplying // by any integer M function isAP(A, B, C) { let new_b = B * 2; // Check if A, B, C are already // in A.P. if (new_b == (A + C)) return true; // Check if multiplying B will // give A.P. if ((A + C) % new_b == 0) return true; if ((A + C) > new_b) return false; // Check if multiplying A will // give A.P. if ((new_b - C) % A == 0) return true; // Check if multiplying C will // give A.P. if ((new_b - A) % C == 0) return true; return false; } // Driver code let A, B, C; A = 30; B = 5; C = 10; // Function call let ans = isAP(A, B, C); // Displaying the answer if (ans) document.write("YES"); else document.write("NO"); // This code is contributed by Potta Lokesh </script> |
Output
YES
Time Complexity: O(1)
Auxiliary Space: O(1)
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