Given four positive numbers A, B, C, D, such that A and B are the first term and the common difference of first Arithmetic sequence respectively, while C and D represent the same for the second Arithmetic sequence respectively as shown below:
First Arithmetic Sequence: A, A + B, A + 2B, A + 3B, …….
Second Arithmetic Sequence: C, C + D, C + 2D, C + 3D, …….
The task is to find the least common element from the above AP sequences. If no such number exists, then print -1.
Examples:
Input: A = 2, B = 20, C = 19, D = 9
Output: 82
Explanation:
Sequence 1: {2, 2 + 20, 2 + 2(20), 2 + 3(20), 2 + 4(20), …..} = {2, 22, 42, 62, 82, …..}
Sequence 2: {19, 19 + 9, 19 + 2(9), 19 + 3(9), 19 + 4(9), 19 + 5(9), 19 + 6(9), 19 + 7(9) …..} = {19, 28, 37, 46, 55, 64, 73, 82, …..}
Therefore, 82 is the smallest common element.
Input: A = 2, B = 18, C = 19, D = 9
Output: -1
Approach:
Since any term of the given two sequences can be expressed as A + x*B and C + y*D, therefore, to solve the problem, we need to find the smallest values of x and y for which the two terms are equal.
Follow the steps below to solve the problem:
- In order to find the smallest value common in both the AP sequences, the idea is to find the smallest integer values of x and y satisfying the following equation:
A + x*B = C + y*D
=> The above equation can be rearranged as
x*B = C – A + y*D
=> The above equation can be further rearranged as
x = (C – A + y*D) / B
- Check if there exists any integer value y such that (C – A + y*D) % B is 0 or not. If it exists, then the smallest number is (C + y*D).
- Check whether (C + y*D) is the answer or not, where y will be in a range (0, B) because from i = B, B+1, …. the remainder values will be repeated.
- If no such number is obtained from the above steps, then print -1.
Below is the implementation of the above approach:
C++
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to find the smallest element// common in both the subsequenceslong smallestCommon(long a, long b, long c, long d){ // If a is equal to c if (a == c) return a; // If a exceeds c if (a > c) { swap(a, c); swap(b, d); } long first_term_diff = (c - a); long possible_y; // Check for the satisfying // equation for (possible_y = 0; possible_y < b; possible_y++) { // Least value of possible_y // satisfying the given equation // will yield true in the below if // and break the loop if ((first_term_diff % b + possible_y * d) % b == 0) { break; } } // If the value of possible_y // satisfying the given equation // lies in range [0, b] if (possible_y != b) { return c + possible_y * d; } // If no value of possible_y // satisfies the given equation return -1;}// Driver Codeint main(){ long A = 2, B = 20, C = 19, D = 9; cout << smallestCommon(A, B, C, D); return 0;} |
Java
// Java program to implement// the above approachimport java.util.*;class GFG{// Function to find the smallest element// common in both the subsequencesstatic int smallestCommon(int a, int b, int c, int d){ // If a is equal to c if (a == c) return a; // If a exceeds c if (a > c) { swap(a, c); swap(b, d); } int first_term_diff = (c - a); int possible_y; // Check for the satisfying // equation for (possible_y = 0; possible_y < b; possible_y++) { // Least value of possible_y // satisfying the given equation // will yield true in the below if // and break the loop if ((first_term_diff % b + possible_y * d) % b == 0) { break; } } // If the value of possible_y // satisfying the given equation // lies in range [0, b] if (possible_y != b) { return c + possible_y * d; } // If no value of possible_y // satisfies the given equation return -1;} static void swap(int x, int y){ int temp = x; x = y; y = temp;} // Driver Codepublic static void main(String[] args){ int A = 2, B = 20, C = 19, D = 9; System.out.print(smallestCommon(A, B, C, D));}}// This code is contributed by PrinciRaj1992 |
Python3
# Python3 program to implement# the above approach# Function to find the smallest element# common in both the subsequencesdef smallestCommon(a, b, c, d): # If a is equal to c if (a == c): return a; # If a exceeds c if (a > c): swap(a, c); swap(b, d); first_term_diff = (c - a); possible_y = 0; # Check for the satisfying # equation for possible_y in range(b): # Least value of possible_y # satisfying the given equation # will yield True in the below if # and break the loop if ((first_term_diff % b + possible_y * d) % b == 0): break; # If the value of possible_y # satisfying the given equation # lies in range [0, b] if (possible_y != b): return c + possible_y * d; # If no value of possible_y # satisfies the given equation return -1;def swap(x, y): temp = x; x = y; y = temp;# Driver Codeif __name__ == '__main__': A = 2; B = 20; C = 19; D = 9; print(smallestCommon(A, B, C, D));# This code is contributed by Rajput-Ji |
C#
// C# program to implement// the above approachusing System;class GFG{ // Function to find the smallest element// common in both the subsequencesstatic int smallestCommon(int a, int b, int c, int d){ // If a is equal to c if (a == c) return a; // If a exceeds c if (a > c) { swap(a, c); swap(b, d); } int first_term_diff = (c - a); int possible_y; // Check for the satisfying // equation for (possible_y = 0; possible_y < b; possible_y++) { // Least value of possible_y // satisfying the given equation // will yield true in the below if // and break the loop if ((first_term_diff % b + possible_y * d) % b == 0) { break; } } // If the value of possible_y // satisfying the given equation // lies in range [0, b] if (possible_y != b) { return c + possible_y * d; } // If no value of possible_y // satisfies the given equation return -1;} static void swap(int x, int y){ int temp = x; x = y; y = temp;} // Driver Codepublic static void Main(String[] args){ int A = 2, B = 20, C = 19, D = 9; Console.Write(smallestCommon(A, B, C, D));}}// This code is contributed by Rajput-Ji |
Javascript
<script>// JavaScript program for the// above approach// Function to find the smallest element// common in both the subsequencesfunction smallestCommon(a, b, c, d){ // If a is equal to c if (a == c) return a; // If a exceeds c if (a > c) { swap(a, c); swap(b, d); } let first_term_diff = (c - a); let possible_y; // Check for the satisfying // equation for (possible_y = 0; possible_y < b; possible_y++) { // Least value of possible_y // satisfying the given equation // will yield true in the below if // and break the loop if ((first_term_diff % b + possible_y * d) % b == 0) { break; } } // If the value of possible_y // satisfying the given equation // lies in range [0, b] if (possible_y != b) { return c + possible_y * d; } // If no value of possible_y // satisfies the given equation return -1;} function swap(x, y){ let temp = x; x = y; y = temp;}// Driver Code let A = 2, B = 20, C = 19, D = 9; document.write(smallestCommon(A, B, C, D));</script> |
Output:
82
Time Complexity: O(B)
Auxiliary Space: O(1)
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