Solution of Arithmetic Number from gfg is added

Author: Ishitamukherjee2004

dsa-solutions/gfg-solutions/Easy problems/Arithmetic-Number.md

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+---
+id: arithmetic-number
+title: Arithmetic Number
+sidebar_label: Arithmetic-Number
+tags:
+  - Intermediate
+  - Array
+  - Mathematical
+  - GeeksforGeeks
+  - CPP
+  - Python
+  - DSA
+description: "This tutorial covers the solution to the Find the Arithmetic Number problem from the GeeksforGeeks."
+---
+## Problem Description
+
+Given three integers  `'A'` denoting the first term of an arithmetic sequence , `'C'` denoting the common difference of an arithmetic sequence and an integer `'B'`. you need to tell whether `'B'` exists in the arithmetic sequence or not. Return `1` if `B` is present in the sequence. Otherwise, returns `0`.
+
+## Examples
+
+**Example 1:**
+
+```
+Input: A = 1, B = 3, C = 2
+Output: 1
+Explaination: 3 is the second term of the 
+sequence starting with 1 and having a common 
+difference 2.
+```
+
+**Example 2:**
+
+```
+Input: A = 1, B = 2, C = 3
+Output: 0
+Explaination: 2 is not present in the sequence.
+```
+
+## Your Task
+
+You do not need to read input or print anything. Your task is to complete the function inSequence() which takes A, B and C and returns 1 if B is present in the sequence. Otherwise, returns 0.
+
+Expected Time Complexity: $O(1)$
+
+Expected Auxiliary Space: $O(1)$
+
+## Constraints
+
+* `-10^9 ≤ A, B, C ≤ 10^9`
+
+## Problem Explanation
+Given three integers  'A' denoting the first term of an arithmetic sequence , 'C' denoting the common difference of an arithmetic sequence and an integer 'B'. you need to tell whether 'B' exists in the arithmetic sequence or not. Return 1 if B is present in the sequence. Otherwise, returns 0.
+
+
+## Code Implementation
+
+<Tabs>
+  <TabItem value="Python" label="Python" default>
+  <SolutionAuthor name="@Ishitamukherjee2004"/>
+
+  ```py
+  def is_present(A, C, B):
+  if (B - A) % C == 0 and (B - A) / C >= 0:
+    return 1
+  return 0
+
+
+  ```
+
+  </TabItem>
+  <TabItem value="C++" label="C++">
+  <SolutionAuthor name="@Ishitamukherjee2004"/>
+
+  ```cpp
+ int isPresent(int A, int C, int B) {
+  if ((B - A) % C == 0 && (B - A) / C >= 0) {
+    return 1;
+  }
+  return 0;
+}
+
+  ```
+
+  </TabItem>
+
+  <TabItem value="Javascript" label="Javascript" default>
+  <SolutionAuthor name="@Ishitamukherjee2004"/>
+
+  ```javascript
+ function isPresent(A, C, B) {
+  if ((B - A) % C === 0 && (B - A) / C >= 0) {
+    return 1;
+  }
+  return 0;
+}
+
+
+  ```
+
+  </TabItem>
+
+  <TabItem value="Typescript" label="Typescript" default>
+  <SolutionAuthor name="@Ishitamukherjee2004"/>
+
+  ```typescript
+function isPresent(A, C, B) {
+  if ((B - A) % C === 0 && (B - A) / C >= 0) {
+    return 1;
+  }
+  return 0;
+}
+
+  ```
+
+  </TabItem>
+
+  <TabItem value="Java" label="Java" default>
+  <SolutionAuthor name="@Ishitamukherjee2004"/>
+
+  ```java
+  public int isPresent(int A, int C, int B) {
+  if ((B - A) % C == 0 && (B - A) / C >= 0) {
+    return 1;
+  }
+  return 0;
+}
+
+
+  ```
+
+  </TabItem>
+</Tabs>
+
+
+## Solution Logic:
+The solution checks if the difference between B and A is a multiple of C (the common difference) and if the result is non-negative. If both conditions are true, it means that B is present in the arithmetic sequence.
+
+
+## Time Complexity
+
+* The time complexity is $O(1)$, because it only involves simple arithmetic operations.
+
+
+## Space Complexity
+
+* The auxiliary space complexity is $O(1)$ because we are not using any extra space proportional to the size of the input array.