---
Difficulty: Basic
Source: 160 Days of Problem Solving (BONUS PROBLEMS 🎁)
Tags:
  - Number Theory
  - Mathematical
  - Algorithms
---

# 🚀 _2. Repetitive Addition of Digits_ 🧠

The problem can be found at the following link: [Problem Link](https://www.geeksforgeeks.org/batch/gfg-160-problems/track/array-bonus-problems/problem/repetitive-addition-of-digits2221)

## 💡 **Problem Description:**

You are given a positive integer `n`. You need to repeatedly add all the digits of `n` to create a new number. Perform this operation until the resultant number has only one digit. Return the final number obtained after performing the given operation.

## 🔍 **Example Walkthrough:**

**Input:**

```
n = 1234
```

**Output:**

```
1
```

**Explanation:**

- Step 1: 1 + 2 + 3 + 4 = 10
- Step 2: 1 + 0 = 1

**Input:**

```
n = 5674
```

**Output:**

```
4
```

**Explanation:**

- Step 1: 5 + 6 + 7 + 4 = 22
- Step 2: 2 + 2 = 4

**Input:**

```
n = 9
```

**Output:**

```
9
```

**Explanation:**  
Since 9 is already a single-digit number, return 9.

### **Constraints:**

- $1 \leq n \leq 10^9$

## 🎯 **My Approach:**

This problem revolves around the **digital root** concept in number theory. The process of repetitive digit addition can be optimized without manually summing digits repeatedly. The result of this process is equivalent to `n % 9` with some edge cases:

1. **Digital Root Observation:**
   - If `n == 0`, the result is `0`.
   - If `n % 9 == 0` and `n > 0`, the result is `9`.
   - Otherwise, the result is `n % 9`.

This allows us to calculate the final single-digit number in constant time.

## 🕒 **Time and Auxiliary Space Complexity**

- **Expected Time Complexity:** O(1), as we compute the result directly using mathematical operations.
- **Expected Auxiliary Space Complexity:** O(1), as no extra space is used.

## 📝 **Solution Code**

## Code (Cpp)

```cpp
class Solution {
public:
    int singleDigit(int n) {
        return (n == 0) ? 0 : (n % 9 == 0 ? 9 : n % 9);
    }
};
```

## Code (Java)

```java
class Solution {
    public int singleDigit(int n) {
        return (n == 0) ? 0 : (n % 9 == 0 ? 9 : n % 9);
    }
}
```

## Code (Python)

```python
class Solution:
    def singleDigit(self, n):
        return 0 if n == 0 else (9 if n % 9 == 0 else n % 9)
```

## 🎯 Contribution and Support:

For discussions, questions, or doubts related to this solution, feel free to connect on LinkedIn: [Any Questions](https://www.linkedin.com/in/patel-hetkumar-sandipbhai-8b110525a/). Let’s make this learning journey more collaborative!

⭐ If you find this helpful, please give this repository a star! ⭐

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