add question no 258

dsa-solutions/lc-solutions/0200-0299/0258-Add-digits.md

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+---
+id: add-digits
+title: Add Digits
+sidebar_label: 0258-Add-digits
+tags: ['Math', 'Simulation', 'NumberTheory' ]
+description: "Given an integer num, repeatedly add all its digits until the result has only one digit, and return it."
+---
+
+## Problem
+
+Given an integer `num`, repeatedly add all its digits until the result has only one digit, and return it..
+
+### Examples
+
+**Example 1:**
+```
+Input: num = 38
+Output: 2
+Explanation: The process is
+38 --> 3 + 8 --> 11
+11 --> 1 + 1 --> 2 
+Since 2 has only one digit, return it.
+```
+**Example 2:**
+```
+Input: num = 0
+Output: 0
+```
+### Constraints
+
+- `0 <= num <= 2^31 - 1`
+
+### Approach
+
+Any number where it's digits add to `9` is always divisible by `9`. `(18, 27, 36, 45, 54, 63, 72, 81, 90, etc.)` Therefore the 'digital root' for any number divisible by `9` is always `9`. You can see this even in larger numbers like 99 because `9 + 9 = 18`, and then `1 + 8 = 9` still, so the root always becomes `9` for any numbers divisible by `9`.
+
+Additionally, 0 always has a digital root of 0 obviously.
+
+The only other cases you need to worry about to find the digital root are when it isn't 0 or 9.
+
+So for any number that isn't 0 and isn't divisible by 9, the root will always `n % 9` for a given number n. (AKA the difference between given number n and the nearest number that is divisible by `9`, since numbers divisible by `9` always have a digital root of `9`).
+For examples: `100 % 9 = 1 (one greater than 99, which is divisible by 9)`.
+`101 % 9 = 2`
+`102 % 9 = 3` and so on.
+
+This explanation/algorithm skips the whole "add digits until there is only 1 remaining", so the description of this problem seems pretty misleading to me since it makes you think the solution will be something unrelated to the optimal one. I guess the point of Leetcode is to learn all of these tricks though.
+
+
+
+### Solution
+
+#### Code in Different Languages
+
+### C++ Solution
+```cpp
+class Solution {
+public:
+    int addDigits(int num) {
+        if(num == 0) return 0;
+        else if(num % 9 == 0) return 9;
+        else return num % 9;
+    }
+};
+```
+### Java Solution
+```java
+class Solution {
+    public int addDigits(int num) {
+        if(num == 0) return 0;
+        else if(num % 9 == 0) return 9;
+        else return num % 9;
+    }
+}
+```
+### Python Solution
+
+```python
+class Solution:
+    def addDigits(self, num: int) -> int:
+        if num == 0 : return 0
+        if num % 9 == 0 : return 9
+        else : return (num % 9)
+```
+### Complexity Analysis
+**Time Complexity:** $O(1)$
+
+
+**Space Complexity:** $O(1)$
+
+
+This solution efficiently add the digits of a number in $o(1)$ space complexity and $o(1)$ time complexity
+
+### References
+**LeetCode Problem:** Add-digits
+