643. Maximum Average Subarray I

Here’s a cleanly formatted version of the problem:

---

### 643. Maximum Average Subarray I (Easy)

**Problem Description:**

You are given an integer array `nums` consisting of `n` elements and an integer `k`.

Find a contiguous subarray whose length is equal to `k` that has the maximum average value and return this value. Any answer with a calculation error less than `10^-5` will be accepted.

---

#### Example 1:

**Input:**  
`nums = [1, 12, -5, -6, 50, 3]`  
`k = 4`

**Output:**  
`12.75000`

**Explanation:**  
Maximum average is calculated as:  
`(12 - 5 - 6 + 50) / 4 = 51 / 4 = 12.75`

---

#### Example 2:

**Input:**  
`nums = [5]`  
`k = 1`

**Output:**  
`5.00000`

---


Q- can you explain the exercise using your own words ?

—> so we are given a list of values of type int and k as that target length 

Our task to find out the a “contigous” , “subarray” of length equal to k 

Where it avg values if the biggest oky 

Q- what is the input/output format ? 

Input : List[int] 
Output : int 

Q- can you go through an example ? 

For example we have this list 

[1,12,-5,-6,50,3]  , k = 4

Q- one question you can ask given this example what are all the sub-contigous-array of length = k

The answer is like this 
Here are all the contiguous subarrays of length `k=4` for the input array `nums = [1, 12, -5, -6, 50, 3]`:

1. `[1, 12, -5, -6]`
2. `[12, -5, -6, 50]`
3. `[-5, -6, 50, 3]`


There are only three such subarrays because the length of the original array is 6, and with a subarray length of 4, there are only `6 - 4 + 1 = 3` possible starting positions for the subarrays.


Q- next we have all the target values we need just for eeach one to compute the avg values for it 

It can look like this 

 “[1, 12, -5, -6]: (1 + 12 + (-5) + (-6)) / 4 = 2 / 4 = 0.5
[12, -5, -6, 50]: (12 + (-5) + (-6) + 50) / 4 = 51 / 4 = 12.75
[-5, -6, 50, 3]: (-5 + (-6) + 50 + 3) / 4 = 42 / 4 = 10.5
So, the average values of the subarrays are 0.5, 12.75, and 10.5, respectively.
“

Q- next we have find all the possible subarrays and next we need to peak the biggest one oky 
Which is here is = 12.75 

We return it 


Q- but as you see this is more experimental solution not based in will defined formal algorithm can … i mean my question can be genelize the steps in the last example yes or no ? if yes how ? using mathematique cs notation 

Here's a neatly formatted version of your text:

---

**Yes, you can generalize the experimental steps you outlined into a formal algorithm using mathematical and computer science (CS) notation.** Formalizing the solution not only makes it more structured and efficient but also easier to implement and understand, especially for larger inputs.

---

### Formal Algorithm: Sliding Window Approach

The Sliding Window technique is optimal for this problem, achieving a time complexity of \(O(n)\), where \(n\) is the number of elements in the input array `nums`. Here's how you can formalize the solution:

---

#### Algorithm Steps:

**1. Initialization:**

- **Compute the Sum of the First Window:**  
\[
\text{current\_sum} = \sum_{i=0}^{k-1} \text{nums}[i]
\]
- **Set the Initial Maximum Sum:**  
\[
\text{max\_sum} = \text{current\_sum}
\]

---

**2. Slide the Window Across the Array:**  
For each index \(i\) from \(k\) to \(n-1\):
- **Update the Current Sum by Sliding the Window:**  
Remove the element exiting the window \((\text{nums}[i - k])\) and add the new element entering the window \((\text{nums}[i])\):
\[
\text{current\_sum} = \text{current\_sum} - \text{nums}[i - k] + \text{nums}[i]
\]
- **Update the Maximum Sum if Necessary:**  
\[
\text{if } \text{current\_sum} > \text{max\_sum}, \text{ then set } \text{max\_sum} = \text{current\_sum}
\]

---

**3. Compute the Maximum Average:**  
After sliding the window through the entire array, calculate the average of the subarray with the maximum sum:
\[
\text{max\_average} = \frac{\text{max\_sum}}{k}
\]

---

**4. Return the Result:**  
\[
\text{return } \text{max\_average}
\]

---

### Pseudocode Implementation:

```python
def findMaxAverage(nums: List[int], k: int) -> float:
    n = len(nums)
    if n == 0 or k > n:
        return 0.0  # or handle error as appropriate

    # Step 1: Initialize current_sum and max_sum
    current_sum = sum(nums[:k])
    max_sum = current_sum

    # Step 2: Slide the window from index k to n-1
    for i in range(k, n):
        current_sum = current_sum - nums[i - k] + nums[i]
        if current_sum > max_sum:
            max_sum = current_sum

    # Step 3: Compute the maximum average
    max_average = max_sum / k

    # Step 4: Return the result
    return max_average
```

---

### Detailed Explanation:

**1. Initialization:**
- `current_sum` holds the sum of the current window of size `k`.
- `max_sum` keeps track of the highest sum encountered so far.

**2. Sliding the Window:**
- For each new element entering the window (`nums[i]`), subtract the element exiting (`nums[i - k]`) from `current_sum` and add the new element.
- Update `max_sum` if the new `current_sum` is greater.

**3. Calculating the Average:**
- After traversing the array, `max_sum` holds the maximum sum of any contiguous subarray of size `k`.
- Divide `max_sum` by `k` to get the average.

**4. Edge Cases:**
- If the array is empty or `k` is larger than the array size, handle accordingly (e.g., return `0.0` or raise an error).

---

### Mathematical Notation Summary:

- **Initial Window Sum:**  
\[
\text{current\_sum} = \sum_{i=0}^{k-1} \text{nums}[i]
\]
- **Sliding the Window:**  
For \(i = k\) to \(n-1\):
\[
\text{current\_sum} = \text{current\_sum} - \text{nums}[i - k] + \text{nums}[i]
\]
\[
\text{max\_sum} = \max(\text{max\_sum}, \text{current\_sum})
\]
- **Final Average:**  
\[
\text{max\_average} = \frac{\text{max\_sum}}{k}
\]

---

### Applying the Algorithm to the Example:

Given:  
\[
\text{nums} = [1, 12, -5, -6, 50, 3], \quad k = 4
\]

**Step-by-Step Execution:**

1. **Initialization:**  
\[
\text{current\_sum} = 1 + 12 + (-5) + (-6) = 2
\]  
\[
\text{max\_sum} = 2
\]

2. **First Slide (i = 4):**  
\[
\text{current\_sum} = 2 - 1 + 50 = 51
\]  
\[
\text{max\_sum} = \max(2, 51) = 51
\]

3. **Second Slide (i = 5):**  
\[
\text{current\_sum} = 51 - 12 + 3 = 42
\]  
\[
\text{max\_sum} = \max(51, 42) = 51
\]

4. **Compute Maximum Average:**  
\[
\text{max\_average} = \frac{51}{4} = 12.75
\]

**Result:**  
\[
\text{return } 12.75
\]

---

### Advantages of the Formal Algorithm:

- **Efficiency:**
  - **Time Complexity:** \(O(n)\) because it traverses the array only once after the initial window sum calculation.
  - **Space Complexity:** \(O(1)\) as it uses a constant amount of extra space.
  
- **Scalability:**  
  Handles large input sizes effectively due to its linear time complexity.

- **Simplicity:**  
  The sliding window approach is intuitive once formalized and is easy to implement in any programming language.

- **Clarity:**  
  Mathematical notation provides a clear and precise description of the algorithm, ensuring correctness.

---

### Conclusion:

By formalizing your experimental steps into a structured sliding window algorithm, you enhance both the understanding and efficiency of the solution. This approach ensures that the algorithm performs optimally for all valid inputs and serves as a solid foundation for implementation in code.

---

Q- imagine your are the interviewer what are the top 5 question you will ask me here 

     # 2269. Find the K-Beauty of a Number

**Difficulty:** Easy

**Topics:** String, Sliding Window

**Hint:** Iterate through all substrings of length `k` and check if they are divisors of `num`.

## Problem Description

The k-beauty of an integer `num` is defined as the number of substrings of `num` (when it is read as a string) that meet the following conditions:

1. It has a length of `k`.
2. It is a divisor of `num`.

Given integers `num` and `k`, return the k-beauty of `num`.

**Note:**

* Leading zeros are allowed.
* 0 is not a divisor of any value.
* A substring is a contiguous sequence of characters in a string.


## Examples

**Example 1:**
   
content_copy Use code with caution.Markdown
Input: num = 240, k = 2
Output: 2
Explanation: The following are the substrings of num of length k:
"24" from "240": 24 is a divisor of 240.
"40" from "240": 40 is a divisor of 240.
Therefore, the k-beauty is 2.
     **Example 2:**
   
content_copy Use code with caution.
Input: num = 430043, k = 2
Output: 2
Explanation: The following are the substrings of num of length k:
"43" from "430043": 43 is a divisor of 430043.
"30" from "430043": 30 is not a divisor of 430043.
"00" from "430043": 0 is not a divisor of 430043.
"04" from "430043": 4 is not a divisor of 430043.
"43" from "430043": 43 is a divisor of 430043.
Therefore, the k-beauty is 2.
     ## Constraints:

* 1 <= num <= 10<sup>9</sup>
* 1 <= k <= num.length (taking num as a string)
   
content_copy Use code with caution.


Q- can you explain it using you own words ? 

First we have a definition here oky 

→ k-beauty : is a substring of num that foolow this rules 

It has a length =”k” 
It’s a divisor of num 
Given the num and k returns the k-beauty numbrs 

Q- what is the input/output format ? 

Input : int  , k:int 
Output : int 

Q- can you go through an example ? 

Input: num = 430043, k = 2


All the sub_string for this input num is 

43 , 30 , 00, 04,43 
From the substrings "43", "30", "00", "04", and "43" of the number 430043:
Only "43" is a divisor of 430043. "00" evaluates to 0, which is not a divisor of anything. The other substrings don't divide evenly.

So from the 5 potential sub_string only 2 verfy the second condtion which is to be divior of input int → that why we return only 2 oky 

Q- can we genelize the ideas ? 

We can leverage the slide window technique , where we know the target window length (which is equal to k ) … and then just update the window oky 

First we compute the current_sub_string = num[:k]
We also intiate k-beauty = 0 to track the number of k-beauty that satisfy the condition  
If num%current_sub_string==0:
	k-beauty +1
We also intiate k-beauty = 0 to track the number of k-beauty that satisfy the condition 
Next from i start in k until n-1 
	We update the current_sub_string=current_sub_string-num[i-k]+num[i] 
	If num%current_sub_string:
		k-beauty+=1
Return k-beauty 

Q- i do know if my logic is 100% correct but my intuition is in the right track  right ? 
If yes make it more formal with mathematical notaion ? and do manual execution in it using the previous example oky