Minimum Size Subarray Sum
Problem Description:
Given an array of positive integers nums and a target integer k, find the minimum length of a subarray whose sum is greater than or equal to k.
Example:
Input: nums = [2,3,1,2,4,3], k = 7
Output: 2
Explanation: The subarray [4,3] has the minimum length under the problem constraint.


Constraints:
1 <= nums.length <= 10^5
1 <= nums[i] <= 10^4
1 <= k <= 10^9
Q- can you explain the problem using you own words ?
→ so we are given a list of integers and target value called k 
We need to find the smallest subarray where it value sum are equal or greater than k 

Q- what is the input-output format? 
Input : List[int] , k:int 
Output: int 
Q-  can you go through an example ? 

First think we need to find out all the subarray for our exmple it may look like this 

“Here are all the subarrays of the array nums = [2, 3, 1, 2, 4, 3]:
Subarrays of length 1:
[2] → Not >= 7
[3] → Not >=7
[1] →
[2]
[4]
[3]


Subarrays of length 2:
[2, 3]
[3, 1]
[1, 2]
[2, 4]
[4, 3] → the right one we the samllest possible one 


Subarrays of length 3:
[2, 3, 1]
[3, 1, 2]
[1, 2, 4]
[2, 4, 3]


Subarrays of length 4:
[2, 3, 1, 2]
[3, 1, 2, 4]
[1, 2, 4, 3]


Subarrays of length 5:
[2, 3, 1, 2, 4]
[3, 1, 2, 4, 3]


Subarrays of length 6:
[2, 3, 1, 2, 4, 3]


There are a total of 21 subarrays (including the array itself). The formula for the number of subarrays in an array of length n is n * (n + 1) / 2. In this case, 6 * (6+1) / 2 = 21.
Q- one question you may ask now how you we end up compute all of this ? 
We can use two iteratore i and j 
Where i from 0 until n  (outer ) 
And j from i+1 until n  ( inner ) 
Compute each time the “currrent” sub_array → sum
And it’s length 
If current_sum==target : 
We check if current_length < global_val 
	We update global_val 
Return it 

Q- give your feedback about my naive solution here so ? 
**Feedback on Your Naive Solution for the Minimum Size Subarray Sum Problem**

Your solution is on the right track in terms of correctness, but there are critical considerations regarding its efficiency and scalability, especially given the problem's constraints. Here's a detailed breakdown:

### 1. **Correctness**

#### Strengths:
- **Comprehensive Enumeration**: By using two nested iterators (`i` and `j`), you are enumerating all possible subarrays, ensuring that every subarray is considered.
- **Condition Checking**: You're correctly checking if the `current_sum` of a subarray is greater than or equal to the target `k` and updating the `global_min_length` accordingly.

#### Considerations:
- Ensure that your condition checks and updates handle **edge cases**, such as when no subarray meets the criteria. You may need to return `0` or another sentinel value based on the problem's requirements.

### 2. **Time Complexity**

#### Current Approach:
- **Nested Loops**: Using two nested loops (`i` and `j`) results in a time complexity of **O(n²)**, where `n` is the length of the `nums` array.

#### Implications:
- **Scalability Issues**: Given the constraint that `nums.length` can be up to 10⁵, an O(n²) solution would result in up to **10¹⁰ operations** in the worst case. This is computationally infeasible and would lead to timeouts or excessive resource consumption.

### 3. **Space Complexity**

#### Current Approach:
- **Auxiliary Space**: If you're only using a few additional variables (like `current_sum`, `global_min_length`), the space complexity remains **O(1)**, which is optimal.

#### Considerations:
- Ensure that you’re not inadvertently using extra space (such as storing all subarrays), which could increase space usage.

### 4. **Optimization Recommendations**

To handle large input sizes efficiently, consider the following optimized approaches:

#### A. **Sliding Window Technique (Optimal Solution)**

**Overview**:
- Use a sliding window to maintain a current subarray that meets the sum condition.
- Expand the window to include more elements until the sum is `>= k`.
- Shrink the window from the left to find the minimal length that still satisfies the condition.

**Advantages**:
- **Time Complexity**: **O(n)** — Each element is visited at most twice (once when the window expands and once when it contracts).
- **Space Complexity**: **O(1)** — Requires only a few additional variables.

**Sample Implementation**:

```python
def min_subarray_len(k, nums):
    n = len(nums)
    min_length = n + 1
    current_sum = 0
    left = 0

    for right in range(n):
        current_sum += nums[right]
        
        while current_sum >= k:
            min_length = min(min_length, right - left + 1)
            current_sum -= nums[left]
            left += 1

    return min_length if min_length <= n else 0
```

**Explanation**:
- **Initialization**: Start with both `left` and `right` pointers at the beginning of the array.
- **Expansion**: Move the `right` pointer to include elements until `current_sum >= k`.
- **Contraction**: Move the `left` pointer to reduce the window size while maintaining `current_sum >= k`, updating `min_length` as you go.
- **Termination**: If `min_length` remains unchanged (i.e., no valid subarray found), return `0`.

#### B. **Binary Search with Prefix Sums**

**Overview**:
- Compute the prefix sums of the array.
- For each position, use binary search to find the smallest subarray that meets or exceeds the target `k`.

**Advantages**:
- **Time Complexity**: **O(n log n)**
- **Space Complexity**: **O(n)**

**Considerations**:
- This approach is slightly more complex but is another efficient way to solve the problem, especially useful in scenarios where the sliding window might not be applicable.

### 5. **Practical Recommendations**
- **Use the Sliding Window Technique**: Given its linear time complexity and constant space usage, it's the most suitable for large datasets as per the problem constraints.
- **Avoid Enumerating All Subarrays**: As demonstrated, the naive brute-force method is not feasible for large `n` due to its quadratic time complexity.
- **Consider Edge Cases**:
  - No subarray meets the sum condition.
  - The entire array sums to less than `k`.
  - Multiple subarrays of the same minimal length satisfy the condition.

### 6. **Final Thoughts**
While the naive approach is conceptually straightforward and ensures correctness by exhaustively checking all possibilities, it's essential to consider efficiency, especially with large input sizes. Optimizing your solution using techniques like the **sliding window** can significantly improve performance, making your solution both efficient and scalable.
Q- give the solution for my native idea 
class Solution:
    def minSubArrayLen(self, target: int, nums: List[int]) -> int:
       
Certainly! Below is the implementation of the naive solution using two nested loops to find the minimum length of a subarray whose sum is greater than or equal to the target `k`.

```python
from typing import List

class Solution:
    def minSubArrayLen(self, target: int, nums: List[int]) -> int:
        n = len(nums)
        min_length = float('inf')

        for i in range(n):
            current_sum = 0
            for j in range(i, n):
                current_sum += nums[j]
                if current_sum >= target:
                    min_length = min(min_length, j - i + 1)
                    break  # No need to extend the subarray further

        return min_length if min_length != float('inf') else 0

# Example usage:
solution = Solution()
print(solution.minSubArrayLen(7, [2, 3, 1, 2, 4, 3]))  # Output: 2
```

### Explanation:

1. **Initialization**:
   - `n` is the length of the input list `nums`.
   - `min_length` is initialized to infinity to ensure any valid subarray length will be smaller.

2. **Outer Loop (`i`)**:
   - Iterates through each starting index of potential subarrays.

3. **Inner Loop (`j`)**:
   - Iterates through each ending index of potential subarrays starting from `i`.
   - `current_sum` accumulates the sum of the subarray from `i` to `j`.

4. **Condition Check**:
   - If `current_sum` is greater than or equal to the target `k`, update `min_length` with the length of the current subarray (`j - i + 1`).
   - Break the inner loop since extending the subarray further won't yield a smaller length.

5. **Return Statement**:
   - If `min_length` remains infinity, it means no valid subarray was found, so return `0`.
   - Otherwise, return `min_length`.

### Considerations:
- This naive solution has a time complexity of **O(n²)**, which is not efficient for large input sizes.
- For large datasets, consider using the optimized sliding window technique as previously discussed.

### Example Usage:
The provided example `solution.minSubArrayLen(7, [2, 3, 1, 2, 4, 3])` correctly outputs `2`, as the subarray `[4, 3]` is the shortest subarray with a sum greater than or equal to `7`.
Q- now let’s try to optimize this solution using slide windows solution ? 

→ beasue this problem se do not have a fixed size windows but a dynamique windows that dynamically update it’s size based on it’s target search 
→ once we have reached the target we return it value 

Q- the first question how to build a dynamique slide window ? 
—> we can do this by keeping track of the lower bound and upper bound and dynamically remove or add a new element oky 

Q- did my approach of slide windows so far true ?