Solution Explanation

This problem asks for the kth smallest subarray sum. A naive approach would involve generating all subarrays, calculating their sums, sorting them, and then selecting the kth element. This is inefficient, with a time complexity that's at least O(n³), where n is the length of the input array nums. A much more efficient solution utilizes binary search and a helper function to count subarrays with sums less than or equal to a given value.

Approach:

1. Binary Search: The core idea is to perform a binary search on the range of possible subarray sums. The minimum possible sum is the minimum element in nums, and the maximum possible sum is the total sum of all elements in nums.

2. Counting Function (f): A crucial helper function f(s) counts the number of subarrays whose sum is less than or equal to s. This function uses a sliding window technique. It iterates through the array, maintaining a window whose sum is less than or equal to s. When the window sum exceeds s, the left side of the window is moved to reduce the sum. The cnt variable keeps track of the total number of such subarrays found.

3. Binary Search Logic: The binary search repeatedly checks a midpoint mid as a potential kth smallest sum. If f(mid) (the count of subarrays with sums ≤ mid) is greater than or equal to k, it means the kth smallest sum is less than or equal to mid, so we search in the lower half. Otherwise, the kth smallest sum must be greater than mid, and we search in the upper half.

4. Result: The binary search continues until l and r converge, at which point l holds the kth smallest subarray sum.

Time Complexity Analysis:

• The binary search takes O(log(sum(nums))) steps.
• The f(s) function takes O(n) time in the worst case (it iterates through the array once).
• Therefore, the overall time complexity is O(n log(sum(nums))). This is significantly better than the naive O(n³) approach.

Space Complexity Analysis:

The space complexity is O(1) because the algorithm uses only a constant amount of extra space.

Code Explanation (Python):

class Solution:
    def kthSmallestSubarraySum(self, nums: List[int], k: int) -> int:
        def f(s):
            t = j = 0  # t: current window sum, j: left pointer of the window
            cnt = 0  # count of subarrays with sum <= s
            for i, x in enumerate(nums):
                t += x
                while t > s:  # Shrink the window if sum exceeds s
                    t -= nums[j]
                    j += 1
                cnt += i - j + 1  # Add the count of subarrays ending at index i
            return cnt
 
        l, r = min(nums), sum(nums)  # Initialize the search range
        # Using bisect_left for efficiency instead of a while loop
        return l + bisect_left(range(l, r + 1), True, key=f)
 

The Python code efficiently uses bisect_left from the bisect module to perform the binary search, making the code more concise and potentially faster than an explicit while loop implementation. The key=f argument tells bisect_left to use the result of the f function as the comparison key during the search.

The other languages (Java, C++, Go) follow a similar logic, implementing the binary search and the counting function explicitly using while loops. The core algorithm remains the same across all languages.