Solution Explanation for Binary Subarrays With Sum

The problem asks to find the number of non-empty subarrays with a sum equal to a given goal. Two efficient solutions are presented below: using a prefix sum and a sliding window approach.

Solution 1: Prefix Sum with Hash Table

This solution leverages the concept of prefix sums and a hash table (dictionary in Python) to efficiently count subarrays.

Approach:

1. Prefix Sum: We maintain a prefix sum s. As we iterate through the array, s accumulates the sum of elements encountered so far.
2. Hash Table: A hash table cnt stores the frequency of each prefix sum encountered. We initialize cnt[0] = 1 because an empty subarray has a sum of 0.
3. Counting Subarrays: For each element v, we update the prefix sum s += v. Then, we check if s - goal exists as a key in cnt. If it does, it means there's a previous prefix sum that, when added to a subarray ending at the current position, results in the goal sum. The number of such subarrays is given by cnt[s - goal]. We add this count to the ans.
4. Updating Frequency: Finally, we increment the frequency of the current prefix sum s in cnt.

Time Complexity: O(N), where N is the length of the input array nums. We iterate through the array once. Hash table operations (insertion and lookup) take O(1) on average.

Space Complexity: O(N) in the worst case, as the hash table might store all unique prefix sums.

Code (Python):

from collections import Counter
 
class Solution:
    def numSubarraysWithSum(self, nums: List[int], goal: int) -> int:
        cnt = Counter({0: 1})  # Initialize with empty subarray sum
        ans = s = 0
        for v in nums:
            s += v
            ans += cnt[s - goal]  # Count subarrays with sum equal to goal
            cnt[s] += 1  # Update frequency of current prefix sum
        return ans

Other languages (Java, C++, Go, JavaScript) follow a very similar structure, utilizing their respective hash table implementations.

Solution 2: Sliding Window Approach

This approach uses two sliding windows to count subarrays with sums greater than or equal to the goal.

Approach:

1. Two Pointers: We maintain two pointers, i1 and i2, representing the start of two sliding windows. i1 tracks the window with a sum strictly greater than goal, and i2 tracks the window with a sum greater than or equal to goal.
2. Iterating and Adjusting Windows: We iterate through the array using a j pointer. For each element, we update the sums s1 (for i1) and s2 (for i2). We adjust i1 and i2 to maintain the conditions on their respective sums.
3. Counting Subarrays: The difference i2 - i1 represents the number of subarrays ending at position j with a sum equal to goal. We add this difference to the ans.

Time Complexity: O(N), as we iterate through the array once. The inner while loops do not increase the time complexity beyond linear.

Space Complexity: O(1). We only use a few constant extra variables.

Code (Python):

class Solution:
    def numSubarraysWithSum(self, nums: List[int], goal: int) -> int:
        i1 = i2 = s1 = s2 = j = ans = 0
        n = len(nums)
        while j < n:
            s1 += nums[j]
            s2 += nums[j]
            while i1 <= j and s1 > goal:
                s1 -= nums[i1]
                i1 += 1
            while i2 <= j and s2 >= goal:
                s2 -= nums[i2]
                i2 += 1
            ans += i2 - i1
            j += 1
        return ans
 

Again, other languages have very similar implementations using the sliding window technique. Both solutions offer efficient and optimized ways to solve this problem, with the prefix sum approach generally being slightly easier to understand. The sliding window approach might offer a marginal performance advantage in specific cases.