Problem Description

Given a binary array, which consists of only 0's and 1's, and a target integer goal, the task is to find the number of subarrays whose sum equals goal. A subarray is defined as any continuous sequence of the array. The binary nature of the array means any sum of its subarray will also be an integer, making the problem about finding subarray sums that match the given integer.

Intuition

The solution employs a two-pointer approach that creates a sliding window of variable size over the array. Two pointers, i1 and i2, are moved along the array to adjust the window size. Pointer j is used to extend the window by moving right. The variables s1 and s2 are used to track the sums of numbers within the window.

• We use two sliding windows: one to find the number of subarrays with sum just over goal (s1 and i1), and another to find the number of subarrays with sum equal to or just over goal (s2 and i2).
• As we move the main pointer j through the array, we increase s1 and s2 by the new element included in the window nums[j].
• Next, we need to shrink the windows if necessary. If s1 is greater than goal, we move i1 to reduce the window and the sum. Similarly, if s2 is greater or equal than goal, we move i2.
• The difference between i2 and i1 gives us the number of new subarrays that have a sum equal to goal considering the new element nums[j].
• We increment ans by this difference since i2 - i1 represents the number of valid subarrays ending at j with sum exactly goal.
• We repeat this process for every element of the array by incrementing j, thus exploring every potential subarray.

Why does this work? The two sliding windows track the lower and upper bounds of sums around our goal. By taking the difference of the counts, we effectively count only those subarrays whose sum is exactly goal. Since we consider each subarray ending at various j positions, we can ensure that all possible subarrays are counted.

With every increment of j, we essentially say, "Given all the subarrays ending at j, count how many have a sum of goal." The sliding of i1 and i2 keeps the window's sum around goal and accurately maintains the count.

This approach is efficient since it requires only a single pass through the array, leading to a time complexity of O(n), where n is the length of the input array nums.

Learn more about Prefix Sum and Sliding Window patterns.

Solution Approach

The solution to the subarray sum problem implements a variation of the two-pointer technique which helps to optimize the search for subarrays that add up to the goal. Here's a step-by-step explanation:

1. Initialize two pairs of pointers i1, i2 and their corresponding sums s1, s2 to 0. These will be used to manage two sliding windows. Also, initialize a pointer j to extend the window and ans to accumulate the number of valid subarrays found.
2. Iterate over the array with the main pointer j. For each element nums[j] being considered:
   • Add nums[j] to both s1 and s2, which represents attempting to add the current element to our current subarray sums.
3. If s1 exceeds the goal, shrink the first window:
   • Subtract nums[i1] from s1 and increment i1 to reduce the window from the left, doing this until s1 is no longer greater than goal.
4. Similarly, if s2 is at least the goal, shrink the second window:
   • Subtract nums[i2] from s2 and increment i2 to reduce the window from the left, doing this until s2 is smaller than goal.
5. After adjusting both windows, calculate the number of new subarrays found:
   • The difference i2 - i1 tells us how many valid subarrays end at j with the desired sum goal because those will be the subarrays contained within the window tracked by i2 but not yet by i1.
   • Add this difference to ans which tallies our result.
6. Repeat steps 2 through 5 as you move the pointer j to the right until you've processed every element in the array.
7. Once the iteration is complete, ans holds the final count of subarrays whose sum is equal to goal.

This code efficiently finds all subarrays with the desired sum in linear time, utilizing O(1) extra space, excluding the input and output. The key ingredients of the solution are the two-pointer technique and a single pass, avoiding unnecessary recomputation.

No complex data structures are needed because the pointers and sums effectively manage the windows of potential subarrays. The algorithm iteratively adjusts these windows to find all valid subarrays in the most optimized manner.

Example Walkthrough

Let’s consider the binary array nums = [1, 0, 1, 0, 1] and the target integer goal = 2. We need to find the number of subarrays with the sum equal to goal.

1. We initialize pointers and sums: i1 = i2 = 0, s1 = s2 = 0, and ans = 0.
2. Start iterating with pointer j from left to right. The main goal is to add nums[j] to both s1 and s2 and then adjust the windows with pointers i1 and i2.

Using the array provided, here's how the solution progresses:

• For j = 0 (the first element is 1):
  • s1 = s2 = 1. No window adjustment needed since neither s1 nor s2 is over the goal yet.
• For j = 1 (the second element is 0):
  • s1 = s2 = 1. Still no adjustment as we are not over the goal.
• For j = 2 (the third element is 1):
  • s1 = s2 = 2. Since s2 >= goal, we increment i2. Now, i2 = 1 and s2 = 1 (s1 remains 2 because we have to exceed goal to adjust i1).
  • i2 - i1 = 1. We found one valid subarray [1, 0, 1] and increment ans by 1.
• For j = 3 (the fourth element is 0):
  • s1 = 2 and s2 = 1. No adjustment required.
• For j = 4 (the fifth element is 1):
  • s1 = s2 = 2. Both sums are equal to the goal.
  • Since s1 == goal, we increment i1. Now, i1 = 1 and s1 = 1.
  • Similarly, i2 also increments because s2 >= goal. Now, i2 = 3 and s2 = 0.
  • i2 - i1 = 2. The two new valid subarrays are [1, 0, 1] and [0, 1] ending at j=4. So we increment ans by 2.

After iterating through the array, we’ve found ans = 3 valid subarrays ([1, 0, 1], [1, 0, 1], and [0, 1]) where the sum matches the goal.

It’s important to understand that the same subarray may be counted at different stages depending on the j position. This method ensures that every unique subarray is considered without double counting, and the result is achieved with a single traversal.

Solution Implementation

Time and Space Complexity

The code provided uses two pointers technique to keep track of subarrays with sums equal to or just above the goal. The time complexity of the code is O(n) because the while loop runs for each element in the array (n elements), and inside the loop, each pointer (i1 and i2) moves forward (never backwards), meaning each element is processed once.

The space complexity of the code is O(1) as the space used does not grow with the input size n. The variables i1, i2, s1, s2, j, ans, and n use a constant amount of space.

Learn more about how to find time and space complexity quickly using problem constraints.