Problem Description

In this problem, we have two binary arrays nums1 and nums2. Both arrays are of the same length and contain only 0s and 1s. Our goal is to find the widest pair of indices (i, j) that satisfy two conditions:

1. i must be less than or equal to j, and
2. The sum of elements from i to j in nums1 must be equal to the sum of elements from i to j in nums2.

The concept of the "widest" pair refers to the pair (i, j) with the biggest difference between i and j. This means that we are looking for the longest subarray where the sums of nums1 and nums2 are equal. The "distance" indicates the number of elements in this subarray, which is calculated as j - i + 1.

We need to return the "distance" of the widest pair of indices. If no pairs meet the condition, we should return 0.

Intuition

To find the solution, we use the cumulative sums of both arrays and keep track of their differences at each index. If the difference between the sums of subarrays from nums1 and nums2 up to the current index i has been seen before at some index j, then the subarrays from j+1 to i in both nums1 and nums2 must be equal.

Here are the steps to solve this problem:

1. Maintain a dictionary d to store the first occurrence of each cumulative sum difference with its index. Initialize this dictionary with {0: -1}, meaning that if the difference is 0 at index 0, it can be considered as a subarray starting from index 0.

2. Iterate through the two arrays nums1 and nums2 simultaneously using enumerate(zip(nums1, nums2)) to get both the index and the elements.

3. Calculate the cumulative difference s by subtracting the element of nums2 from the element of nums1 and adding it to the previous cumulative difference.

4. Check if this difference s has occurred before by looking it up in the dictionary d. If it exists, it means that there is a subarray that starts from the stored index plus 1 and ends at the current index i which has equal sums in nums1 and nums2. Update ans with the maximum distance found so far, which is i - d[s].

5. If the difference s does not exist in the dictionary, it means this is the first time we've seen this cumulative sum difference, so we add the current index i to the dictionary with the key being the sum difference s.

6. After traversing through the whole arrays, the variable ans will contain the distance of the widest pair of indices. Return ans.

Learn more about Prefix Sum patterns.

Solution Approach

The solution to this problem utilizes a dictionary data structure for efficient lookups and an iterative approach to find the widest pair where the subarrays have equal sums. Here's a deep dive into how the Solution class implements this:

1. Dictionary for Cumulative Difference Tracking:

   • The dictionary d is used to store cumulative sum differences as keys and their first occurring indices as values. It's initialized with {0: -1} to handle cases where the subarray starts from the first element.

2. Running Sum Difference Calculation:

   • As we loop through each element of both arrays, we calculate a running sum difference s by adding nums1[i] - nums2[i] to the current sum difference. This represents the difference between the cumulative sums up to the current index i.

3. Lookup and Widest Pair Tracking:

   • During each iteration, we check if the current sum difference s is found in dictionary d.
     • If it exists, it indicates we have previously seen this sum difference at an earlier index, meaning the sums of subarrays from the previous index to the current index in both nums1 and nums2 are equal.
     • The distance of the current widest pair is calculated by subtracting the earlier index where the sum difference was the same from the current index, i.e., i - d[s].
     • We update ans to the maximum of the current distance and the existing widest distance found so far.
   • If the sum difference s is not found in d, it implies this difference is encountered the first time at index i, so we add s as a key to d with the current index i as its value. This will potentially serve as the starting index for a future widest pair.

4. Returning the Result:

   • After the loop is completed, ans holds the distance of the widest pair found. This value is returned as the output of the function.

The algorithm's complexity is primarily dictated by the single traversal of the input arrays, leading to an overall time complexity of O(n) where n is the number of elements in the input arrays. The space complexity is O(k), where k is the number of unique sum differences, which can be at most n in the worst case.

The solution is efficient because it avoids nested loops by using the cumulative sum difference, which allows us to identify equal sum subarrays in linear time.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach using two binary arrays nums1 and nums2.

Initial State:

• Let nums1 = [1, 0, 0, 1] and nums2 = [0, 1, 1, 0].
• Initialize the dictionary d with {0: -1} to handle the case where the sum difference starts from index 0.

Step-by-Step Solution:

1. Start iterating over both arrays. Initialize a variable s to keep track of the cumulative sum difference, and ans to store the maximum width found so far.

2. Iteration 0: The elements at index 0 are nums1[0] = 1 and nums2[0] = 0.

   • The sum difference s becomes 1 - 0 = 1.
   • This difference is not in d, so we add it: d[1] = 0.

3. Iteration 1: The elements at index 1 are nums1[1] = 0 and nums2[1] = 1.

   • The sum difference s becomes 1 + (0 - 1) = 0, which is already in d.
   • The width between d[0] + 1 = 0 and index 1 is 1 - 0 + 1 = 2.
   • Update ans to 2.

4. Iteration 2: The elements at index 2 are nums1[2] = 0 and nums2[2] = 1.

   • The sum difference s remains 0 (since 0 - 0 + 0 - 1 = -1 and we add this to the previous s which was 1, making the new s back to 0).
   • We have already seen 0 in d.
   • The width is 2 - (-1) + 1 = 4.
   • Update ans to 4, since this is larger than the previous value of ans.

5. Iteration 3: The elements at index 3 are nums1[3] = 1 and nums2[3] = 0.

   • The sum difference s becomes 0 + (1 - 0) = 1, which is already in d.
   • We do not update ans because the current distance 3 - d[1] = 3 is not greater than the current ans (which is 4).

6. After traversing all items, ans is 4, which represents the distance of the widest pair of indices where the sum of elements in nums1 is equal to the sum of elements in nums2.

The solution finds the widest subarray efficiently, utilizing a dictionary to track the first occurrence of each cumulative sum difference and iterating over the arrays only once.

Solution Implementation

Time and Space Complexity

The given Python function widestPairOfIndices seeks to find the widest pair of indices (i, j) such that sum(nums1[k]) for k in range(0, i + 1) is equal to sum(nums2[k]) for k in range(0, i + 1). It accomplishes this by using a dictionary d to keep track of the differences between the prefixes of the two lists and when each difference was first seen.

Time Complexity:

The time complexity of the function can be analyzed as follows:

• There is a single loop that iterates over the elements of nums1 and nums2, which both have the same length, say n.
• Within each iteration, the function performs constant-time operations: a calculation involving a and b, an if condition check, a dictionary lookup, and an update to the dictionary (which on average is constant time for a hash table) and variable for the answer ans.
• These operations within the loop are conducted in constant time, regardless of the size of the input.

As the loop runs n times where n is the length of the input lists, and each iteration is done in constant time, the time complexity of the function is O(n).

Space Complexity:

The space complexity of the function depends on the auxiliary space used, which is mainly for storing the dictionary d:

• In the worst-case scenario, the dictionary could store a distinct entry for each element in the input lists if all prefix sums were unique. Thus, in the worst case, it could have up to n entries.
• Aside from this, the space used by variables s and ans is constant, as well as a few other intermediate variables.

Therefore, the worst-case space complexity is O(n) where n is the length of the input lists.

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