Solution Explanation: Widest Pair of Indices With Equal Range Sum

This problem asks to find the widest pair of indices (i, j) in two binary arrays nums1 and nums2 such that the sum of elements in nums1 from index i to j is equal to the sum of elements in nums2 from index i to j. The "widest" pair is the one with the largest distance (j - i + 1).

The most efficient approach uses prefix sums and a hash table. Here's a breakdown:

1. Prefix Sum Difference Array:

Instead of calculating the range sums for nums1 and nums2 separately for each pair (i, j), we create a new array diff where diff[i] = nums1[i] - nums2[i]. Now, the condition "sum of elements in nums1 from i to j equals sum of elements in nums2 from i to j" translates to:

sum(diff[i...j]) == 0

This simplifies the problem to finding the longest subarray in diff with a sum of 0.

2. Hash Table for Prefix Sums:

We use a hash table (dictionary in Python) to store the prefix sums of diff. The keys are the prefix sums, and the values are the indices where those prefix sums first occur.

We initialize the hash table with prefixSum = 0 at index -1. This handles the case where the entire array from the start has a sum of 0.

We iterate through diff:

• For each element diff[i], we update the current prefixSum: prefixSum += diff[i].
• We check if this prefixSum already exists in the hash table. If it does, it means there's a subarray between the previous occurrence of prefixSum and the current index i that sums to 0. We update the maxDistance accordingly.
• If prefixSum is not in the hash table, we add it with its current index i.

3. Time and Space Complexity:

• Time Complexity: O(n), where n is the length of the input arrays. We iterate through the arrays once.
• Space Complexity: O(n) in the worst case, to store the prefix sums in the hash table. If all prefix sums are unique, the hash table will grow linearly with the input size.

Code Implementation (Python):

def widestPairOfIndices(nums1, nums2):
    diff = [a - b for a, b in zip(nums1, nums2)]
    prefix_sums = {0: -1}  # Initialize with prefix sum 0 at index -1
    max_distance = 0
    current_sum = 0
    for i, num in enumerate(diff):
        current_sum += num
        if current_sum in prefix_sums:
            distance = i - prefix_sums[current_sum]
            max_distance = max(max_distance, distance)
        else:
            prefix_sums[current_sum] = i
    return max_distance
 

This Python code directly implements the algorithm described above. Other languages (Java, C++, Go, etc.) would follow a similar structure, using their respective hash table implementations (HashMap, unordered_map, map). The core logic remains the same.