The "Widest Pair of Indices With Equal Range Sum" problem gives you two binary arrays of the same length. You need to find the maximum distance between two indices i and j (where i <= j) such that the sum of elements from i to j in the first array is equal to the sum of elements from i to j in the second array. Essentially, you are looking for the longest subarray where both arrays have the same total.
Microsoft and other top-tier companies ask this to test your ability to simplify complex requirements. Comparing two different arrays might seem like it requires O(N^2) checks. However, by transforming the problem into finding the same prefix sum difference twice, you can solve it in O(N). This evaluates your skill in using Hash Tables and Prefix Sums for optimization.
The primary pattern is Prefix Sums combined with a Hash Map. Let diff[i] be the difference between the prefix sums of the two arrays up to index i. If diff[i] == diff[j], then the sum of elements between i+1 and j must be the same in both arrays. By storing the first occurrence of each difference in a hash map, you can calculate the distance to any subsequent occurrence of the same difference.
Array A: [0, 1, 0], Array B: [1, 1, 0].
{ -1: 0 }.i, the subarray from the very beginning (index 0) to i is a valid candidate. You should initialize your map with {0: -1}.j - i or j - i + 1.Whenever you are asked to compare sums of two arrays or find subsegments with equal counts/sums, try subtracting the values or tracking their difference. This often reduces the problem to "Find two indices where the cumulative value is the same," which is a classic Hash Map pattern.
| Title | Difficulty | Topics | LeetCode |
|---|---|---|---|
| Stable Subarrays With Equal Boundary and Interior Sum | Medium | Solve | |
| Count of Interesting Subarrays | Medium | Solve | |
| Maximum Size Subarray Sum Equals k | Medium | Solve | |
| Make Sum Divisible by P | Medium | Solve | |
| Subarray Sums Divisible by K | Medium | Solve |