Difference of Number of Distinct Values on Diagonals - Solution & Explanation

Problem Overview: You are given an m x n grid. For every cell (i, j), compute the absolute difference between the number of distinct values on the top-left diagonal before that cell and the number of distinct values on the bottom-right diagonal after it. The result is stored in a matrix of the same size.

Approach 1: Diagonal Traversal with Precomputation (O(m*n) time, O(m*n) space)

Each cell belongs to exactly one top-left → bottom-right diagonal. Traverse every diagonal and compute two pieces of information: the count of distinct values seen before each position and the count of distinct values after it. Use a hash set while walking the diagonal from top-left to bottom-right to track distinct values encountered so far, storing prefix counts. Then walk the same diagonal in reverse to collect suffix distinct counts. The result for each cell is abs(prefixDistinct - suffixDistinct). This approach processes each element a constant number of times, giving O(m*n) time. Extra arrays for prefix/suffix information lead to O(m*n) space.

Diagonal traversal is a common pattern in matrix problems where elements share structural relationships. Using a hash table or set ensures distinct checks remain O(1) on average.

Approach 2: Simultaneous Diagonal Pass Method (O(m*n) time, O(min(m,n)) space)

Instead of storing prefix and suffix arrays, process each diagonal in a single controlled pass using two hash sets. Move along the diagonal and maintain a set of values already seen (top-left side). Before updating it, compute the number of distinct values remaining on the bottom-right side by maintaining another set or pre-counted structure. As you advance, shift elements from the future set to the past set and update the answer in place. Because each diagonal length is at most min(m, n), the auxiliary memory remains bounded by the diagonal size rather than the whole matrix.

This method keeps the same O(m*n) runtime but reduces auxiliary memory usage. It relies heavily on fast set operations and careful ordering of updates while iterating along the diagonal.

Problems like this combine array traversal with diagonal indexing tricks. The key insight: treat each diagonal independently instead of recalculating distinct counts from every cell.

Recommended for interviews: The diagonal traversal with precomputation approach is the safest explanation during interviews. It clearly demonstrates how you decompose the matrix into diagonals and track distinct values using hash sets. Once the interviewer sees the O(m*n) logic, mentioning the simultaneous-pass optimization shows deeper understanding of space tradeoffs.