My first instinct was brute force, check every subarray and track max/min.
Use a sliding window (two pointers) to maintain a window where the difference between the maximum and minimum elements is at most the limit. Efficiently track the max and min within the window using monotonic deques, expanding the right pointer and shrinking the left pointer when the condition is violated. Keep track of the maximum window length seen.
Pro tip: Emphasize that the monotonic deque approach gives O(n) time, which is optimal, and discuss how it handles duplicates and edge cases like empty arrays or limit=0. Mention that this is a common pattern for problems involving subarray constraints.
Restate the problem in your own words and confirm edge cases: empty array, single element, negative numbers, and limit=0. Ask if the array can be modified or if extra space is allowed.
Mention that a brute force approach would check all subarrays, compute max and min for each, resulting in O(n^2) or O(n^3) time. This shows you understand the baseline and the need for optimization.
Explain that you'll maintain a window [left, right] and use two deques: one for maximums (decreasing order) and one for minimums (increasing order). Expand right, update deques, and while max-min > limit, shrink from left and update deques accordingly.
Trace the algorithm on a small example, such as [8,2,4,7] with limit=4, to demonstrate how the window expands and contracts and how the deques are updated. This validates your approach.
State that each element is added and removed from each deque at most once, giving O(n) time and O(n) space in the worst case. Discuss edge cases like empty array, limit negative, or all elements equal.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
Use dynamic programming to compute the size of the largest square ending at each cell, then return the square of the maximum size. Explain the recurrence: dp[i][j] = min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1]) + 1 if matrix[i][j] == 1, else 0. Optimize space to O(n) by keeping only the previous row.
Pro tip: After presenting the DP solution, mention that you can optimize space to O(n) by using a single array and a variable to track the previous diagonal value. This shows you think about efficiency beyond the basic solution.
Ask clarifying questions: Is the matrix guaranteed to be rectangular? Can it be empty? What are the constraints on dimensions? Confirm that the square must be axis-aligned and contain only 1s.
Mention that a brute force approach would check every possible square, which is O(m*n*min(m,n)^2) or worse. This sets the stage for optimization.
Explain that the largest square ending at (i,j) depends on the largest squares ending at (i-1,j), (i,j-1), and (i-1,j-1). The recurrence is dp[i][j] = min(three neighbors) + 1 if cell is 1, else 0.
Write code for the DP solution, then optimize space to O(n) by using a 1D array and tracking the previous diagonal value. Analyze time complexity O(m*n) and space O(n).
Walk through a small example, test edge cases like empty matrix, all 0s, all 1s, and single row/column. Verify the recurrence and return the area (max side squared).
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.