Matrix Traversal and Logic Flashcards
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Read the first 7 Matrix Traversal and Logic flashcards as text
You apply a 'game of life' update where cells change state simultaneously based on neighbor counts. To do this in-place, you use encoded states (e.g., 2 = was dead, now alive). Why is encoding necessary?
Answer: So that reading a cell's original state is still possible after partially updating the matrix
Encoding original state into intermediate values lets later-processed cells determine what their neighbors' states were before any updates.
You have an N×N matrix and want to rotate it 180 degrees. Which sequence of operations achieves this?
Answer: Rotate 90° clockwise twice using transpose and row-reversal
Two successive 90° clockwise rotations (each via transpose + row-reversal) produce a 180° rotation.
In the 'matrix chain multiplication' problem, dp[i][j] stores the minimum number of scalar multiplications to compute the product of matrices i through j. What is the time complexity of the standard DP solution?
Answer: O(N^3)
The DP fills an N×N table where each entry requires O(N) work to evaluate all split points, giving O(N^3) total.
Given a binary matrix, you want the side length of the largest square containing only 1s. The DP recurrence is dp[i][j] = min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1]) + 1 when matrix[i][j] == 1. What does dp[i-1][j-1] represent?
Answer: The largest square whose bottom-right corner is at (i-1, j-1)
dp[i-1][j-1] is the size of the largest all-1 square with its bottom-right corner at position (i-1, j-1).
When performing an anti-diagonal traversal of a matrix (from top-right to bottom-left), elements on the same anti-diagonal share which property?
Answer: Their row + column sum is constant
Cells on the same anti-diagonal satisfy row + col = constant, grouping them by this sum enables traversal.
You want to check if a matrix is symmetric (equal to its transpose). What is the minimum number of element comparisons needed for an N×N matrix?
Answer: N*(N-1)/2
Only upper (or lower) triangular elements excluding the diagonal need to be compared against their mirror: N*(N-1)/2 comparisons.
In a 'flood fill' algorithm starting from cell (r, c) with a new color, what condition must be checked before enqueuing a neighbor to avoid infinite loops?
Answer: The neighbor must have the original color and not yet be in the queue/visited
Only cells matching the original color and not yet visited should be enqueued; checking both conditions prevents revisiting and incorrect fills.