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The key move is realizing you only care about m = min(rows, cols) and then it collapses to m*(m+1)/2.
First, clarify the problem: for each query (rows, cols), consider the smaller square sub-rectangle of size min(rows, cols) x min(rows, cols) anchored at the top-left corner. The count of cells on or above the main diagonal in an n x n square is n(n+1)/2. So for each query, compute n = min(rows, cols) and return n(n+1)/2. This yields an O(1) per query solution after O(1) preprocessing.
Pro tip: Mention that the formula n(n+1)/2 counts the cells on or above the diagonal because it's the sum of the first n integers. Also, note that using 64-bit integers is crucial to avoid overflow for large n, and that the answer is independent of the larger dimension.
Confirm that the sub-rectangle is the top-left square of size min(rows, cols) and that 'on or above the main diagonal' includes the diagonal cells. Ask if there are any constraints on the number of queries or the maximum dimensions.
For an n x n square, the number of cells on or above the main diagonal is the sum of the first n positive integers: 1 + 2 + ... + n = n(n+1)/2. Explain why this works by counting row by row.
For each query (rows, cols), compute n = min(rows, cols). Then the answer is n*(n+1)//2. This is O(1) per query, so total time O(Q).
If n = 0, the answer is 0. For large n (e.g., up to 10^9), use 64-bit integers (long long in C++ or Python's arbitrary precision) to avoid overflow. Mention that the formula works for any non-negative integer n.
Time complexity: O(Q) for Q queries, O(1) per query. Space complexity: O(1) extra space. This is optimal since each query must be read.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.