The Range Addition II problem gives you an m×n matrix initialized to zeros and a list of operations: each operation adds 1 to the rectangle [0,0] → [r-1,c-1]. Find the maximum value in the matrix after all operations. The key insight: the maximum is always the count of operations that include position (0,0). The array and math interview pattern is demonstrated.
IXL asks this easy problem to test whether candidates recognize that the maximum value cell is always (0,0) — since every operation includes it, and any cell farther from the origin is included in fewer operations. The answer is simply the number of operations, bounded by min-row and min-column.
Mathematical observation. The maximum value = number of operations that cover cell (0,0). Since all operations start at (0,0), every operation covers (0,0). The cell (0,0) is covered by ALL operations. The maximum = total operations. But actually: operations with r=0 or c=0 cover only row 0 or column 0... Wait: each operation covers [0..a-1][0..b-1]. So position (0,0) is always covered. Maximum = count of all operations.
Actually the overlap at (0,0) is always the total number of operations. The answer is len(ops) if ops is non-empty, else 0. But more precisely: the maximum count is at (0,0) = intersection of all operation rectangles: min(ops[i][0]) * min(ops[i][1]) cells share the maximum value.
Answer = count(ops). Wait — the maximum VALUE is len(ops), and it appears at min_row × min_col cells.
m=3, n=3, ops=[[2,2],[3,3]]. All operations include (0,0). Max value = 2 (both ops cover it). Max value cells = min(2,3)×min(2,3) = 4. But the question asks for the MAX VALUE = 2.
Range Addition II demonstrates that mathematical insight beats simulation. The maximum cell is always at (0,0) because it's included in every operation. Max value = number of operations. This observation reduces O(mnops) simulation to O(1). Practice identifying similar "always at the origin" invariants in range operations.
