598. Range Addition II

The idea of this solution is pretty simple, the positions with the max value will be in the sub-matrix of range 0 to minimum y pos, and 0 to minimum x pos.

If you don’t understand look at the example bellow:

input: m = 6, n = 5, ops = [[4, 3], [2, 6], [3, 5]] (Note: They give the input 1-indexed, instead of 0-indexed)

As you can see, all rectangles top left corner is (0, 0) (I added an orange dot at (0, 0)). And the sub-matrix that has the maximum values is surrounded by the yellow dotted line.

We have four parts of the dotted line, the top, bottom, right, and left.

  • The top line is on the x-axis, so we don’t have to worry about that.
  • The left line is on the y-axis, so we also don’t have to worry about that.
  • The right line is the minimum x value in all the ops because the minimum is the max position that we know we have all the operations in.
  • The bottom line is the minimum y value in ops, and it is that for the same reason as the right line.
func maxCount(m int, n int, ops [][]int) int {
    x, y := m, n
    for _, op := range ops {
        x = min(op[0], x)
        y = min(op[1], y)
    }
    return x * y
}

func min(a, b int) int {
    if a > b {
        return b
    }
    return a
}
Go