Wind-Blown Leaves Remaining in a Grid
Abridge● EasyNEW GRADOAProblem statement
A garden is a grid. grid[i][j] is the number of leaves in the cell at row i and column j. Row 0 is the top row and column 0 is the left column.
The string winds lists gusts of wind in order. Each gust moves every leaf still in the grid exactly one cell:
Umoves leaves one row up.Dmoves leaves one row down.Lmoves leaves one column left.Rmoves leaves one column right.
All leaves move at the same time during a gust. A leaf that is moved outside the grid is gone for good, even if a later gust blows the other way.
Return the total number of leaves left in the grid after all gusts.
Interviewer Follow-up
Why should each gust build a new grid instead of updating the grid in place? What can go wrong with an in-place update? This follow-up is for discussion; the tests check only the returned total.
Function
remainingLeaves(grid: int[][], winds: String) → longExamples
Example 1
grid = [[1,2,3],[4,5,6]]winds = "R"return = 12Every leaf moves one column right. The leaves in the last column, 3 and 6, leave the grid. The rest stay: 1 + 2 + 4 + 5 = 12.
Example 2
grid = [[0,5,0],[0,0,0],[7,0,0]]winds = "DL"return = 5After D, the 7 leaves in the bottom row are blown out, and the 5 leaves move to row 1, column 1. After L, they move to row 1, column 0, which is still inside. So 5 leaves remain.
Example 3
grid = [[4,1],[2,3]]winds = "LR"return = 4The first wind L blows the left column (4 and 2) out of the grid. The second wind R moves the other leaves back to the right column. Leaves that left do not come back, so 1 + 3 = 4 remain.
Constraints
1 <= grid.length <= 501 <= grid[i].length <= 50, and every row has the same length.0 <= grid[i][j] <= 10^60 <= winds.length <= 2000- Each character of
windsisU,D,LorR.