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| 1 | +You are given a 2D 0-indexed integer array dimensions. |
| 2 | + |
| 3 | +For all indices i, 0 <= i < dimensions.length, dimensions[i][0] represents the length and dimensions[i][1] represents the width of the rectangle i. |
| 4 | + |
| 5 | +Return the area of the rectangle having the longest diagonal. If there are multiple rectangles with the longest diagonal, return the area of the rectangle having the maximum area. |
| 6 | + |
| 7 | + ------------------------------------------------------------------------------------------------ |
| 8 | + |
| 9 | +class Solution: |
| 10 | + def areaOfMaxDiagonal(self, dimensions: List[List[int]]) -> int: |
| 11 | + maxdiag = 0 |
| 12 | + maxres = 0 |
| 13 | + |
| 14 | + print(dimensions) |
| 15 | + for i in range(len(dimensions)): |
| 16 | + w,h = dimensions[i] |
| 17 | + |
| 18 | + |
| 19 | + diag = (w*w + h*h)**0.5 |
| 20 | + cur_res = w*h |
| 21 | + |
| 22 | + dimensions[i].extend([diag,cur_res]) |
| 23 | + |
| 24 | + |
| 25 | + dimensions.sort(key = lambda x: (x[2], x[3])) |
| 26 | + |
| 27 | + res = dimensions[-1][3] |
| 28 | + |
| 29 | + return res |
| 30 | + |
| 31 | +-------------------------------------------------- |
| 32 | + |
| 33 | +#other solution |
| 34 | + |
| 35 | +class Solution: |
| 36 | + def areaOfMaxDiagonal(self, dimensions: list[list[int]]) -> int: |
| 37 | + max_area, max_diag = 0, 0 |
| 38 | + |
| 39 | + for l, w in dimensions: |
| 40 | + curr_diag = l * l + w * w |
| 41 | + if curr_diag > max_diag or (curr_diag == max_diag and l * w > max_area): |
| 42 | + max_diag = curr_diag |
| 43 | + max_area = l * w |
| 44 | + |
| 45 | + return max_area |
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