|
| 1 | +# Time: O(1) |
| 2 | +# Space: O(1) |
| 3 | + |
| 4 | +class Solution(object): |
| 5 | + def maximumNumberOfOnes(self, width, height, sideLength, maxOnes): |
| 6 | + """ |
| 7 | + :type width: int |
| 8 | + :type height: int |
| 9 | + :type sideLength: int |
| 10 | + :type maxOnes: int |
| 11 | + :rtype: int |
| 12 | + """ |
| 13 | + if width < height: |
| 14 | + width, height = height, width |
| 15 | + |
| 16 | + # 1. split matrix by SxS tiles |
| 17 | + # 2. split each SxS tile into four parts |
| 18 | + # (r, c), (r, S-c), (S-r, c), (S-r, S-c) |
| 19 | + # 3. for each count of tile part in matrix is |
| 20 | + # (R+1)*(C+1), (R+1)*C, R*(C+1), R*C (already in descending order) |
| 21 | + # 4. fill one into matrix by tile part of which count is in descending order |
| 22 | + # until number of ones in a tile comes to maxOnes |
| 23 | + # |
| 24 | + # ps. area of a tile and its count in matrix are as follows: |
| 25 | + # |
| 26 | + # |<---- c ---->|<-- S-c -->| |
| 27 | + # ^ | | |
| 28 | + # | | | |
| 29 | + # r (R+1)*(C+1) | (R+1)*C | |
| 30 | + # | | | |
| 31 | + # v | | |
| 32 | + # --------------------------- |
| 33 | + # ^ | | |
| 34 | + # | | | |
| 35 | + # S-r R*(C+1) | R*C | |
| 36 | + # | | | |
| 37 | + # v | | |
| 38 | + # --------------------------- |
| 39 | + # |
| 40 | + |
| 41 | + R, r = divmod(height, sideLength) |
| 42 | + C, c = divmod(width, sideLength) |
| 43 | + assert(R <= C) |
| 44 | + area_counts = [(r*c, (R+1)*(C+1)), \ |
| 45 | + (r*(sideLength-c), (R+1)*C), \ |
| 46 | + ((sideLength-r)*c, R*(C+1)), \ |
| 47 | + ((sideLength-r)*(sideLength-c), R*C)] |
| 48 | + result = 0 |
| 49 | + for area, count in area_counts: |
| 50 | + area = min(maxOnes, area) |
| 51 | + result += count*area |
| 52 | + maxOnes -= area |
| 53 | + if not maxOnes: |
| 54 | + break |
| 55 | + return result |
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