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5 | 5 |
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6 | 6 | public class _764 {
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7 | 7 | public static class Solution1 {
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8 |
| - /** |
9 |
| - * Brute force |
10 |
| - * <p> |
11 |
| - * Time: O(N^3) |
12 |
| - * Space: O(mines.length) |
13 |
| - */ |
14 |
| - public int orderOfLargestPlusSign(int N, int[][] mines) { |
15 |
| - Set<Integer> banned = new HashSet<>(); |
16 |
| - for (int[] mine : mines) { |
17 |
| - banned.add(mine[0] * N + mine[1]); |
18 |
| - } |
19 |
| - int result = 0; |
20 |
| - for (int row = 0; row < N; row++) { |
21 |
| - for (int col = 0; col < N; col++) { |
22 |
| - int k = 0; |
23 |
| - while (k <= row && row < N - k && k <= col && col < N - k |
24 |
| - && !banned.contains((row - k) * N + col) |
25 |
| - && !banned.contains((row + k) * N + col) |
26 |
| - && !banned.contains(row * N + col - k) |
27 |
| - && !banned.contains(row * N + col + k)) { |
28 |
| - k++; |
29 |
| - } |
30 |
| - result = Math.max(result, k); |
31 |
| - } |
32 |
| - } |
33 |
| - return result; |
34 |
| - } |
35 |
| - } |
36 |
| - |
37 |
| - public static class Solution2 { |
38 | 8 | /**
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39 | 9 | * Dp
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40 | 10 | * <p>
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@@ -82,4 +52,81 @@ public int orderOfLargestPlusSign(int N, int[][] mines) {
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82 | 52 | return result;
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83 | 53 | }
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84 | 54 | }
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| 55 | + |
| 56 | + public static class Solution2 { |
| 57 | + /** |
| 58 | + * credit: https://leetcode.com/problems/largest-plus-sign/discuss/113314/JavaC%2B%2BPython-O(N2)-solution-using-only-one-grid-matrix |
| 59 | + */ |
| 60 | + public int orderOfLargestPlusSign(int n, int[][] mines) { |
| 61 | + int[][] grid = new int[n][n]; |
| 62 | + for (int i = 0; i < n; i++) { |
| 63 | + for (int j = 0; j < n; j++) { |
| 64 | + grid[i][j] = n; |
| 65 | + } |
| 66 | + } |
| 67 | + for (int i = 0; i < mines.length; i++) { |
| 68 | + grid[mines[i][0]][mines[i][1]] = 0; |
| 69 | + } |
| 70 | + for (int i = 0; i < n; i++) { |
| 71 | + for (int j = 0, k = n - 1, l = 0, r = 0, u = 0, d = 0; j < n; j++, k--) { |
| 72 | + grid[i][j] = Math.min(grid[i][j], l = (grid[i][j] == 0 ? 0 : l + 1));//left direction |
| 73 | + grid[i][k] = Math.min(grid[i][k], r = (grid[i][k] == 0 ? 0 : r + 1));//right direction |
| 74 | + grid[j][i] = Math.min(grid[j][i], u = (grid[j][i] == 0 ? 0 : u + 1));//upwards |
| 75 | + grid[k][i] = Math.min(grid[k][i], d = (grid[k][i] == 0 ? 0 : d + 1));//downwards |
| 76 | + } |
| 77 | + } |
| 78 | + int result = 0; |
| 79 | + for (int i = 0; i < n; i++) { |
| 80 | + for (int j = 0; j < n; j++) { |
| 81 | + result = Math.max(result, grid[i][j]); |
| 82 | + } |
| 83 | + } |
| 84 | + return result; |
| 85 | + } |
| 86 | + |
| 87 | + /** |
| 88 | + * break the above into FOUR separate loops to go over four directions for easier understanding |
| 89 | + */ |
| 90 | + public int orderOfLargestPlusSign_initialVersion(int n, int[][] mines) { |
| 91 | + int[][] grid = new int[n][n]; |
| 92 | + for (int i = 0; i < n; i++) { |
| 93 | + for (int j = 0; j < n; j++) { |
| 94 | + grid[i][j] = n; |
| 95 | + } |
| 96 | + } |
| 97 | + for (int i = 0; i < mines.length; i++) { |
| 98 | + grid[mines[i][0]][mines[i][1]] = 0; |
| 99 | + } |
| 100 | + for (int i = 0; i < n; i++) { |
| 101 | + for (int j = 0, l = 0; j < n; j++) { |
| 102 | + grid[i][j] = Math.min(grid[i][j], l = (grid[i][j] == 0 ? 0 : l + 1)); |
| 103 | + } |
| 104 | + } |
| 105 | + |
| 106 | + for (int i = 0; i < n; i++) { |
| 107 | + for (int j = 0, k = n - 1, r = 0; j < n; j++, k--) { |
| 108 | + grid[i][k] = Math.min(grid[i][k], r = (grid[i][k] == 0 ? 0 : r + 1)); |
| 109 | + } |
| 110 | + } |
| 111 | + |
| 112 | + for (int i = 0; i < n; i++) { |
| 113 | + for (int j = 0, k = n - 1, u = 0; j < n; j++, k--) { |
| 114 | + grid[j][i] = Math.min(grid[j][i], u = (grid[j][i] == 0 ? 0 : u + 1)); |
| 115 | + } |
| 116 | + } |
| 117 | + |
| 118 | + for (int i = 0; i < n; i++) { |
| 119 | + for (int j = 0, k = n - 1, d = 0; j < n; j++, k--) { |
| 120 | + grid[k][i] = Math.min(grid[k][i], d = (grid[k][i] == 0 ? 0 : d + 1)); |
| 121 | + } |
| 122 | + } |
| 123 | + int result = 0; |
| 124 | + for (int i = 0; i < n; i++) { |
| 125 | + for (int j = 0; j < n; j++) { |
| 126 | + result = Math.max(result, grid[i][j]); |
| 127 | + } |
| 128 | + } |
| 129 | + return result; |
| 130 | + } |
| 131 | + } |
85 | 132 | }
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