By Long Luo
This article is the solution 3 Approaches: BFS(Dijkstra), Binary Search, Union Find of Problem 1631. Path With Minimum Effort .
Here shows 3 Approaches to slove this problem: BFS(Dijkstra), Binary Search, Union Find.
BFS(Dijkstra)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
| public int minimumEffortPath(int[][] heights) {
if (heights == null || heights.length == 0 || heights[0].length == 0) {
return 0;
}
int rows = heights.length;
int cols = heights[0].length;
int[][] dirs = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
PriorityQueue<int[]> pq = new PriorityQueue<>((edge1, edge2) -> edge1[2] - edge2[2]);
pq.offer(new int[]{0, 0, 0});
int[] dist = new int[rows * cols];
Arrays.fill(dist, Integer.MAX_VALUE);
dist[0] = 0;
boolean[][] vis = new boolean[rows][cols];
while (!pq.isEmpty()) {
int[] edge = pq.poll();
int x = edge[0];
int y = edge[1];
int d = edge[2];
if (vis[x][y]) {
continue;
}
if (x == rows - 1 && y == cols - 1) {
break;
}
vis[x][y] = true;
for (int[] dir : dirs) {
int nextX = x + dir[0];
int nextY = y + dir[1];
if (nextX >= 0 && nextX < rows && nextY >= 0 && nextY < cols
&& Math.max(d, Math.abs(heights[nextX][nextY] - heights[x][y])) < dist[nextX * cols + nextY]) {
dist[nextX * cols + nextY] = Math.max(d, Math.abs(heights[nextX][nextY] - heights[x][y]));
pq.offer(new int[]{nextX, nextY, dist[nextX * cols + nextY]});
}
}
}
return dist[rows * cols - 1];
}
|
Analysis
- Time Complexity: \(O(mn \log (mn))\).
- Space Complexity: \(O(mn)\).
Binary Search
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
| public int minimumEffortPath(int[][] heights) {
int[][] dirs = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
int rows = heights.length;
int cols = heights[0].length;
int left = 0;
int right = 1_000_000 - 1;
while (left < right) {
int mid = left + (right - left) / 2;
boolean[][] visited = new boolean[rows][cols];
Queue<int[]> queue = new LinkedList<>();
queue.offer(new int[]{0, 0});
visited[0][0] = true;
while (!queue.isEmpty()) {
int[] curPos = queue.poll();
for (int[] dir : dirs) {
int nextX = curPos[0] + dir[0];
int nextY = curPos[1] + dir[1];
if (nextX >= 0 && nextX < rows && nextY >= 0 && nextY < cols && !visited[nextX][nextY]
&& Math.abs(heights[nextX][nextY] - heights[curPos[0]][curPos[1]]) <= mid) {
visited[nextX][nextY] = true;
queue.offer(new int[]{nextX, nextY});
}
}
}
if (visited[rows - 1][cols - 1]) {
right = mid;
} else {
left = mid + 1;
}
}
return left;
}
|
Analysis
- Time Complexity: \(O(mn \log C)\).
- Space Complexity: \(O(mn)\).
Union Find
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
| public int minimumEffortPath(int[][] heights) {
int rows = heights.length;
int cols = heights[0].length;
List<int[]> edges = new ArrayList<>();
for (int i = 0; i < rows; i++) {
for (int j = 0; j < cols; j++) {
int index = i * cols + j;
if (i > 0) {
edges.add(new int[]{index - cols, index, Math.abs(heights[i][j] - heights[i - 1][j])});
}
if (j > 0) {
edges.add(new int[]{index - 1, index, Math.abs(heights[i][j] - heights[i][j - 1])});
}
}
}
Collections.sort(edges, (edge1, edge2) -> edge1[2] - edge2[2]);
UnionFind uf = new UnionFind(rows * cols);
int ans = 0;
for (int[] edge : edges) {
int x = edge[0];
int y = edge[1];
int d = edge[2];
uf.union(x, y);
if (uf.isConnected(0, rows * cols - 1)) {
ans = d;
break;
}
}
return ans;
}
class UnionFind {
int[] parent;
int[] size;
int count;
public UnionFind(int n) {
count = n;
parent = new int[n];
size = new int[n];
for (int i = 0; i < n; i++) {
parent[i] = i;
size[i] = 1;
}
}
public void union(int x, int y) {
int rootX = find(x);
int rootY = find(y);
if (rootX == rootY) {
return;
}
if (size[rootX] < size[rootY]) {
parent[rootX] = rootY;
size[rootY] += size[rootX];
} else {
parent[rootY] = rootX;
size[rootX] += size[rootY];
}
count--;
}
public int find(int x) {
while (x != parent[x]) {
parent[x] = parent[parent[x]];
x = parent[x];
}
return x;
}
public boolean isConnected(int x, int y) {
int rootX = find(x);
int rootY = find(y);
return rootX == rootY;
}
}
|
Analysis
- Time Complexity: \(O(mn \log (mn))\).
- Space Complexity: \(O(mn)\).
All suggestions are welcome. If you have any query or suggestion please comment below. Please upvote👍 if you like💗 it. Thank you:-)
Explore More Leetcode Solutions. 😉😃💗