[LeetCode][1631. Path With Minimum Effort] 3 Approaches: BFS(Dijkstra), Binary Search, Union Find

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)

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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: \(O(mnlog(mn))\).
  • Space Complexity: \(O(mn)\).

Binary Search

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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: \(O(mnlogC)\).
  • Space Complexity: \(O(mn)\).

Union Find

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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: \(O(mnlog(mn))\).
  • Space Complexity: \(O(mn)\).

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