01 Matrix

Author: jeantimex

README.md

@@ -266,6 +266,7 @@ A collection of JavaScript problems and solutions for studying algorithms.
 - [Shortest Path in Maze II](src/bfs/shortest-path-in-maze-ii.js)
 - [Number of Islands](src/bfs/number-of-islands.js)
 - [Pacific Atlantic Water Flow](src/bfs/pacific-atlantic-water-flow.js)
+- [01 Matrix](src/bfs/01-matrix.js)
 
 ### Depth First Search
 
@@ -305,6 +306,7 @@ A collection of JavaScript problems and solutions for studying algorithms.
 - [Matrix Chain Multiplication](src/dynamic-programming/matrix-chain-multiplication.js)
 - [Burst Balloons](src/dynamic-programming/burst-balloons.js)
 - [Guess Number Higher or Lower II](src/dynamic-programming/guess-number-higher-or-lower-ii.js)
+- [01 Matrix](src/dynamic-programming/01-matrix.js)
 
 ### Greedy
 

src/bfs/01-matrix.js

@@ -0,0 +1,86 @@
+/**
+ * 01 Matrix
+ *
+ * Given a matrix consists of 0 and 1, find the distance of the nearest 0 for each cell.
+ *
+ * The distance between two adjacent cells is 1.
+ * Example 1:
+ * Input:
+ *
+ * 0 0 0
+ * 0 1 0
+ * 0 0 0
+ *
+ * Output:
+ *
+ * 0 0 0
+ * 0 1 0
+ * 0 0 0
+ *
+ * Example 2:
+ *
+ * Input:
+ *
+ * 0 0 0
+ * 0 1 0
+ * 1 1 1
+ *
+ * Output:
+ *
+ * 0 0 0
+ * 0 1 0
+ * 1 2 1
+ *
+ * Note:
+ *
+ * - The number of elements of the given matrix will not exceed 10,000.
+ * - There are at least one 0 in the given matrix.
+ * - The cells are adjacent in only four directions: up, down, left and right.
+ */
+
+/**
+ * BFS Solution
+ *
+ * @param {number[][]} matrix
+ * @return {number[][]}
+ */
+const updateMatrix = matrix => {
+  const m = matrix.length;
+  const n = matrix[0].length;
+  const dist = Array(m)
+    .fill()
+    .map(() => Array(n).fill(Number.MAX_SAFE_INTEGER));
+
+  const queue = [];
+
+  for (let i = 0; i < m; i++) {
+    for (let j = 0; j < n; j++) {
+      if (matrix[i][j] === 0) {
+        dist[i][j] = 0;
+        queue.push([i, j]);
+      }
+    }
+  }
+
+  const dirs = [[-1, 0], [1, 0], [0, -1], [0, 1]];
+
+  while (queue.length > 0) {
+    const curr = queue.shift();
+
+    for (let dir of dirs) {
+      const x = curr[0] + dir[0];
+      const y = curr[1] + dir[1];
+
+      if (x >= 0 && x < m && y >= 0 && y < n) {
+        if (dist[x][y] > dist[curr[0]][curr[1]] + 1) {
+          dist[x][y] = dist[curr[0]][curr[1]] + 1;
+          queue.push([x, y]);
+        }
+      }
+    }
+  }
+
+  return dist;
+};
+
+export { updateMatrix };

src/dynamic-programming/01-matrix.js

@@ -0,0 +1,87 @@
+/**
+ * 01 Matrix
+ *
+ * Given a matrix consists of 0 and 1, find the distance of the nearest 0 for each cell.
+ *
+ * The distance between two adjacent cells is 1.
+ * Example 1:
+ * Input:
+ *
+ * 0 0 0
+ * 0 1 0
+ * 0 0 0
+ *
+ * Output:
+ *
+ * 0 0 0
+ * 0 1 0
+ * 0 0 0
+ *
+ * Example 2:
+ *
+ * Input:
+ *
+ * 0 0 0
+ * 0 1 0
+ * 1 1 1
+ *
+ * Output:
+ *
+ * 0 0 0
+ * 0 1 0
+ * 1 2 1
+ *
+ * Note:
+ *
+ * - The number of elements of the given matrix will not exceed 10,000.
+ * - There are at least one 0 in the given matrix.
+ * - The cells are adjacent in only four directions: up, down, left and right.
+ */
+
+/**
+ * Dynamic Programming Solution
+ *
+ * @param {number[][]} matrix
+ * @return {number[][]}
+ */
+const updateMatrix = matrix => {
+  const m = matrix.length;
+  const n = matrix[0].length;
+  const dist = Array(m)
+    .fill()
+    .map(() => Array(n).fill(Number.MAX_SAFE_INTEGER));
+
+  for (let i = 0; i < m; i++) {
+    for (let j = 0; j < n; j++) {
+      if (matrix[i][j] === 0) {
+        dist[i][j] = 0;
+      } else {
+        if (i > 0) {
+          dist[i][j] = Math.min(dist[i][j], dist[i - 1][j] + 1);
+        }
+        if (j > 0) {
+          dist[i][j] = Math.min(dist[i][j], dist[i][j - 1] + 1);
+        }
+      }
+    }
+  }
+
+  for (let i = m - 1; i >= 0; i--) {
+    for (let j = n - 1; j >= 0; j--) {
+      if (matrix[i][j] === 0) {
+        dist[i][j] = 0;
+      } else {
+        if (i < m - 1) {
+          dist[i][j] = Math.min(dist[i][j], dist[i + 1][j] + 1);
+        }
+        if (j < n - 1) {
+          dist[i][j] = Math.min(dist[i][j], dist[i][j + 1] + 1);
+        }
+      }
+    }
+  }
+
+  return dist;
+};
+
+export { updateMatrix };