Time: 1407 ms (5.03%), Space: 46.7 MB (15.50%) - LeetHub

Author: ayusshrathore

0542-01-matrix/0542-01-matrix.java

@@ -0,0 +1,41 @@
+class Solution { // DFS, TC: O(n*m)
+    private int n, m;
+    private int[][] dir = {{-1,0}, {1,0}, {0,-1}, {0, 1}};
+
+    public int[][] updateMatrix(int[][] mat) {
+        this.m = mat.length;
+        this.n = mat[0].length;
+        boolean[][] visited = new boolean[m][n];
+
+        for (int i = 0; i < m; i++) {
+            for (int j = 0; j < n; j++) {
+                if(mat[i][j] == 0 && !visited[i][j]) {
+                    solve(i, j, visited, mat, 0);
+                }
+            }
+        }
+        return mat;
+    }
+
+    private void solve (int i, int j, boolean[][] visited, int[][] mat, int lvl) {
+        if (i < 0 || i >= m || j < 0 || j >= n || (mat[i][j]==0 && lvl != 0)) { // out of bounds check or if cell contains zero & it is not the same cell where we started
+            return;
+        }
+        
+        if (visited[i][j] && mat[i][j] <= lvl) { // if cell is already been visited & distance so far is greater than already calculated one, so no need to calculate further
+            return;
+        }
+
+        visited[i][j] = true;
+
+        mat[i][j] = lvl;
+
+        for (int[] d : dir) {
+            int _m = i + d[0];
+            int _n = j + d[1];
+            
+            solve(_m, _n, visited, mat, lvl+1);
+        }
+
+    }
+}