Adding all the solutions

01Matrix.java

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+// TimeComplexity: O(m*n)
+// SpaceComplexity: O(m*n)
+// Explanation: I start by marking all 1s as -1 to indicate unvisited cells and enqueue all 0-cells. 
+// Then, I perform BFS layer by layer, updating each neighbor cell with its distance from the nearest 0. 
+// I increase the distance after each BFS layer to ensure all cells get the minimum distance from a 0. 
+// This guarantees every cell is updated once and efficiently.
+class Solution {
+    public int[][] updateMatrix(int[][] mat) {
+        int m = mat.length;
+        int n = mat[0].length;
+        Queue<Integer> q = new LinkedList<>();
+        int[][] dirs = {{1,0},{-1,0},{0,1},{0,-1}};
+        for(int i =0;  i<m ; i++) {
+            for(int j=0; j<n; j++) {
+                if(mat[i][j] == 1){
+                    mat[i][j] = -1;
+                }
+                if(mat[i][j] == 0) {
+                    q.add(i);
+                    q.add(j);
+                }
+            }
+        }
+        int dist =1;
+        while(!q.isEmpty()) {
+            int size = q.size();
+            for(int i=0; i<size; i=i+2) {
+                int r= q.poll();
+                int c = q.poll();
+                for(int[] dir : dirs ){
+                    int nr = r+dir[0];
+                    int nc = c+dir[1];
+                    if(nr>=0 && nr<m && nc>=0 && nc<n){
+                        if(mat[nr][nc] == -1){
+                            mat[nr][nc] = dist;
+                            q.add(nr);
+                            q.add(nc);
+                        }
+                    }
+                }
+            }
+            dist++;
+        }
+        return mat;
+    }
+}

FloodFill.java

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+
+// Solution -1
+
+// TimeComplexity: O(m*n);
+// SpaceComplexity: O(m*n)
+// Explanation: I traverse the image using DFS starting from the given cell. I color each cell and recursively visit its neighbors that have the original color. 
+// This ensures all connected cells are updated exactly once.
+class Solution {
+    int[][] dirs = {{0,-1}, {0,1}, {1,0}, {-1,0}};
+    public int[][] floodFill(int[][] image, int sr, int sc, int color) {
+        if(image[sr][sc] == color) return image;
+        dfs(sr,sc,image,color, image[sr][sc]);
+        return image;     
+    }
+
+    private void dfs(int sr, int sc, int[][] image, int color,int initColor) {
+        image[sr][sc] = color;
+        for(int i =0; i<dirs.length; i ++) {
+            int r = sr+ dirs[i][0];
+            int c = sc+ dirs[i][1];
+            if(r>=0 && r<image.length && c>=0 && c<image[0].length) {
+                if(image[r][c] == initColor) {
+                    image[r][c] = color;
+                    dfs(r,c,image,color, initColor);
+                }
+            }
+        }
+
+    }
+}
+
+
+
+// Solution-2
+
+// TimeComplexity: O(m*n);
+// SpaceComplexity: O(m*n)
+// Explanation: I traverse the image using BFS starting from the given cell. I color each cell and enqueue its neighbors with the original color. 
+// This ensures all connected cells are updated efficiently, layer by layer.
+// class Solution {
+//     public int[][] floodFill(int[][] image, int sr, int sc, int color) {
+//         if(image[sr][sc] == color) return image;
+//         int[][] dirs = {{0,-1}, {0,1}, {1,0}, {-1,0}};
+//         int initColor = image[sr][sc];
+//         image[sr][sc] = color;
+//         Queue<Integer> q = new LinkedList<>();
+//         q.add(sr);
+//         q.add(sc);
+//         while(!q.isEmpty()) {
+//             int currR = q.poll();
+//             int currC = q.poll();
+//             for(int i =0; i<dirs.length; i ++) {
+//                 int r = currR+ dirs[i][0];
+//                 int c = currC+ dirs[i][1];
+//                 if(r>=0 && r<image.length && c>=0 && c<image[0].length) {
+//                     if(image[r][c] == initColor) {
+//                         image[r][c] = color;
+//                         q.add(r);
+//                         q.add(c);
+//                     }
+//                 }
+//             }
+//         }
+//         return image;
+
+//     }
+// }