From d307a3c8d8d44f18ca2fbd2232f0176da8fa81f1 Mon Sep 17 00:00:00 2001
From: jeetusingh247 <jeetusingh11010@gmail.com>
Date: Thu, 19 Feb 2026 19:09:54 +0530
Subject: [PATCH] solved

---
 01_matrix.java        | 67 +++++++++++++++++++++++++++++++++++++++++++
 Flood_Fill_Image.java | 42 +++++++++++++++++++++++++++
 Sample.java           |  7 -----
 3 files changed, 109 insertions(+), 7 deletions(-)
 create mode 100644 01_matrix.java
 create mode 100644 Flood_Fill_Image.java
 delete mode 100644 Sample.java

diff --git a/01_matrix.java b/01_matrix.java
new file mode 100644
index 00000000..7d82a12e
--- /dev/null
+++ b/01_matrix.java
@@ -0,0 +1,67 @@
+/*
+Approach 01: Since we have connected components or neighbours to explore
+here we can use BFS with queue approach.
+
+For Brute Force we can do BFS for each and every cell or bucket
+
+Time: m sqr n sqr(not optimal) --> time limit exceeded
+Space: mxn
+
+To Optimize this:
+Instead of starting from 1's we can start from 0's
+so we need not to repeat BFS and we can compute all ans together.
+ */
+
+// Time: O(mxn)
+// Space: O(mxn)
+
+// Code works fine over leetcode
+
+
+class Solution {
+    public int[][] updateMatrix(int[][] mat) {
+
+        int m = mat.length;          
+        int n = mat[0].length;       
+
+        Queue<int[]> q = new LinkedList<>();  // queue for BFS
+        boolean[][] visited = new boolean[m][n];  // track processed cells
+
+        // Push all 0 cells into queue 
+        for (int i = 0; i < m; i++) {
+            for (int j = 0; j < n; j++) {
+                if (mat[i][j] == 0) {
+                    q.offer(new int[]{i, j});   // source cell
+                    visited[i][j] = true;       // mark visited immediately
+                }
+            }
+        }
+
+        int[][] dirs = {{1,0}, {-1,0}, {0,1}, {0,-1}}; 
+
+        // BFS traversal
+        while (!q.isEmpty()) {
+
+            int[] curr = q.poll();  
+            int r = curr[0];
+            int c = curr[1];
+
+            // Explore all 4 neighbours
+            for (int[] d : dirs) {
+
+                int nr = r + d[0];   
+                int nc = c + d[1];  
+
+                // Check boundary and ensure not already visited
+                if (nr >= 0 && nc >= 0 && nr < m && nc < n && !visited[nr][nc]) {
+
+                    mat[nr][nc] = mat[r][c] + 1;  // distance 
+                    visited[nr][nc] = true;       
+                    q.offer(new int[]{nr, nc});   // push neighbor into queue
+                }
+            }
+        }
+
+        return mat;   
+    }
+}
\ No newline at end of file
diff --git a/Flood_Fill_Image.java b/Flood_Fill_Image.java
new file mode 100644
index 00000000..64c6dcf4
--- /dev/null
+++ b/Flood_Fill_Image.java
@@ -0,0 +1,42 @@
+/* 
+Approach: as we want to recolor all the connected pixels that have same color
+As we already have four directions, it looks like graph traversal problem
+Hence we can use depth-first-search, since we have all connected components here
+*/  
+
+// Time : O(mxn) --> in worst case scenario every pixel in the image visited only once
+// Space : O(mxn) --> as recursion stack in the worst case (entire grid filled with the same color).
+
+// Yes the code successfully ran over leetcode.
+
+class Solution {
+    int[][] dirs;
+    int m, n;
+    public int[][] floodFill(int[][] image, int sr, int sc, int color) {
+        this.dirs = new int[][]{{-1,0}, {1,0}, {0,1}, {0,-1}};
+        this.m = image.length;
+        this.n = image[0].length;
+
+        int oldColor = image[sr][sc]; // store original color of start pixel
+        if(oldColor == color) return image; // check if the original color is equal to already new color
+
+        dfs(image, sr, sc, color, oldColor); // perform dfs and recolor the current pixel
+
+        return image;
+
+    }
+
+    private void dfs(int[][] image, int i, int j, int color, int oldColor) {
+
+        if(i < 0 || j < 0 || i == m || j == n || image[i][j] != oldColor) return;
+
+        image[i][j] = color; // recolor the current pixel
+
+        for(int[] dir : dirs) {
+            int r = dir[0] + i;
+            int c = dir[1] + j;
+
+            {dfs(image, r, c, color, oldColor);} // recursively call dfs to check all connected components
+        }
+    }
+}
\ No newline at end of file
diff --git a/Sample.java b/Sample.java
deleted file mode 100644
index f5c45b5f..00000000
--- a/Sample.java
+++ /dev/null
@@ -1,7 +0,0 @@
-// Time Complexity :
-// Space Complexity :
-// Did this code successfully run on Leetcode :
-// Any problem you faced while coding this :
-
-
-// Your code here along with comments explaining your approach
\ No newline at end of file