From db37f45f7186a0a9a1048465650b7e5ede33b4be Mon Sep 17 00:00:00 2001
From: Khevna Khona <khkhona@microsoft.com>
Date: Sun, 25 Jan 2026 16:58:15 -0500
Subject: [PATCH] completed dfs - 1

---
 01Matrix.kt  | 44 ++++++++++++++++++++++++++++++++++++++++++++
 FloodFill.kt | 41 +++++++++++++++++++++++++++++++++++++++++
 2 files changed, 85 insertions(+)
 create mode 100644 01Matrix.kt
 create mode 100644 FloodFill.kt

diff --git a/01Matrix.kt b/01Matrix.kt
new file mode 100644
index 00000000..acec0a86
--- /dev/null
+++ b/01Matrix.kt
@@ -0,0 +1,44 @@
+// Given a 01 matrix, we need to find the distance of the nearest 0 for each cell. we use bfs to solve this problem, we start from all the 0 cells and explore their neighbors.
+// If the neighbor is unvisited (value -1), we update its distance and add it to the queue. We continue this process until all cells are processed.
+// Time complexity is O(m*n) where m is number of rows and n is number of columns
+// Space complexity is O(m*n)
+class Solution {
+    fun updateMatrix(mat: Array<IntArray>): Array<IntArray> {
+        val dirs = arrayOf(
+            intArrayOf(-1,0),
+            intArrayOf(1, 0),
+            intArrayOf(0,-1),
+            intArrayOf(0, 1)
+        )
+        val m = mat.size
+        val n = mat[0].size
+
+        val q = ArrayDeque<IntArray>()
+
+        for(i in 0 until m) {
+            for (j in 0 until n) {
+                if (mat[i][j] == 0) {
+                    q.addLast(intArrayOf(i,j))
+                } else {
+                    mat[i][j] *= -1
+                }
+            }
+        }
+
+        while(q.isNotEmpty()) {
+            val curr = q.removeFirst()
+
+            for(dir in dirs) {
+                val r = curr[0] + dir[0]
+                val c = curr[1] + dir[1]
+
+                if (r >= 0 && c>= 0 && r < m && c < n && mat[r][c] == -1) {
+                    q.addLast(intArrayOf(r, c))
+                    mat[r][c] = mat[curr[0]][curr[1]] + 1
+                }
+            }
+        }
+
+        return mat
+    }
+}
\ No newline at end of file
diff --git a/FloodFill.kt b/FloodFill.kt
new file mode 100644
index 00000000..34e9f9a0
--- /dev/null
+++ b/FloodFill.kt
@@ -0,0 +1,41 @@
+// In this problen, we need to change the color of certain area in the image. we use bfs to solve this problem. 
+// We start from the given pixel and explore all its 4 directionally connected pixels that have the same color as the starting pixel.
+// We change their color to the new color and continue this process until all connected pixels are processed. 
+// Time complexity is O(N) where N is the number of pixels in the image
+// Space complexity is O(N)
+
+class Solution {
+    fun floodFill(image: Array<IntArray>, sr: Int, sc: Int, color: Int): Array<IntArray> {
+        val dirs = arrayOf(
+            intArrayOf(-1, 0),
+            intArrayOf(1, 0),
+            intArrayOf(0, -1),
+            intArrayOf(0, 1)
+        )
+
+        val m = image.size
+        val n = image[0].size
+        val oldColor = image[sr][sc]
+        if(oldColor == color) return image
+        val q : ArrayDeque<IntArray> = ArrayDeque()
+
+        
+
+        q.addLast(intArrayOf(sr,sc)) 
+        image[sr][sc] = color
+
+        while(q.isNotEmpty()) {
+            val curr = q.removeFirst()
+            for(dir in dirs) {
+                val r = curr[0] + dir[0]
+                val c = curr[1] + dir[1]
+
+                if(r >= 0 && c>= 0 && r < m && c < n && image[r][c] == oldColor) {
+                    image[r][c] = color
+                    q.addLast(intArrayOf(r,c))
+                }
+            }
+        }
+        return image
+    }
+}
\ No newline at end of file