From f5723dce42c9c7445f79d33dd5113121781a34d7 Mon Sep 17 00:00:00 2001
From: Purva <ptand@uic.edu>
Date: Thu, 1 Jan 2026 06:08:24 -0600
Subject: [PATCH] done

---
 problem1.py | 37 +++++++++++++++++++++++++++++++++++++
 problem2.py | 35 +++++++++++++++++++++++++++++++++++
 2 files changed, 72 insertions(+)
 create mode 100644 problem1.py
 create mode 100644 problem2.py

diff --git a/problem1.py b/problem1.py
new file mode 100644
index 00000000..06956a40
--- /dev/null
+++ b/problem1.py
@@ -0,0 +1,37 @@
+"""
+I used dfs approach to solve this question as it ask us to go deep and then perform action, so keeping the boundary conditions, we change the particular cell into the start tcolor
+and move ahead and then follow the same for all 4 directions. This way everything is covered and we can return the final image.
+TC is O(m x n) and SC is also O(m x n)
+"""
+
+class Solution(object):
+    def floodFill(self, image, sr, sc, color):
+        """
+        :type image: List[List[int]]
+        :type sr: int
+        :type sc: int
+        :type color: int
+        :rtype: List[List[int]]
+        """
+        rows, cols = len(image), len(image[0])
+        start = image[sr][sc]
+
+        if start == color:
+            return image
+
+        def dfs(r, c):
+
+            if (r < 0 or r >= rows or 
+                c < 0 or c >= cols or
+                image[r][c] != start):
+                return
+
+            image[r][c] = color
+
+            dfs(r - 1, c)
+            dfs(r, c - 1)
+            dfs(r + 1, c)
+            dfs(r, c + 1)
+
+        dfs(sr, sc)
+        return image
\ No newline at end of file
diff --git a/problem2.py b/problem2.py
new file mode 100644
index 00000000..5383b6a3
--- /dev/null
+++ b/problem2.py
@@ -0,0 +1,35 @@
+"""
+Instead of performing BFS on the 1, we apply it on 0 and then transform all other into -1, so when all 0 positions are appended to the queue, we explore al 4 directions, and 
+if we find a -1 and are within bounds, then we take its og value and add one to it as our distance. then again append this to the queue and once all this is done, we ret the image
+TC will be O(mxn) and SC will be o(mxn) in worst case
+"""
+
+
+class Solution(object):
+    def updateMatrix(self, mat):
+        """
+        :type mat: List[List[int]]
+        :rtype: List[List[int]]
+        """
+        rows, cols = len(mat), len(mat[0])
+        q = collections.deque()
+
+        for r in range(rows):
+            for c in range(cols):
+                if mat[r][c] == 0:
+                    q.append((r, c))
+                else:
+                    mat[r][c] = -1
+
+        dirs = [(0, 1), (1, 0), (0, -1), (-1, 0)]
+        while q:
+            r, c = q.popleft()
+            for dr, dc in dirs:
+                nr, nc = r + dr, c + dc
+                if (0 <= nr < rows and
+                    0 <= nc < cols and
+                    mat[nr][nc] == -1): 
+                    mat[nr][nc] = mat[r][c] + 1
+                    q.append((nr, nc))
+        return mat
+