no message

Author: Chris Wu

problems/shortest-bridge.py

@@ -1,50 +1,70 @@
-import collections
+"""
+First, identify one of the island by marking its value by 2 (island2). [0]
+
+Second ([1]), use BFS to go through another island (island1). When we encounter 0 we put it on `q2`, otherwise `q`.
+When q is finished, it means that the island1 is completely explored (all `(i, j)` is in the `visied`).
+Now we make put `q2` to the `q` and clear `q2`. [2]
+
+When q is finished, it means that all of the one-tile outward `(i, j)` is explored (in the `visited`).
+Now we make put `q2` to the `q` and clear `q2`.
+
+When q is finished, it means that all of the two-tile outward `(i, j)` is explored (in the `visited`).
+Now we make put `q2` to the `q` and clear `q2`.
+
+...
+
+We keep on doing this until we find the island2. [3]
+
+The time complexity is `O(N)`, `N` is the number of element in `A`.
+The space complexity is `O(N)`, too. Since we might put most of the `(i, j)` in the `visited`.
+
+Of course, we can optimize the space, by changing the value in the `A`.
+The code would be a little bit harder the read and understand.
+I challenge you to try it out!
+"""
 class Solution(object):
     def shortestBridge(self, A):
-        def findFirst():
+        def findFirst(target):
             for i in xrange(M):
                 for j in xrange(N):
-                    if A[i][j]==1: return (i, j)
+                    if A[i][j]==target: return (i, j)
 
         M, N = len(A), len(A[0])
-        opt = float('inf')
 
-        q1 = collections.deque([findFirst()])
-        while q1:
-            i, j = q1.popleft()
+        #[0]
+        q = collections.deque([findFirst(1)])
+        while q:
+            i, j = q.popleft()
             if i<0 or i>=M: continue
             if j<0 or j>=N: continue
             if A[i][j]==2 or A[i][j]==0: continue
             if A[i][j]==1: A[i][j] = 2
-            q1.extend([(i+1, j), (i-1, j), (i, j+1), (i, j-1)])
-
-        i0, j0 = findFirst()
-        q2 = collections.deque([(i0, j0, 0)])
-        while q2:
-            i, j, dis = q2.popleft()
-            if A[i][j]==3 or A[i][j]==4:
-                continue
-            if A[i][j]==2:
-                opt = min(opt, dis)
-                continue
-
-            if A[i][j]==1: A[i][j] = 3
-            if A[i][j]==0: A[i][j] = 4
-            for ni, nj in [(i+1, j), (i-1, j), (i, j+1), (i, j-1)]:
-                if ni<0 or ni>=M: continue
-                if nj<0 or nj>=N: continue
-                if A[ni][nj]==1:
-                    q2.append((ni, nj, 0))
-                elif A[ni][nj]==0:
-                    q2.append((ni, nj, dis+1))
-                elif A[ni][nj]==2:
-                    q2.append((ni, nj, dis))
+            q.extend([(i+1, j), (i-1, j), (i, j+1), (i, j-1)])
 
-        return opt
+        #[1]
+        q.append(findFirst(1))
+        q2 = []
+        tile = 0
+        visited = set()
+        while q or q2:
+            if not q: #[2]
+                q.extend(q2)
+                q2 = []
+                tile+=1
 
-A = [[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0],[0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,0,0,0,0,0],[0,0,0,0,0,0,0,0,0,0,0,1,1,0,1,1,0,0,0,0],[0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,0,1,1,0],[0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,0,0],[0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,0,0,0],[0,0,0,0,0,0,0,0,0,1,0,1,1,0,0,1,0,0,0,0],[0,0,0,0,0,0,0,0,0,0,0,1,0,1,1,0,1,0,0,0],[0,0,0,0,0,0,0,0,0,0,0,1,0,0,1,0,1,0,0,0],[0,0,0,0,0,0,0,0,0,0,0,0,1,0,1,1,1,1,1,0],[0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,0],[0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,0],[0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,0],[0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,0]]
-print Solution().shortestBridge(A)
+            i, j = q.popleft()
+            if (i, j) in visited: continue
+            visited.add((i, j))
 
+            if A[i][j]==2: return tile #[3]
 
+            for ni, nj in [(i+1, j), (i-1, j), (i, j+1), (i, j-1)]:
+                if ni<0 or ni>=M: continue
+                if nj<0 or nj>=N: continue
+                if A[ni][nj]==0:
+                    q2.append((ni, nj))
+                else:
+                    q.append((ni, nj))
 
+        return tile #should not comes here