Merge pull request #18 from alberthu16/day24

Day24: Strings Rearrangement (backtracking)

Author: Albert Hu

README.md

@@ -32,5 +32,6 @@ Challenges by day (4 days missed):
 [Day 21](./day21) - (CodeFights) Helping Stephan  
 [Day 22](./day22) - Permutations of a list  
 [Day 23](./day23) - Strings Rearrangement  
+[Day 24](./day24) - Strings Rearrangement (backtracking)  
 
 

day23/README.md

@@ -41,3 +41,18 @@ Any better way to approach this problem? I think so..
 
 https://en.wikipedia.org/wiki/Hamiltonian_path_problem
 
+also check out this minified and slightly worse runtime version LOL (180 characters)
+
+```python
+from itertools import permutations as p
+
+def stringsRearrangement(i):
+    return any(n(x) for x in p(i))
+
+def n(i):
+    return len(i) <= 1 or (d(i[0], i[1]) and n(i[1:]))
+
+def d(w, a):
+    return sum(i != j for i, j in zip(w, a)) == 1
+```
+

day24/README.md

@@ -0,0 +1,30 @@
+Today's challenge is a continuation of [yesterday's](../day23)!
+
+## Ideas
+
+I wanted to take a different approach to this question; instead of doing a
+full brute-force exploration on all combinations of the inputArray,
+construct a graph and then do DFS from every node to find a possible 
+Hamiltonian path. The existence of a path would verify whether there is
+a desired arrangement of the strings.
+
+I can construct the graph in `O(N<sup>2</sup>)` time where `N` is the number of 
+elements in the inputArray, and also the number of vertices in my graph. I check
+every possible pair of strings to see whether they differ by exactly one place,
+and add an edge in the graph between the two if so.
+
+I can then run DFS from each node in `N * O(N) = O(N<sup>2</sup>)` time (this
+analysis doesn't seem right to me actually) to complete the algorithm.
+
+Overall, it comes out to a runtime of `O(n<sup>2</sup>)`. The space complexity
+scales in proportion to the number of vertices and edges I need to keep track
+of in my graph. Vertices increase with every additional element in the inputArray.
+Edges increase when there are higher frequencies of words in the inputArray
+that are closer to each other in edit distance.
+
+## Code
+
+[Python](./stringsRearrangementBacktracking.py) (unfinished)
+
+## Follow up
+

day24/stringsRearrangementBacktracking.py

@@ -0,0 +1,92 @@
+def differByOne(word, anotherWord):
+    return sum(c1 != c2 for c1, c2 in zip(word, anotherWord)) == 1
+
+def tuplelizeDuplicates(inputArray):
+    dups = {elem:0 for elem in inputArray}
+    outputTuples = []
+    for elem in inputArray:
+        if elem in dups:
+            outputTuples.append((elem, dups[elem]))
+            dups[elem] += 1
+    return outputTuples
+
+def createGraph(inputTuples):
+    g = {elem:set() for elem in inputTuples}
+
+    size = len(inputTuples)
+
+    for i in xrange(size):
+        for j in xrange(size):
+
+            if i == j:
+                continue
+
+            v1 = inputTuples[i]
+            v2 = inputTuples[j]
+
+            if differByOne(v1[0], v2[0]):
+                g[v1].add(v2)
+                g[v2].add(v1)
+    return g
+
+def derp(graph, startTuple, visited=set()):
+    # do dfs
+    for vertexTuple in graph[startTuple]:
+        print vertexTuple
+        if vertexTuple not in visited:
+            visited.add(vertexTuple)
+            if derp(graph, vertexTuple, visited):
+                return True
+    print visited
+    
+    if len(visited) == len(graph):
+        return True
+    return False
+
+def hamiltonianPath(graph):
+    if len(graph) == 0:
+        return True
+
+    print graph
+    for node in graph:
+        if derp(graph, node):
+            return True
+
+    return False
+
+def testDifferByOne():
+    assert not differByOne("", "")
+    assert not differByOne("a", "a")
+    assert not differByOne("aaa", "aaa")
+    assert not differByOne("abcdeff", "abcedff")
+    assert differByOne("a", "b")
+    assert differByOne("abc", "abb")
+    assert differByOne("abc", "bbc")
+    assert differByOne("abcdefg", "abcdefz")
+
+def testTuplelizeDuplicates():
+    assert tuplelizeDuplicates(["qq", "qq", "qq"]) == [("qq", 0), ("qq", 1), ("qq", 2)]
+
+def testCreateGraph():
+    assert createGraph(tuplelizeDuplicates(["aba", "bbb", "bab"])) == {("aba", 0): set([]), ("bbb", 0): set([("bab", 0)]), ("bab", 0): set([("bbb", 0)])}
+    assert createGraph(tuplelizeDuplicates(["qq", "qq", "qq"])) == {('qq', 1): set([]), ('qq', 0): set([]), ('qq', 2): set([])}
+    assert createGraph(tuplelizeDuplicates(["ab", "ad", "ef", "eg"])) == {('ab', 0): set([('ad', 0)]), ('ef', 0): set([('eg', 0)]), ('ad', 0): set([('ab', 0)]), ('eg', 0): set([('ef', 0)])}
+
+def testHamiltonianPath():
+    assert hamiltonianPath(createGraph([]))
+    assert not hamiltonianPath(createGraph(["aba", "bbb", "bab"]))
+    assert hamiltonianPath(createGraph(["ab", "bb", "aac"]))
+    assert not hamiltonianPath(createGraph(["qq", "qq", "qq"]))
+    assert hamiltonianPath(createGraph(["aaa", "aba", "aaa", "aba", "aaa"]))
+    assert not hamiltonianPath(createGraph(["ab", "ad", "ef", "eg"]))
+    assert hamiltonianPath(createGraph(["abc", "abx", "axx", "abx", "abc"]))
+    assert hamiltonianPath(createGraph(["f", "g", "a", "h"]))
+
+def main():
+    testDifferByOne()
+    testTuplelizeDuplicates()
+    testCreateGraph()
+    testHamiltonianPath()
+
+if __name__ == "__main__":
+    main()