Add solution for challenge #63

Author: Mauricio Klein

README.md

@@ -62,3 +62,4 @@
 - [x] [Challenge 60: Group anagrams](challenge-60/)
 - [x] [Challenge 61: Full Binary Tree](challenge-61/)
 - [x] [Challenge 62: Palindrome with maximum "k" deletions](challenge-62/)
+- [x] [Challenge 63: Jump to the end](challenge=63/)

challenge-63/Makefile

@@ -0,0 +1,4 @@
+test:
+	python test_*.py
+
+.PHONY: test

challenge-63/README.md

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+# Jump to the end
+
+This problem was asked by Facebook.
+
+## Description
+
+Starting at index 0, for an element n at index i, you are allowed to jump at most n indexes ahead. 
+
+Given a list of numbers, find the minimum number of jumps to reach the end of the list.
+
+## Example
+
+```
+Input:
+  Input: [3, 2, 5, 1, 1, 9, 3, 4]
+  Output: 2 (3 -> 5 -> 4)
+```

challenge-63/__init__.py

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+"""
+  Time  complexity: O(n^2)
+  Space complexity: O(n^2) (for the recursion tree)
+"""
+def jumps_to_end(nums):
+  return helper(nums, 0, 0)
+
+def helper(nums, acc_jumps, index):
+  N = len(nums)
+
+  if index == N - 1:
+    return acc_jumps
+
+  if index >= N:
+    return float('inf')
+
+  best = float('inf')
+
+  for i in range(1, nums[index] + 1):
+    best = min(best, helper(nums, acc_jumps + 1, index + i))
+
+  return best

challenge-63/solver.py

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+import unittest
+from solver import jumps_to_end
+
+class TestSolver(unittest.TestCase):
+  def test_jumps_to_end(self):
+    self.assertEqual(jumps_to_end([3, 2, 5, 1, 1, 9, 3, 4]), 2) # (3 -> 5 -> 4)
+    self.assertEqual(jumps_to_end([1, 1, 1, 1, 1]         ), 4) # (1 -> 1 -> 1 -> 1 -> 1)
+    self.assertEqual(jumps_to_end([2, 2, 0, 1, 1]         ), 3) # (2 -> 2 -> 1 -> 1)
+    self.assertEqual(jumps_to_end([2]                     ), 0) # (2)
+
+    self.assertEqual(jumps_to_end([2, 0, 0, 1, 1]), float('inf')) # (No solution possible)
+
+if __name__ == "__main__":
+    unittest.main()