Create minimum-adjacent-swaps-to-reach-the-kth-smallest-number.py

Author: kamyu104

Python/minimum-adjacent-swaps-to-reach-the-kth-smallest-number.py

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+# Time:  O((k + n) * n)
+# Space: O(n)
+
+class Solution(object):
+    def getMinSwaps(self, num, k):
+        """
+        :type num: str
+        :type k: int
+        :rtype: int
+        """
+        def next_permutation(nums, begin, end):
+            def reverse(nums, begin, end):
+                left, right = begin, end-1
+                while left < right:
+                    nums[left], nums[right] = nums[right], nums[left]
+                    left += 1
+                    right -= 1
+
+            k, l = begin-1, begin
+            for i in reversed(xrange(begin, end-1)):
+                if nums[i] < nums[i+1]:
+                    k = i
+                    break
+            else:
+                reverse(nums, begin, end)
+                return False
+            for i in reversed(xrange(k+1, end)):
+                if nums[i] > nums[k]:
+                    l = i
+                    break
+            nums[k], nums[l] = nums[l], nums[k]
+            reverse(nums, k+1, end)
+            return True
+        
+        new_num = list(num)
+        while k:
+            next_permutation(new_num, 0, len(new_num))
+            k -= 1
+        result = 0
+        for i in xrange(len(new_num)):
+            if new_num[i] == num[i]:
+                continue
+            #   // greedily move the one with the least cost from new_num to num without missing optimal cost
+            for j in xrange(i+1, len(new_num)):
+                if new_num[j] == num[i]:
+                    break
+            result += j-i
+            for j in reversed(xrange(i+1, j+1)):
+                new_num[j], new_num[j-1] = new_num[j-1], new_num[j]
+        return result