1850. Minimum Adjacent Swaps to Reach the Kth Smallest Number

# 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

Solution from kamyu104/LeetCode-Solutions · MIT