LC-0031 Medium LeetCode

31. Next Permutation

Read the full problem statement on LeetCode.

Difficulty: medium Acceptance: 43% Topics: Array, Two Pointers

Reading material

Arrays & Hashing 12 min Two Pointers 9 min

Reference solution (spoiler · python)

# Time:  O(n)
# Space: O(1)

class Solution(object):
    def nextPermutation(self, nums):
        """
        :type nums: List[int]
        :rtype: None Do not return anything, modify nums in-place instead.
        """
        k, l = -1, 0
        for i in reversed(xrange(len(nums)-1)):
            if nums[i] < nums[i+1]:
                k = i
                break
        else:
            nums.reverse()
            return

        for i in reversed(xrange(k+1, len(nums))):
            if nums[i] > nums[k]:
                l = i
                break
        nums[k], nums[l] = nums[l], nums[k]
        nums[k+1:] = nums[:k:-1]
        

# Time:  O(n)
# Space: O(1)
class Solution2(object):
    def nextPermutation(self, nums):
        """
        :type nums: List[int]
        :rtype: None Do not return anything, modify nums in-place instead.
        """
        k, l = -1, 0
        for i in xrange(len(nums)-1):
            if nums[i] < nums[i+1]:
                k = i

        if k == -1:
            nums.reverse()
            return

        for i in xrange(k+1, len(nums)):
            if nums[i] > nums[k]:
                l = i
        nums[k], nums[l] = nums[l], nums[k]
        nums[k+1:] = nums[:k:-1]

Solution from kamyu104/LeetCode-Solutions · MIT