#31: Next Permutation

Author: papilo-cloud

Two_Pointers/next_permutation.py

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+# A permutation of an array of integers is an arrangement of its members
+# into a sequence or linear order.
+
+# For example, for arr = [1,2,3], the following are all the permutations
+# of arr: [1,2,3], [1,3,2], [2, 1, 3], [2, 3, 1], [3,1,2], [3,2,1].
+# The next permutation of an array of integers is the next lexicographically 
+# greater permutation of its integer. More formally, if all the permutations
+# of the array are sorted in one container according to their lexicographical 
+# order, then the next permutation of that array is the permutation that follows
+# it in the sorted container. If such arrangement is not possible, the array 
+# must be rearranged as the lowest possible order (i.e., sorted in ascending order).
+
+# For example, the next permutation of arr = [1,2,3] is [1,3,2].
+# Similarly, the next permutation of arr = [2,3,1] is [3,1,2].
+# While the next permutation of arr = [3,2,1] is [1,2,3] because [3,2,1] does not
+# have a lexicographical larger rearrangement.
+# Given an array of integers nums, find the next permutation of nums.
+
+# The replacement must be in place and use only constant extra memory.
+
+
+# Example 1:
+# Input: nums = [1,2,3]
+# Output: [1,3,2]
+
+# Example 2:
+# Input: nums = [3,2,1]
+# Output: [1,2,3]
+
+# Example 3:
+# Input: nums = [1,1,5]
+# Output: [1,5,1]
+ 
+
+# Constraints:
+# 1 <= nums.length <= 100
+# 0 <= nums[i] <= 100
+
+
+from typing import List
+class Solution:
+    def nextPermutation(self, nums: List[int]) -> None:
+        """
+        Do not return anything, modify nums in-place instead.
+        """
+        n = len(nums)
+        j = -1
+
+        for i in range(n - 2, -1, -1):
+            if nums[i] < nums[i + 1]:
+                j = i
+                break
+        
+        if j == -1:
+            nums.reverse()
+            return
+
+        for i in range(n - 1, j, -1):
+            if nums[i] > nums[j]:
+                nums[i], nums[j] = nums[j], nums[i]
+                break
+
+        l, r = j + 1, n - 1
+        while l < r:
+            nums[l], nums[r] = nums[r], nums[l]
+            l += 1
+            r -= 1