# Python solution with detailed explanation

• Solution

Permutation Sequence https://leetcode.com/problems/permutation-sequence/

Recursive Solution

• Imagine N = 4 and K = 14
• Permutations starting with 1: (2,3,4)! = 6. Total Perms ending with 1 as first element = 6.
• Permutations starting with 2: (1,3,4)! = 6. Total Perms ending with 2 as first element = 12.
• Permutations starting with 3: (2,1,4)! = 6. Total Perms ending with 3 as first element = 18.
• So the 14th permutation should start with 3. The problem reduces to finding the second permutation with input as [1,2 4]. nums = [1,2,4] and K=2.
• Permutations starting with 1: (2,4)! = 2. Total Perms ending with 1 as first element = 2.
• Permutations starting with 2: (1,4)! = 2. Total Perms ending with 2 as first element = 4.
• So the 2nd permutation should start with 1. The problem reduced to finding the second permutation with input as [2,4]. nums = [2,4] and K = 2.
• Permutations starting with 2: (4)! = 1. Total Perms ending with 2 as first element = 1
• Permutations starting with 4: (2)! = 1. Total Perms ending with 4 as first element = 2
• So the 2nd permutation should start with 4. The problem reduces to finding the first permutation with input as [2]. nums = [2] and K = 1. This is the base condition: K=1 just return since nums has the right permutation
``````class Solution(object):
def factorial(self, n):
fact, cache = 1, {0:1}
for i in range(1, n+1):
fact *= i
cache[i] = fact
return cache

def helper(self, nums, i, k, cache):
if k == 1:
return
count = 0
for j in range(i, len(nums)):
if count + cache[len(nums)-i-1] < k:
count += cache[len(nums)-i-1]
else:
nums[i], nums[j] = nums[j], nums[i]
nums[i+1:] = sorted(nums[i+1:])
break
self.helper(nums, i+1, k-count, cache)
return

def getPermutation(self, n, k):
"""
:type n: int
:type k: int
:rtype: str
"""
cache, nums = self.factorial(n), [x for x in range(1,n+1)]
self.helper(nums, 0, k, cache)
return "".join([str(x) for x in nums])
``````

Iterative Implementation

• Here is an iterative implementation of the above idea.
``````class Solution(object):
def factorial(self, n):
fact, cache = 1, {0:1}
for i in range(1, n+1):
fact *= i
cache[i] = fact
return cache

def getPermutation(self, n, k):
"""
:type n: int
:type k: int
:rtype: str
"""
cache, count, so_far = self.factorial(n), 0, [x for x in range(1,n+1)]
for i in range(n):
for j in range(i, n):
if count + cache[n-i-1] >= k:
so_far[i], so_far[j] = so_far[j], so_far[i]
so_far = so_far[:i+1] + sorted(so_far[i+1:])
break
else:
count = count + cache[n-i-1]
return "".join([str(x) for x in so_far])
``````

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