Leetcode #1850: Minimum Adjacent Swaps to Reach the Kth Smallest Number
In this guide, we solve Leetcode #1850 Minimum Adjacent Swaps to Reach the Kth Smallest Number in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given a string num, representing a large integer, and an integer k. We call some integer wonderful if it is a permutation of the digits in num and is greater in value than num.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Greedy, Two Pointers, String
Intuition
The constraints hint that we can reason about two ends of the data at once, which is perfect for a two-pointer scan.
Moving one pointer at a time keeps the invariant intact and avoids nested loops.
Approach
Place pointers at the left and right ends and move them based on the comparison or target condition.
This yields a clean linear pass after any required sorting.
Steps:
- Set left and right pointers.
- Move a pointer based on the condition.
- Update the best answer while scanning.
Example
Input: num = "5489355142", k = 4
Output: 2
Explanation: The 4th smallest wonderful number is "5489355421". To get this number:
- Swap index 7 with index 8: "5489355142" -> "5489355412"
- Swap index 8 with index 9: "5489355412" -> "5489355421"
Python Solution
class Solution:
def getMinSwaps(self, num: str, k: int) -> int:
def next_permutation(nums: List[str]) -> bool:
n = len(nums)
i = n - 2
while i >= 0 and nums[i] >= nums[i + 1]:
i -= 1
if i < 0:
return False
j = n - 1
while j >= 0 and nums[j] <= nums[i]:
j -= 1
nums[i], nums[j] = nums[j], nums[i]
nums[i + 1 : n] = nums[i + 1 : n][::-1]
return True
s = list(num)
for _ in range(k):
next_permutation(s)
d = [[] for _ in range(10)]
idx = [0] * 10
n = len(s)
for i, c in enumerate(num):
j = ord(c) - ord("0")
d[j].append(i)
arr = [0] * n
for i, c in enumerate(s):
j = ord(c) - ord("0")
arr[i] = d[j][idx[j]]
idx[j] += 1
return sum(arr[j] > arr[i] for i in range(n) for j in range(i))
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.