How do you solve Lexicographically Smallest String After a Swap on LeetCode?

Lexicographically Smallest String After a Swap (LeetCode #3216) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Swapping only adjacent digits with the same parity allows for targeted optimizations.

What is the time complexity of Lexicographically Smallest String After a Swap?

The optimal solution for Lexicographically Smallest String After a Swap runs in O(n log n) time and uses O(n) space. The time complexity is O(n log n) due to the sorting of the even and odd lists, while the space complexity is O(n) because we store the digits in separate lists.

What is the brute force solution for Lexicographically Smallest String After a Swap?

The brute force approach for Lexicographically Smallest String After a Swap is Brute Force, with O(n²) time complexity and O(1) space complexity. We can try every possible swap of adjacent digits that have the same parity and see which results in the smallest string. This is straightforward but can be inefficient. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Lexicographically Smallest String After a Swap?

Lexicographically Smallest String After a Swap uses the following concepts: String, Greedy. The recommended patterns to study are: Greedy, Sorting, Two Pointers.

What are the interview tips for Lexicographically Smallest String After a Swap?

Always clarify the constraints and requirements of the problem. Think about edge cases, such as strings that are already optimal. Discuss your thought process and alternative approaches with the interviewer.

What are common mistakes when solving Lexicographically Smallest String After a Swap?

Not considering all adjacent pairs for swaps. Failing to account for parity when swapping.

#3216

Lexicographically Smallest String After a Swap

Easy

StringGreedyGreedySortingTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

Instead of checking all pairs, we can group the digits by parity and sort them. This allows us to swap the smallest digits in each group to form the smallest possible string.

⚙️

Algorithm

3 steps

1Step 1: Create two lists, one for even digits and one for odd digits, and populate them with the corresponding digits from the string.
2Step 2: Sort both lists to have the smallest digits at the front.
3Step 3: Reconstruct the string by replacing the digits in the original string with the smallest available digit from the corresponding sorted list based on parity.

solution.py13 lines

1def smallestStringAfterSwap(s):
2    evens = sorted([c for c in s if int(c) % 2 == 0])
3    odds = sorted([c for c in s if int(c) % 2 != 0])
4    result = []
5    even_index, odd_index = 0, 0
6    for c in s:
7        if int(c) % 2 == 0:
8            result.append(evens[even_index])
9            even_index += 1
10        else:
11            result.append(odds[odd_index])
12            odd_index += 1
13    return ''.join(result)

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting of the even and odd lists, while the space complexity is O(n) because we store the digits in separate lists.

1Swapping only adjacent digits with the same parity allows for targeted optimizations.
2Sorting helps in quickly finding the smallest possible arrangement.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.