What is the time complexity of Check If String Is Transformable With Substring Sort Operations?

The optimal solution for Check If String Is Transformable With Substring Sort Operations runs in O(n) time and uses O(n) space. The time complexity is O(n) because we traverse the strings linearly. The space complexity is O(n) due to the storage used for counting characters in the anagram check.

What is the brute force solution for Check If String Is Transformable With Substring Sort Operations?

The brute force approach for Check If String Is Transformable With Substring Sort Operations is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible substring of 's' and sorting it to see if we can eventually match 't'. This is straightforward but inefficient, as it doesn't leverage any patterns in the strings. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Check If String Is Transformable With Substring Sort Operations?

Check If String Is Transformable With Substring Sort Operations uses the following concepts: String, Greedy, Sorting. The recommended patterns to study are: Two Pointers, Sorting, Greedy Algorithms.

What are the interview tips for Check If String Is Transformable With Substring Sort Operations?

Always start by checking for basic conditions like anagram status before diving into complex logic. Think about how sorting affects the order of elements and how you can leverage that in your solution. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Check If String Is Transformable With Substring Sort Operations?

Failing to check if 's' and 't' are anagrams before proceeding with transformations. Not maintaining the relative order of characters while checking for matches.

#1585

Check If String Is Transformable With Substring Sort Operations

Hard

StringGreedySortingTwo PointersSortingGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

The optimal solution leverages the fact that we can only sort characters in 's' to match the order in 't' while maintaining their relative positions. We can use a two-pointer technique to check if we can match 't' by traversing 's'.

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Algorithm

3 steps

1Step 1: Check if 's' and 't' are anagrams (same characters in different order).
2Step 2: Use two pointers to traverse both strings simultaneously.
3Step 3: For each character in 't', check if it can be found in 's' while maintaining the order and relative positions.

solution.py9 lines

1def canTransform(s, t):
2    if sorted(s) != sorted(t):
3        return False
4    s_pointer, t_pointer = 0, 0
5    while t_pointer < len(t):
6        if s_pointer < len(s) and s[s_pointer] == t[t_pointer]:
7            t_pointer += 1
8        s_pointer += 1
9    return t_pointer == len(t)

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Complexity note: The time complexity is O(n) because we traverse the strings linearly. The space complexity is O(n) due to the storage used for counting characters in the anagram check.

1Sorting a substring does not change the overall character counts, so we must first check if both strings are anagrams.
2The order of characters can be maintained by ensuring we can find each character of 't' in 's' in the correct sequence.

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