How do you solve Check If a String Can Break Another String on LeetCode?

Check If a String Can Break Another String (LeetCode #1433) 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: Sorting allows for direct comparison of characters.

What is the brute force solution for Check If a String Can Break Another String?

The brute force approach for Check If a String Can Break Another String is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all permutations of both strings and checking if any permutation of one string can break the other. This is straightforward but inefficient due to the factorial growth of permutations. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Check If a String Can Break Another String?

Check If a String Can Break Another String uses the following concepts: String, Greedy, Sorting. The recommended patterns to study are: Sorting, Greedy.

What are the interview tips for Check If a String Can Break Another String?

Explain your thought process clearly when discussing approaches. Always consider edge cases, such as strings with identical characters. Practice writing clean and efficient code under time constraints.

What are common mistakes when solving Check If a String Can Break Another String?

Failing to sort both strings before comparison. Not considering both directions of breaking (s1 to s2 and s2 to s1).

#1433

Check If a String Can Break Another String

Medium

StringGreedySortingSortingGreedy

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)

The optimal solution leverages sorting and a greedy approach. By sorting both strings, we can directly compare their characters in a single pass to determine if one can break the other.

⚙️

Algorithm

4 steps

1Step 1: Sort both strings s1 and s2.
2Step 2: Check if sorted s1 can break sorted s2 by comparing characters at each index.
3Step 3: Check if sorted s2 can break sorted s1 in the same manner.
4Step 4: Return true if either condition is satisfied; otherwise, return false.

solution.py11 lines

1# Full working Python code
2
3def can_break(s1, s2):
4    s1_sorted = sorted(s1)
5    s2_sorted = sorted(s2)
6    def can_break_helper(a, b):
7        return all(x >= y for x, y in zip(a, b))
8    return can_break_helper(s1_sorted, s2_sorted) or can_break_helper(s2_sorted, s1_sorted)
9
10# Example usage
11print(can_break('abc', 'xya'))  # Output: True

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step, which is the most time-consuming operation. The space complexity is O(n) because we create sorted copies of the strings.

1Sorting allows for direct comparison of characters.
2Greedy approach is effective when checking conditions across sorted arrays.

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