Leetcode #1433: Check If a String Can Break Another String
In this guide, we solve Leetcode #1433 Check If a String Can Break Another String 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
Given two strings: s1 and s2 with the same size, check if some permutation of string s1 can break some permutation of string s2 or vice-versa. In other words s2 can break s1 or vice-versa.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Greedy, String, Sorting
Intuition
A locally optimal choice leads to a globally optimal result for this structure.
That means we can commit to decisions as we scan without backtracking.
Approach
Sort or preprocess if needed, then repeatedly take the best available local choice.
Maintain the minimal state necessary to validate the greedy decision.
Steps:
- Sort or preprocess as needed.
- Iterate and pick the best local option.
- Track the current solution.
Example
Input: s1 = "abc", s2 = "xya"
Output: true
Explanation: "ayx" is a permutation of s2="xya" which can break to string "abc" which is a permutation of s1="abc".
Python Solution
class Solution:
def checkIfCanBreak(self, s1: str, s2: str) -> bool:
cs1 = sorted(s1)
cs2 = sorted(s2)
return all(a >= b for a, b in zip(cs1, cs2)) or all(
a <= b for a, b in zip(cs1, cs2)
)
Complexity
The time complexity is O(n log n). The space complexity is O(1) to O(n).
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.