Is Check If a String Can Break Another String easy or hard?

The problem is rated Medium because the key insight is recognizing that sorting both strings reveals whether a valid permutation exists. Once that idea is understood, the implementation becomes straightforward. The frequency-count optimization adds a small extra layer of algorithmic thinking.

Check If a String Can Break Another String Python/Java solution

In Python or Java, the simplest implementation converts each string to a character array, sorts them, and checks pairwise comparisons in a loop. If all characters in one sorted string are greater than or equal to the other, return true. The same logic works in C++, JavaScript, and C# with identical O(n log n) complexity.

How to solve Check If a String Can Break Another String in O(n)?

Use two frequency arrays of size 26 to count characters in both strings. Iterate from 'a' to 'z' while maintaining cumulative counts that simulate the sorted order. If the cumulative count of one string never falls behind the other, that string can break the other. This avoids sorting and reduces time complexity to O(n) with O(1) space.

What is the best approach for Check If a String Can Break Another String?

The most common solution sorts both strings and compares them character by character. After sorting, check whether s1[i] >= s2[i] for every index or vice versa. This greedy comparison works because sorting places characters in the best possible order. The approach runs in O(n log n) time with O(1) or O(n) extra space depending on the language.

Is Check If a String Can Break Another String asked at Google/Amazon/Meta?

Problems involving string ordering, greedy comparisons, and frequency counting commonly appear in interviews at companies like Amazon, Google, and Meta. Variants often test whether one sequence dominates another after reordering. The question evaluates understanding of sorting strategies and prefix comparisons.

What data structure is used in Check If a String Can Break Another String?

The core solutions use arrays and basic string manipulation. The sorting approach relies on sorted arrays of characters, while the optimized method uses fixed-size frequency arrays of length 26. Both approaches iterate linearly through the characters to verify the dominance condition.

What is the time complexity of Check If a String Can Break Another String?

The typical sorting solution runs in O(n log n) time because both strings must be sorted before comparison. After sorting, a single O(n) pass verifies whether one string dominates the other. With a frequency counting optimization, the problem can be solved in O(n) time using a fixed 26-character alphabet.

Description Hints 1 Solution Video 8:21

1433. Check If a String Can Break Another String

Medium70.9% acceptance

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.

A string x can break string y (both of size n) if x[i] >= y[i] (in alphabetical order) for all i between 0 and n-1.

Example 1:

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".

Example 2:

Input: s1 = "abe", s2 = "acd"
Output: false 
Explanation: All permutations for s1="abe" are: "abe", "aeb", "bae", "bea", "eab" and "eba" and all permutation for s2="acd" are: "acd", "adc", "cad", "cda", "dac" and "dca". However, there is not any permutation from s1 which can break some permutation from s2 and vice-versa.

Example 3:

Input: s1 = "leetcodee", s2 = "interview"
Output: true

Constraints:

s1.length == n
s2.length == n
1 <= n <= 10^5
All strings consist of lowercase English letters.

Topics

Hints open one at a time

Hint 1

Sort both strings and then check if one of them can break the other.

8:21

Still stuck?Read 3 approaches, or watch a walkthrough.

Open solution

LanguageLoading editor…

1class Solution:
2    def checkIfCanBreak(self, s1: str, s2: str) -> bool:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2Case 3

Input

"abc"
"xya"

Input

"abe"
"acd"

Input

"leetcodee"
"interview"

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