#1433

Medium

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Check If a String Can Break Another String

This problem checks whether one string can break another using permutations and a greedy approach for comparison.

Problem Statement

Given two strings s1 and s2, both of equal length, determine if one string can break the other. A string x can break string y if for all indices i, the characters of x are greater than or equal to the corresponding characters in y, considering alphabetical order. This condition must hold true for every character in both strings.

In other words, we need to check if one permutation of string s1 can break a permutation of string s2, or if vice-versa is true. The simplest approach involves sorting both strings and comparing their characters sequentially to see if one string consistently breaks the other.

Examples

Example 1

Input: s1 = "abc", s2 = "xya"

Output: true

"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

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

Example details omitted.

Constraints

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

Solution Approach

Sort both strings

Sort both strings in non-decreasing order. This allows us to easily compare the characters at corresponding positions in both strings.

Greedy comparison

After sorting, perform a greedy check to see if each character in one string is greater than or equal to the corresponding character in the other string. If this condition holds true for all characters, the string can break the other.

Invariant validation

By validating the invariant that one string’s characters should always be greater than or equal to the other, the solution can efficiently determine whether a valid permutation exists. This invariant ensures a consistent comparison between the two strings.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the sorting step, which is O(n log n), where n is the length of the string. The comparison step is O(n), so the overall time complexity is O(n log n). Space complexity is O(n) due to the storage requirements for the sorted strings.

What Interviewers Usually Probe

The candidate effectively applies sorting and greedy techniques.
The candidate recognizes the greedy choice property and validates it appropriately.
The candidate can explain the invariant used for string comparison after sorting.

Common Pitfalls or Variants

Common pitfalls

Overlooking the need to sort the strings before comparison.
Failing to handle edge cases, such as strings with different characters or lengths.
Assuming a brute force approach to check all permutations without recognizing the efficiency of sorting.

Follow-up variants

Check if one string can break another with case-sensitive comparison.
Extend the problem to check if two strings of different lengths can break each other with additional constraints.
Implement a solution using dynamic programming to verify the breaking condition without sorting.

FAQ

What is the time complexity of the solution?

The time complexity is O(n log n) due to the sorting step, followed by an O(n) comparison step.

How do I handle edge cases in this problem?

Edge cases such as strings with different characters or lengths can be handled by ensuring both strings are of the same length and sorting both before comparing.

Can you explain the greedy approach used in this problem?

The greedy approach sorts both strings and compares their characters in order. The solution checks if one string’s characters are always greater than or equal to the corresponding characters of the other string.

How does sorting both strings help in solving this problem?

Sorting both strings in non-decreasing order ensures that the characters are compared in the right order, making it easier to check if one string can break the other.

What is the primary pattern used to solve the problem?

The primary pattern used is a greedy choice combined with invariant validation. This ensures an efficient comparison of the strings.

terminal

Solution

Solution 1

#### Python3

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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)
        )

Continue Practicing

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