Leetcode 1433: Check If a String Can Break Another String

Input: The input consists of two strings `s1` and `s2`, both containing only lowercase English letters.

Example: s1 = 'cat', s2 = 'abc'

Constraints:

• 1 <= s1.length == s2.length <= 10^5

• All strings consist of lowercase English letters.

Output: The output is a boolean indicating whether one permutation of `s1` can break some permutation of `s2`, or vice-versa.

Example: true

Constraints:

• The output is 'true' if one permutation can break the other, otherwise 'false'.

Goal: To determine if one permutation of `s1` can break some permutation of `s2` or vice-versa.

Steps:

• Step 1: Sort both strings `s1` and `s2`.

• Step 2: Compare the characters of the sorted strings. If at any position, a character from `s1` is less than the corresponding character from `s2`, return 'false'.

• Step 3: If no such position is found, return 'true'.

Goal: The solution should efficiently handle strings of length up to 10^5.

Steps:

• Both strings are of equal length.

• The strings consist only of lowercase English letters.

Assumptions:

• Both input strings are valid and consist of lowercase English letters.

• The strings are of equal length.

• Input: s1 = 'cat', s2 = 'abc'

• Explanation: After sorting both strings, we have 'act' and 'abc'. Since each character in 'act' is greater than or equal to the corresponding character in 'abc', 's1' can break 's2'.

• Input: s1 = 'dog', s2 = 'god'

• Explanation: After sorting both strings, we have 'dgo' and 'dgo'. Since the strings are identical, neither can break the other.

Approach: We will compare the sorted versions of both strings to determine if one string can break the other. Sorting ensures that we are comparing the lexicographically ordered characters.

Observations:

• Sorting the strings ensures that we can directly compare their characters in lexicographical order.

• By comparing sorted strings character by character, we can determine if one can break the other.

Steps:

• Step 1: Sort both `s1` and `s2`.

• Step 2: Compare each character of the sorted strings. If any character of `s1` is smaller than the corresponding character of `s2`, return false.

• Step 3: If the comparison holds for all characters, return true.

Empty Inputs:

• There are no empty inputs as the problem guarantees non-empty strings.

Large Inputs:

• Ensure the solution handles strings up to the length of 10^5 efficiently.

Special Values:

• Consider cases where the strings are identical (e.g., 'abc' and 'abc').

Constraints:

• The solution should be efficient enough to handle strings of length up to 10^5.

bool checkIfCanBreak(string s1, string s2) {
    sort(s1.begin(), s1.end());
    sort(s2.begin(), s2.end());
    int n = s1.size();
    string a = s1 < s2 ? s2: s1;
    string b = s1 < s2 ? s1: s2;   
    // cout << a << "\n"<< b;
    for(int i = 0; i < n; i++) {
        if(a[i] < b[i]) return false;
    }
    return true;
}

1 : Function Definition

bool checkIfCanBreak(string s1, string s2) {

Defines the function `checkIfCanBreak`, which takes two strings `s1` and `s2` as input and determines whether one can break the other.

2 : Sorting

    sort(s1.begin(), s1.end());

Sorts the string `s1` in lexicographical order.

3 : Sorting

    sort(s2.begin(), s2.end());

Sorts the string `s2` in lexicographical order.

4 : Variable Initialization

    int n = s1.size();

Stores the size of string `s1` in the variable `n`. Both strings are assumed to have the same length.

5 : String Comparison

    string a = s1 < s2 ? s2: s1;

Assigns `a` to be the lexicographically larger of `s1` and `s2`.

6 : String Comparison

    string b = s1 < s2 ? s1: s2;

Assigns `b` to be the lexicographically smaller of `s1` and `s2`.

7 : Loop

    for(int i = 0; i < n; i++) {

Begins a loop that iterates over each character of the strings `a` and `b`.

8 : Conditional Check

        if(a[i] < b[i]) return false;

Compares the characters at the same position in `a` and `b`. If any character in `a` is less than the corresponding character in `b`, returns `false`.

9 : Return Statement

    return true;

If the loop completes without returning `false`, it means that every character in `a` is greater than or equal to the corresponding character in `b`, so it returns `true`.

Best Case: O(n log n) for sorting the strings.

Average Case: O(n log n) for sorting and O(n) for the comparison.

Worst Case: O(n log n) due to the sorting step.

Description: The time complexity is dominated by the sorting step.

Best Case: O(n), as we need to store the sorted versions of the strings.

Worst Case: O(n) for storing the sorted strings.

Description: The space complexity is linear, as we are storing the sorted versions of both strings.

Leetcode 1433: Check If a String Can Break Another String

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