1433. Check If a String Can Break Another String

Problem Description

The problem involves determining if there is a permutation of one string (s1) that can "break" a permutation of another string (s2), or vice-versa. The concept of one string breaking another is defined such that, for each index i from 0 to n-1, where n is the size of the strings, the character from the first string at index i is greater than or equal to the character from the second string at the same index when both strings are arranged in alphabetical order.

For example, if s1 is "abc" and s2 is "xya", one valid permutation of s1 that can break s2 is "cba" (since "cba" ≥ "axy"), and hence the output would be True.

To solve this problem, we need to determine if such permutations exist for s1 and s2.

Intuition

A logical approach to solving this problem is to first sort both strings alphabetically. Once we sort them, we would have the smallest characters located at the start and the largest at the end in each string. If s1 can break s2, after sorting, every character in s1 should be greater than or equal to the corresponding character in s2 at each index. Similarly, if s2 can break s1, then every character in s2 should be greater than or equal to the corresponding character in s1 at each index.

The intuition behind sorting is that we are aligning characters in the order of their ranks (based on their alphabetical order), and we are doing a one-to-one comparison between the characters of both strings at corresponding positions. If all characters in s1 are greater than or equal to all corresponding characters in s2, then we conclude s1 can break s2. Otherwise, if all characters in s2 are greater than or equal to all corresponding characters in s1, we conclude that s2 can break s1. If neither of these conditions hold, then no permutation can break the other.

The reason we can use sorting and then linear comparison in this problem is because of the transitive property of the "can break" relationship, which states that if a >= b and b >= c, then a >= c. Thus, we can get away with just sorting the strings and comparing them once rather than looking at every possible permutation of both strings, which would be very inefficient.

Learn more about Greedy and Sorting patterns.

Solution Approach

The implementation of the solution follows a simple yet efficient approach, utilizing the Python language's native libraries and features. Here's a step-by-step breakdown:

Sort Both Strings: The sorted() function is used to sort s1 and s2 alphabetically. This is a critical step that gets the strings ready for comparison. Sorting is done using an efficient sorting algorithm, typically Timsort in Python, which has a complexity of O(n log n), where n is the number of elements in the string.
```
cs1 = sorted(s1)
cs2 = sorted(s2)
```
Compare Sorted Strings: The zip() function is used to aggregate elements from both sorted lists into pairs, which makes it convenient to compare corresponding characters of cs1 and cs2.
Use List Comprehension with all() Function: This approach uses a list comprehension paired with the all() function to check the break condition. The all() function returns True if all elements in an iterable are True. There are two conditions checked here:
- s1 can break s2: This is checked by all(a >= b for a, b in zip(cs1, cs2)). It compares each corresponding pair of characters between the two sorted strings. If every character a from cs1 is greater than or equal to character b from cs2 for all indexed pairs, then s1 can indeed break s2.
- s2 can break s1: Similarly, all(a <= b for a, b in zip(cs1, cs2)) checks if s2 can break s1. For this to be true, every character a from cs1 must be less than or equal to character b from cs2.

Return the Result: The final return statement combines the two conditions with an or operator. If either s1 can break s2 or s2 can break s1, it returns True; otherwise, it returns False.

return all(a >= b for a, b in zip(cs1, cs2)) or all(
    a <= b for a, b in zip(cs1, cs2)
)

This solution effectively leverages Python's built-in functions and language constructs to produce a concise and readable piece of code. By using sorting and linear comparison, it avoids unnecessary complexity and delivers an optimal solution with a time complexity of O(n log n) due to the sorting step (which is the most time-consuming operation here) and a space complexity of O(n) due to the storage requirements for the sorted lists cs1 and cs2.

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Example Walkthrough

Let's take a small example to illustrate the solution approach described above. Suppose we have two strings s1 = "abe" and s2 = "acd". We want to determine if a permutation of s1 can break s2 or if a permutation of s2 can break s1.

Following the solution steps:

Sort Both Strings:
- We sort both strings using Python's sorted() function.
- cs1 after sorting s1: "abe" becomes "abe".
- cs2 after sorting s2: "acd" becomes "acd".
Since both the strings are already in alphabetical order, the sorted versions remain the same.
Compare Sorted Strings:
- We use zip() to pair up characters from cs1 and cs2.
- This gives us the following pairs: ('a', 'a'), ('b', 'c'), ('e', 'd').
Use List Comprehension with all() Function:
- s1 can break s2:
  - We check if all characters in cs1 are greater than or equal to cs2 using the first condition of the solution: all(a >= b for a, b in zip(cs1, cs2)).
  - Comparing the pairs: 'a' >= 'a' (True), 'b' >= 'c' (False), 'e' >= 'd' (True).
  - Since not all comparisons are True, the first condition evaluates to False.
- s2 can break s1:
  - Similarly, we check if all characters in cs2 are greater than or equal to cs1 using the second condition: all(a <= b for a, b in zip(cs1, cs2)).
  - Comparing the pairs: 'a' <= 'a' (True), 'b' <= 'c' (True), 'e' <= 'd' (False).
  - Since not all comparisons are True, the second condition also evaluates to False.
Return the Result:
- The final result is obtained by combining the two conditions with an or: False or False.
- Since both conditions are False, the final result is False.
- Therefore, for the strings s1 = "abe" and s2 = "acd", no permutation of s1 can break s2 and no permutation of s2 can break s1.

In this example, we can see how the solution approach methodically determines whether one string can break another by leveraging Python's powerful built-in functions and efficient list comprehension. The result is obtained without having to manually check every possible permutation, thus optimizing the time and space complexity of the solution.

Solution Implementation

1class Solution:
2    def checkIfCanBreak(self, s1: str, s2: str) -> bool:
3        # Sort both strings to compare them lexicographically
4        sorted_s1 = sorted(s1)
5        sorted_s2 = sorted(s2)
6
7        # Check if sorted_s1 can "break" sorted_s2. This is true if for every index 'i',
8        # the character in sorted_s1 at index 'i' is greater than or equal to the character in sorted_s2 at the same index.
9        can_s1_break_s2 = all(char_s1 >= char_s2 for char_s1, char_s2 in zip(sorted_s1, sorted_s2))
10
11        # Check if sorted_s2 can "break" sorted_s1. This is similar to the previous check
12        # but in the opposite direction: for every index 'i',
13        # the character in sorted_s2 at index 'i' is greater than or equal to the character in sorted_s1.
14        can_s2_break_s1 = all(char_s2 >= char_s1 for char_s1, char_s2 in zip(sorted_s1, sorted_s2))
15
16        # Return True if either sorted_s1 can break sorted_s2 or sorted_s2 can break sorted_s1.
17        return can_s1_break_s2 or can_s2_break_s1
18

1class Solution {
2
3    /**
4     * Checks if one of the strings can "break" the other by comparing characters
5     * after sorting both strings.
6     * 
7     * @param s1 The first input string.
8     * @param s2 The second input string.
9     * @return True if one string can break the other, false otherwise.
10     */
11    public boolean checkIfCanBreak(String s1, String s2) {
12        // Convert the strings to character arrays for sorting.
13        char[] sortedS1 = s1.toCharArray();
14        char[] sortedS2 = s2.toCharArray();
15      
16        // Sort the character arrays.
17        Arrays.sort(sortedS1);
18        Arrays.sort(sortedS2);
19      
20        // Check if sortedS1 can break sortedS2 or if sortedS2 can break sortedS1.
21        return canBreak(sortedS1, sortedS2) || canBreak(sortedS2, sortedS1);
22    }
23
24    /**
25     * Helper method to check if the first character array can "break" the second.
26     * 
27     * @param array1 The first character array.
28     * @param array2 The second character array.
29     * @return True if array1 can break array2, false otherwise.
30     */
31    private boolean canBreak(char[] array1, char[] array2) {
32        // Iterate through the arrays and compare characters.
33        for (int i = 0; i < array1.length; ++i) {
34            // If any character of array1 is smaller than its counterpart in array2,
35            // then array1 cannot break array2.
36            if (array1[i] < array2[i]) {
37                return false;
38            }
39        }
40      
41        // If all characters in array1 are greater than or equal to their counterparts
42        // in array2, array1 can break array2.
43        return true;
44    }
45}
46

1class Solution {
2public:
3    // Function that checks if one string can break another after sorting
4    bool checkIfCanBreak(string s1, string s2) {
5        // Sort both strings
6        sort(s1.begin(), s1.end());
7        sort(s2.begin(), s2.end());
8      
9        // Check if either string can break the other
10        return canBreak(s1, s2) || canBreak(s2, s1);
11    }
12
13private:
14    // Helper function to check if s1 can break s2
15    bool canBreak(const string& s1, const string& s2) {
16        // Iterate through both strings
17        for (int i = 0; i < s1.size(); ++i) {
18            // If any character in s1 is less than the character in s2 at the same position,
19            // s1 cannot break s2
20            if (s1[i] < s2[i]) {
21                return false;
22            }
23        }
24        // If all characters in s1 are greater than or equal to those in s2 at the same positions,
25        // s1 can break s2
26        return true;
27    }
28};
29

1/**
2 * This function checks if one string can "break" another by comparing their sorted characters.
3 * A string s1 can break s2 if in all indices i, the character from s1 is greater than or equal to the character from s2, after both strings are sorted.
4 * @param s1 First input string
5 * @param s2 Second input string
6 * @returns true if either s1 can break s2 or s2 can break s1, false otherwise
7 */
8function checkIfCanBreak(s1: string, s2: string): boolean {
9    // Convert both strings to arrays of characters and sort them alphabetically
10    const sortedChars1: string[] = Array.from(s1).sort();
11    const sortedChars2: string[] = Array.from(s2).sort();
12
13    /**
14     * This helper function checks if characters of the first array can "break" the second array.
15     * @param chars1 Array of sorted characters from the first string
16     * @param chars2 Array of sorted characters from the second string
17     * @returns true if all characters in chars1 are greater than or equal to chars2, false otherwise
18     */
19    const canBreak = (chars1: string[], chars2: string[]): boolean => {
20        for (let i = 0; i < chars1.length; i++) {
21            if (chars1[i] < chars2[i]) {
22                return false;
23            }
24        }
25        return true;
26    };
27
28    // Return true if either s1 can break s2 or s2 can break s1
29    return canBreak(sortedChars1, sortedChars2) || canBreak(sortedChars2, sortedChars1);
30}
31

Time and Space Complexity

The given Python code defines a method checkIfCanBreak, which takes two strings s1 and s2 as input and checks if one string can "break" the other by comparing the sorted versions of both strings. Here's the complexity analysis of the code:

Time Complexity:

The function performs the following operations:

Sorts string s1 - This has a time complexity of O(n log n), where n is the length of s1.
Sorts string s2 - Similarly, this also has a time complexity of O(n log n), where n is the length of s2 (assuming s1 and s2 are of approximately equal length for simplicity).
Two calls to the all() function with generator expressions containing zip(cs1, cs2) - Each call iterates over the zipped lists and compares the elements, which takes O(n) time as there are n pairs to check.

Assuming n is the length of the longer string, the total time complexity should be 2 * O(n log n) + 2 * O(n). However, since O(n) is dominated by O(n log n) in terms of asymptotic complexity, it is ignored in the final complexity expression. Thus, the time complexity of the function is O(n log n).

Space Complexity:

The function uses extra space for the following:

Storing the sorted version of s1 - This takes O(n) space.
Storing the sorted version of s2 - This also takes O(n) space.

Therefore, the space complexity of the function is O(n), where n is the length of the longer string.

In the given code, no additional space complexity is incurred apart from storing the sorted strings since the generator expressions used in all() functions don’t create an additional list, but rather computes the expressions on the fly.

Learn more about how to find time and space complexity quickly using problem constraints.

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