tags: leetcode medium Hash Table String Sorting

1657. Determine if Two Strings Are Close

Description

Two strings are considered close if you can attain one from the other using the following operations:

• Operation 1: Swap any two existing characters.
  • For example, abcde -> aecdb
• Operation 2: Transform every occurrence of one existing character into another existing character, and do the same with the other character.
  • For example, aacabb -> bbcbaa (all a’s turn into b’s, and all b’s turn into a’s)
    You can use the operations on either string as many times as necessary.

Given two strings, word1 and word2, return true if word1 and word2 are close, and false otherwise.

Examples

Example 1:

Input: word1 = “abc”, word2 = “bca”
Output: true
Explanation: You can attain word2 from word1 in 2 operations.
Apply Operation 1: “abc” -> “acb”
Apply Operation 1: “acb” -> “bca”

Example 2:

Input: word1 = “a”, word2 = “aa”
Output: false
Explanation: It is impossible to attain word2 from word1, or vice versa, in any number of operations.

Example 3:

Input: word1 = “cabbba”, word2 = “abbccc”
Output: true
Explanation: You can attain word2 from word1 in 3 operations.
Apply Operation 1: “cabbba” -> “caabbb”
Apply Operation 2: “caabbb” -> “baaccc”
Apply Operation 2: “baaccc” -> “abbccc”

Constraints:

• $1 \leq word1.length, word2.length \leq 10^5$
• word1 and word2 contain only lowercase English letters.

Code

#define ALP_SIZE 26

// Compare function
int compFunc(const void * a, const void * b){
    return *(int *)b - *(int *)a;
}

bool closeStrings(char * word1, char * word2){
    // Compare length of words
    int length = strlen(word1);
    if(length != strlen(word2))
        return false;
    // alph for counting number of each alphabet
    // check for collecting basis set of words
    int check1[ALP_SIZE] = {0}, check2[ALP_SIZE] = {0};
    int alph1[ALP_SIZE] = {0}, alph2[ALP_SIZE] = {0};

    // set check & count alph
    for(int i = 0 ; i < length ; i++){
        check1[word1[i] - 'a'] = 1;
        alph1[word1[i] - 'a']++;
    }
    for(int i = 0 ; i < length ; i++){
        check2[word2[i] - 'a'] = 1;
        alph2[word2[i] - 'a']++;
    }

    // check difference of basis
    for(int i = 0; i < ALP_SIZE; i++){
        if(check1[i] != check2[i])
            return false;
    }

    // sort & check the alphs
    qsort(alph1, ALP_SIZE, sizeof(int), compFunc);
    qsort(alph2, ALP_SIZE, sizeof(int), compFunc);
    for(int i = 0; i < ALP_SIZE; i++) { 
        if(alph1[i] != alph2[i])
            return false;
    }

    return true;
}

Complexity

Space	Time
$O(1)$	$O(Nlog(N))$

Result

• Runtime: 43 ms, faster than 75.00%
• Memory Usage: 9.7 MB, less than 75.00%

Notes

• Quick sort :

    void qsort(void *base, size_t nitems, size_t size, int (*compar)(const void *, const void*))
  

  • In c <stdlib.h>, sort for assign series.
  • *base: The pointer to the series be sorted.
  • nitems: How many elements need to be sorted.
  • size: The size of each element in the series.
  • *compar: The pointer of function for comparing during sorting.