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To check whether two strings are close, you must determine if they can be made identical using the allowed operations. These operations permit swapping characters or changing all instances of one character to another, as long as they are existing characters in the particular string. Therefore, for two strings to be close:
Thus, comparing the character sets and their frequency distributions will be sufficient to determine if the strings are close.
The time complexity is O(n + m), where n and m are the lengths of the strings, due to counting characters and comparing sets. Sorting the frequencies has a time complexity of O(k log k), where k is the number of unique characters. The space complexity is O(1) since we only store counts for at most 26 characters.
1function closeStrings(word1, word2) {
2 if (word1.length !== word2.length) return false;
3 const count1 = {}, count2 = {};
4 for (const char of word1) {
5 count1[char] = (count1[char] || 0) + 1;
6 }
7 for (const char of word2) {
8 count2[char] = (count2[char] || 0) + 1;
9 }
10 const keys1 = Object.keys(count1).sort();
11 const keys2 = Object.keys(count2).sort();
12 if (keys1.join('') !== keys2.join('')) return false;
13 const values1 = Object.values(count1).sort((a, b) => a - b);
14 const values2 = Object.values(count2).sort((a, b) => a - b);
15 return values1.join('') === values2.join('');
16}The JavaScript solution is similar to the Python one. It uses two objects to count character frequencies in the two strings. It compares the unique character sets by sorting their keys and checks if they are identical. Then, it sorts the character frequency values and checks for equality. It returns true if both checks align, offering efficient detection of closeness.
Alternative to directly sorting the character frequencies to compare them, we can map frequencies to counts using an intermediate structure, effectively hashing the frequency occurrences. This approach ensures that both overall character sets and their frequency distributions align.
The time complexity remains O(n + m) due to creating and comparing maps, with a space complexity of O(1) due to counting up to 26 character frequencies.
1import java.util.HashMap;
2import
Here, we first count the characters like before and then extract the keys into sets and compare them for equality. We use a HashMap to count the frequency of frequencies themselves, comparing these two frequency maps. Equal keys signify a valid transformation is possible between the two frequency distributions.