Close Strings

Problem Statement

Leetcode: 1657. Determine if Two Strings Are Close

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.

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 <= word1.length, word2.length <= 105
• word1 and word2 contain only lowercase English letters.

Intuition

Intuition 1:

If we can swap positions (operation 1), the order of letter does not matter cabbba is same as aabbbc. We only care about their frequency.

Intuition 2:

All letters present in word1 must be present in word2 and no other letters should be present in word2.

Intuition 3:

Given the frquencies of letters, we can arrange then in any order by operation 2. aabbbc - a(2), b(3), c(1) can be achieved from abbccc - a(1), b(2), c(3) Think about the grid game where we swap a piece with its neighbors - all arrangements are possible

We can sort frequencies and compare, if there is a mismatch, strings are not close

Complexity

• Time complexity: O(n)

• Space complexity: Constant

Code

class Solution {
public:
    bool closeStrings(string word1, string word2) {

        // calculate frequencies
        vector<int> fr1(26, 0), fr2(26, 0);
        for (char a: word1) {
            fr1[a - 'a']++;
        }
        for (char a: word2) {
            fr2[a - 'a']++;
        }

        // Check letter mismatch - order matters
        for (int i = 0; i < 26; i++) {
            // if only one of them is zero
            if ((fr1[i] != 0 && fr2[i] == 0)) {
                return false;
            } else if (fr2[i] != 0 && fr1[i] == 0) {
                return false;
            }
        }

        // Check frequency mismatch - order does not matter
        sort(fr1.begin(), fr1.end());
        sort(fr2.begin(), fr2.end());

        for (int i = 0; i < 26; i++) {
            if (fr1[i] != fr2[i]) {
                return false;
            }
        }

        return true;
    }
};