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2839. Check if Strings Can be Made Equal With Operations I

Description

You are given two strings s1 and s2, both of length 4, consisting of lowercase English letters.

You can apply the following operation on any of the two strings any number of times:

  • Choose any two indices i and j such that j - i = 2, then swap the two characters at those indices in the string.

Return true if you can make the strings s1 and s2 equal, and false otherwise.

 

Example 1:

Input: s1 = "abcd", s2 = "cdab"
Output: true
Explanation: We can do the following operations on s1:
- Choose the indices i = 0, j = 2. The resulting string is s1 = "cbad".
- Choose the indices i = 1, j = 3. The resulting string is s1 = "cdab" = s2.

Example 2:

Input: s1 = "abcd", s2 = "dacb"
Output: false
Explanation: It is not possible to make the two strings equal.

 

Constraints:

  • s1.length == s2.length == 4
  • s1 and s2 consist only of lowercase English letters.

Solutions

Solution 1: Counting

We observe the operation in the problem, and find that if the parity of the two indices $i$ and $j$ of the string is the same, then their order can be changed by swapping.

Therefore, we can count the occurrence times of the characters at odd indices and even indices in the two strings. If the counting results of the two strings are the same, then we can make the two strings equal through the operation.

The time complexity is $O(n + |\Sigma|)$, and the space complexity is $O(|\Sigma|)$. Here, $n$ is the length of the string, and $\Sigma$ is the character set.

  • class Solution {
    public boolean canBeEqual(String s1, String s2) {
        int[][] cnt = new int[2][26];
        for (int i = 0; i < s1.length(); ++i) {
            ++cnt[i & 1][s1.charAt(i) - 'a'];
            --cnt[i & 1][s2.charAt(i) - 'a'];
        }
        for (int i = 0; i < 26; ++i) {
            if (cnt[0][i] != 0 || cnt[1][i] != 0) {
                return false;
            }
        }
        return true;
    }
}

  • class Solution {
public:
    bool canBeEqual(string s1, string s2) {
        vector<vector<int>> cnt(2, vector<int>(26, 0));
        for (int i = 0; i < s1.size(); ++i) {
            ++cnt[i & 1][s1[i] - 'a'];
            --cnt[i & 1][s2[i] - 'a'];
        }
        for (int i = 0; i < 26; ++i) {
            if (cnt[0][i] || cnt[1][i]) {
                return false;
            }
        }
        return true;
    }
};

  • class Solution:
    def canBeEqual(self, s1: str, s2: str) -> bool:
        return sorted(s1[::2]) == sorted(s2[::2]) and sorted(s1[1::2]) == sorted(
            s2[1::2]
        )


  • func canBeEqual(s1 string, s2 string) bool {
	cnt := [2][26]int{}
	for i := 0; i < len(s1); i++ {
		cnt[i&1][s1[i]-'a']++
		cnt[i&1][s2[i]-'a']--
	}
	for i := 0; i < 26; i++ {
		if cnt[0][i] != 0 || cnt[1][i] != 0 {
			return false
		}
	}
	return true
}

  • function canBeEqual(s1: string, s2: string): boolean {
    const cnt: number[][] = Array.from({ length: 2 }, () => Array.from({ length: 26 }, () => 0));
    for (let i = 0; i < s1.length; ++i) {
        ++cnt[i & 1][s1.charCodeAt(i) - 97];
        --cnt[i & 1][s2.charCodeAt(i) - 97];
    }
    for (let i = 0; i < 26; ++i) {
        if (cnt[0][i] || cnt[1][i]) {
            return false;
        }
    }
    return true;
}

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