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1347. Minimum Number of Steps to Make Two Strings Anagram

Description

You are given two strings of the same length s and t. In one step you can choose any character of t and replace it with another character.

Return the minimum number of steps to make t an anagram of s.

An Anagram of a string is a string that contains the same characters with a different (or the same) ordering.

 

Example 1:

Input: s = "bab", t = "aba"
Output: 1
Explanation: Replace the first 'a' in t with b, t = "bba" which is anagram of s.

Example 2:

Input: s = "leetcode", t = "practice"
Output: 5
Explanation: Replace 'p', 'r', 'a', 'i' and 'c' from t with proper characters to make t anagram of s.

Example 3:

Input: s = "anagram", t = "mangaar"
Output: 0
Explanation: "anagram" and "mangaar" are anagrams.

 

Constraints:

  • 1 <= s.length <= 5 * 104
  • s.length == t.length
  • s and t consist of lowercase English letters only.

Solutions

Solution 1: Counting

We can use a hash table or an array \(\textit{cnt}\) to count the occurrences of each character in the string \(\textit{s}\). Then, we traverse the string \(\textit{t}\). For each character, we decrement its count in \(\textit{cnt}\). If the decremented value is less than \(0\), it means that this character appears more times in the string \(\textit{t}\) than in the string \(\textit{s}\). In this case, we need to replace this character and increment the answer by one.

After the traversal, we return the answer.

The time complexity is \(O(m + n)\), and the space complexity is \(O(|\Sigma|)\), where \(m\) and \(n\) are the lengths of the strings \(\textit{s}\) and \(\textit{t}\), respectively, and \(|\Sigma|\) is the size of the character set. In this problem, the character set consists of lowercase letters, so \(|\Sigma| = 26\).

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class Solution:
    def minSteps(self, s: str, t: str) -> int:
        cnt = Counter(s)
        ans = 0
        for c in t:
            cnt[c] -= 1
            ans += cnt[c] < 0
        return ans

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class Solution {
    public int minSteps(String s, String t) {
        int[] cnt = new int[26];
        for (char c : s.toCharArray()) {
            cnt[c - 'a']++;
        }
        int ans = 0;
        for (char c : t.toCharArray()) {
            if (--cnt[c - 'a'] < 0) {
                ans++;
            }
        }
        return ans;
    }
}

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class Solution {
public:
    int minSteps(string s, string t) {
        int cnt[26]{};
        for (char c : s) {
            ++cnt[c - 'a'];
        }
        int ans = 0;
        for (char c : t) {
            if (--cnt[c - 'a'] < 0) {
                ++ans;
            }
        }
        return ans;
    }
};

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func minSteps(s string, t string) (ans int) {
    cnt := [26]int{}
    for _, c := range s {
        cnt[c-'a']++
    }
    for _, c := range t {
        cnt[c-'a']--
        if cnt[c-'a'] < 0 {
            ans++
        }
    }
    return
}

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function minSteps(s: string, t: string): number {
    const cnt: number[] = Array(26).fill(0);
    for (const c of s) {
        ++cnt[c.charCodeAt(0) - 97];
    }
    let ans = 0;
    for (const c of t) {
        if (--cnt[c.charCodeAt(0) - 97] < 0) {
            ++ans;
        }
    }
    return ans;
}

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/**
 * @param {string} s
 * @param {string} t
 * @return {number}
 */
var minSteps = function (s, t) {
    const cnt = Array(26).fill(0);
    for (const c of s) {
        ++cnt[c.charCodeAt(0) - 97];
    }
    let ans = 0;
    for (const c of t) {
        if (--cnt[c.charCodeAt(0) - 97] < 0) {
            ++ans;
        }
    }
    return ans;
};

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