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 $|\Sigma| = 26$.

Python3

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

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

TypeScript

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;
}

JavaScript

/**
 * @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;
};