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2186. Minimum Number of Steps to Make Two Strings Anagram II

SourceWeekly Contest 282 Q2DifficultyMediumRating1253

Description

You are given two strings s and t. In one step, you can append any character to either s or t.

Return the minimum number of steps to make s and t anagrams of each other.

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

 

Example 1:

Input: s = "leetcode", t = "coats"
Output: 7
Explanation: 
- In 2 steps, we can append the letters in "as" onto s = "leetcode", forming s = "leetcodeas".
- In 5 steps, we can append the letters in "leede" onto t = "coats", forming t = "coatsleede".
"leetcodeas" and "coatsleede" are now anagrams of each other.
We used a total of 2 + 5 = 7 steps.
It can be shown that there is no way to make them anagrams of each other with less than 7 steps.

Example 2:

Input: s = "night", t = "thing"
Output: 0
Explanation: The given strings are already anagrams of each other. Thus, we do not need any further steps.

 

Constraints:

  • 1 <= s.length, t.length <= 2 * 105
  • s and t consist of lowercase English letters.

Solutions

Solution 1

Thinking

A step inserts one letter into either string. The minimum number of insertions is the \(L_1\) distance between the two frequency vectors — the letters each side still lacks.

Subtract \(t\)’s counts from \(s\)’s and sum absolute values.

One pass over the two strings suffices.

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class Solution:
    def minSteps(self, s: str, t: str) -> int:
        cnt = Counter(s)
        for c in t:
            cnt[c] -= 1
        return sum(abs(v) for v in cnt.values())

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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'];
        }
        for (char c : t.toCharArray()) {
            --cnt[c - 'a'];
        }
        int ans = 0;
        for (int v : cnt) {
            ans += Math.abs(v);
        }
        return ans;
    }
}

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

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

func abs(x int) int {
    if x < 0 {
        return -x
    }
    return x
}

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function minSteps(s: string, t: string): number {
    let cnt = new Array(128).fill(0);
    for (const c of s) {
        ++cnt[c.charCodeAt(0)];
    }
    for (const c of t) {
        --cnt[c.charCodeAt(0)];
    }
    let ans = 0;
    for (const v of cnt) {
        ans += Math.abs(v);
    }
    return ans;
}

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/**
 * @param {string} s
 * @param {string} t
 * @return {number}
 */
var minSteps = function (s, t) {
    let cnt = new Array(26).fill(0);
    for (const c of s) {
        ++cnt[c.charCodeAt() - 'a'.charCodeAt()];
    }
    for (const c of t) {
        --cnt[c.charCodeAt() - 'a'.charCodeAt()];
    }
    let ans = 0;
    for (const v of cnt) {
        ans += Math.abs(v);
    }
    return ans;
};

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