Welcome to Subscribe On Youtube

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

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

  • 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;
        }
    }
    
  • 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;
        }
    };
    
  • 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())
    
    
  • 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
    }
    
  • 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;
    }
    
    
  • /**
     * @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;
    };
    
    

All Problems

All Solutions