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| 1 | +# 1347. Minimum Number of Steps to Make Two Strings Anagram |
| 2 | + |
| 3 | +- Difficulty: Medium. |
| 4 | +- Related Topics: Hash Table, String, Counting. |
| 5 | +- Similar Questions: Determine if Two Strings Are Close, Minimum Number of Steps to Make Two Strings Anagram II. |
| 6 | + |
| 7 | +## Problem |
| 8 | + |
| 9 | +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**. |
| 10 | + |
| 11 | +Return **the minimum number of steps** to make `t` an anagram of `s`. |
| 12 | + |
| 13 | +An **Anagram** of a string is a string that contains the same characters with a different (or the same) ordering. |
| 14 | + |
| 15 | + |
| 16 | +Example 1: |
| 17 | + |
| 18 | +``` |
| 19 | +Input: s = "bab", t = "aba" |
| 20 | +Output: 1 |
| 21 | +Explanation: Replace the first 'a' in t with b, t = "bba" which is anagram of s. |
| 22 | +``` |
| 23 | + |
| 24 | +Example 2: |
| 25 | + |
| 26 | +``` |
| 27 | +Input: s = "leetcode", t = "practice" |
| 28 | +Output: 5 |
| 29 | +Explanation: Replace 'p', 'r', 'a', 'i' and 'c' from t with proper characters to make t anagram of s. |
| 30 | +``` |
| 31 | + |
| 32 | +Example 3: |
| 33 | + |
| 34 | +``` |
| 35 | +Input: s = "anagram", t = "mangaar" |
| 36 | +Output: 0 |
| 37 | +Explanation: "anagram" and "mangaar" are anagrams. |
| 38 | +``` |
| 39 | + |
| 40 | + |
| 41 | +**Constraints:** |
| 42 | + |
| 43 | + |
| 44 | + |
| 45 | +- `1 <= s.length <= 5 * 104` |
| 46 | + |
| 47 | +- `s.length == t.length` |
| 48 | + |
| 49 | +- `s` and `t` consist of lowercase English letters only. |
| 50 | + |
| 51 | + |
| 52 | + |
| 53 | +## Solution |
| 54 | + |
| 55 | +```javascript |
| 56 | +/** |
| 57 | + * @param {string} s |
| 58 | + * @param {string} t |
| 59 | + * @return {number} |
| 60 | + */ |
| 61 | +var minSteps = function(s, t) { |
| 62 | + var map = Array(26).fill(0); |
| 63 | + var a = 'a'.charCodeAt(0); |
| 64 | + for (var i = 0; i < t.length; i++) { |
| 65 | + map[t.charCodeAt(i) - a]++; |
| 66 | + } |
| 67 | + for (var j = 0; j < s.length; j++) { |
| 68 | + map[s.charCodeAt(j) - a]--; |
| 69 | + } |
| 70 | + var res = 0; |
| 71 | + for (var k = 0; k < 26; k++) { |
| 72 | + res += Math.abs(map[k]); |
| 73 | + } |
| 74 | + return res / 2; |
| 75 | +} |
| 76 | +``` |
| 77 | + |
| 78 | +**Explain:** |
| 79 | + |
| 80 | +nope. |
| 81 | + |
| 82 | +**Complexity:** |
| 83 | + |
| 84 | +* Time complexity : O(n). |
| 85 | +* Space complexity : O(1). |
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