Given two equal-size strings 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.
Example 4:
Input: s = "xxyyzz", t = "xxyyzz" Output: 0
Example 5:
Input: s = "friend", t = "family" Output: 4
Constraints:
1 <= s.length <= 50000s.length == t.lengths and t contain lower-case English letters only.struct Solution;
impl Solution {
fn min_steps(s: String, t: String) -> i32 {
let mut count: Vec<i32> = vec![0; 26];
for c in s.chars() {
let i = (c as u32 - 'a' as u32) as usize;
count[i] += 1;
}
for c in t.chars() {
let i = (c as u32 - 'a' as u32) as usize;
count[i] -= 1;
}
count.iter().map(|x| x.abs()).sum::<i32>() / 2
}
}
#[test]
fn test() {
let s = "bab".to_string();
let t = "aba".to_string();
let res = 1;
assert_eq!(Solution::min_steps(s, t), res);
let s = "leetcode".to_string();
let t = "practice".to_string();
let res = 5;
assert_eq!(Solution::min_steps(s, t), res);
let s = "anagram".to_string();
let t = "mangaar".to_string();
let res = 0;
assert_eq!(Solution::min_steps(s, t), res);
let s = "xxyyzz".to_string();
let t = "xxyyzz".to_string();
let res = 0;
assert_eq!(Solution::min_steps(s, t), res);
let s = "friend".to_string();
let t = "family".to_string();
let res = 4;
assert_eq!(Solution::min_steps(s, t), res);
}