Formatted question description: https://leetcode.ca/all/1347.html
1347. Minimum Number of Steps to Make Two Strings Anagram (Medium)
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 <= 50000
s.length == t.length
s
andt
contain lower-case English letters only.
Related Topics:
String
Solution 1.
// OJ: https://leetcode.com/problems/minimum-number-of-steps-to-make-two-strings-anagram/
// Time: O(N)
// Space: O(1)
class Solution {
public:
int minSteps(string s, string t) {
int cnt[26] = {0}, ans = 0;
for (char c : s) cnt[c - 'a']++;
for (char c : t) cnt[c - 'a']--;
for (int n : cnt) {
if (n > 0) ans += n;
}
return ans;
}
};
Java
-
class Solution { public int minSteps(String s, String t) { int length = s.length(); int[] counts = new int[26]; for (int i = 0; i < length; i++) { char c1 = s.charAt(i), c2 = t.charAt(i); counts[c1 - 'a']++; counts[c2 - 'a']--; } int steps = 0; for (int i = 0; i < 26; i++) { if (counts[i] > 0) steps += counts[i]; } return steps; } }
-
// OJ: https://leetcode.com/problems/minimum-number-of-steps-to-make-two-strings-anagram/ // Time: O(N) // Space: O(1) class Solution { public: int minSteps(string s, string t) { int cnt[26] = {0}, ans = 0; for (char c : s) cnt[c - 'a']++; for (char c : t) cnt[c - 'a']--; for (int n : cnt) { if (n > 0) ans += n; } return ans; } };
-
print("Todo!")