A string s
is called good if there are no two different characters in s
that have the same frequency.
Given a string s
, return the minimum number of characters you need to delete to make s
good.
The frequency of a character in a string is the number of times it appears in the string. For example, in the string "aab"
, the frequency of 'a'
is 2
, while the frequency of 'b'
is 1
.
Example 1:
Input: s = "aab"
Output: 0
Explanation: s
is already good.
Example 2:
Input: s = "aaabbbcc" Output: 2 Explanation: You can delete two 'b's resulting in the good string "aaabcc". Another way it to delete one 'b' and one 'c' resulting in the good string "aaabbc".
Example 3:
Input: s = "ceabaacb" Output: 2 Explanation: You can delete both 'c's resulting in the good string "eabaab". Note that we only care about characters that are still in the string at the end (i.e. frequency of 0 is ignored).
Constraints:
1 <= s.length <= 105
s
contains only lowercase English letters.
Solution: Hashtable
The deletion order doesn’t matter, we can process from ‘a’ to ‘z’. Use a hashtable to store the “final frequency” so far, for each char, decrease its frequency until it becomes unique in the final frequency hashtable.
Time complexity: O(n + 26^2)
Space complexity: O(26)
C++
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 |
class Solution { public: int minDeletions(string s) { vector<int> freq(26); for (char c : s) ++freq[c - 'a']; unordered_set<int> seen; int ans = 0; for (int f : freq) { while (f && !seen.insert(f).second) { --f; ++ans; } } return ans; } }; |
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