Problem:
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:
class Solution {
public int minDeletions(String str) {
// create array containing frequency of each character
int[] freqArr = new int[26];
for(char c: str.toCharArray()) {
freqArr[c - 'a']++;
}
// sort frequency array
Arrays.sort(freqArr);
// traverse in reverse order and calculate deletions as freqArr[i] - maxAllowedFreq
int deletions = 0;
int maxAllowedFreq = str.length();
for(int i=freqArr.length - 1; i>=0; i--) {
if(freqArr[i] > maxAllowedFreq) {
deletions += (freqArr[i] - maxAllowedFreq);
freqArr[i] = maxAllowedFreq;
}
maxAllowedFreq = freqArr[i] - 1;
if(maxAllowedFreq < 0) maxAllowedFreq = 0;
}
return deletions;
}
}
Time Complexity: O(N)
where N is the length of the given string. Sorting can be ignored as the array size will always be 26.
Space Complexity: O(1)
as array size is always 26, hence it can be ignored.