**Unique String Frequencies - Microsoft Top Interview Questions**

### Problem Statement :

Given a lowercase alphabet string s, return the minimum number of characters that we need to delete such that the frequency of each character occurs a unique number of times. Constraints n ≤ 100,000 where n is the length of s Example 1 Input s = "aabb" Output 1 Explanation Both "a" and "b" occur twice, so we can remove say "a" once to make the frequencies unique. Example 2 Input s = "abbccc" Output 0 Explanation The string already has unique frequencies for "a", "b" and "c": [1, 2, 3].

### Solution :

` ````
Solution in C++ :
int solve(string s) {
unordered_map<char, int> freq;
for (auto& c : s) freq[c]++;
vector<int> freqCounts;
for (auto& p : freq) freqCounts.push_back(p.second);
sort(freqCounts.rbegin(), freqCounts.rend());
int ret = 0;
int lastFreq = 1e9;
for (int freq : freqCounts) {
// freq must be at most lastFreq
if (freq > lastFreq) {
int rem = freq - lastFreq;
ret += rem;
freq -= rem;
}
lastFreq = max(0, freq - 1);
}
return ret;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public int solve(String s) {
// how many times does char "i" appear?
int[] occ = new int[26];
for (char c : s.toCharArray()) occ[c - 'a'] += 1;
// how many letters appear "i" times?
int[] freq = new int[s.length() + 1];
for (int i = 0; i < 26; i++) {
if (occ[i] > 0)
freq[occ[i]] += 1;
}
// greedy algorithim
int ans = 0;
for (int i = s.length(); i >= 1; i--) {
if (freq[i] > 1) {
ans += (freq[i] - 1);
freq[i - 1] += (freq[i] - 1);
}
}
return ans;
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, s):
freq = Counter(s)
deletions = 0
unique = set()
for f in freq.values():
if f in unique:
while f in unique and f > 0:
f -= 1
deletions += 1
unique.add(f)
return deletions
```

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