2671. Frequency Tracker
Explanation
To solve this problem, we can use a HashMap to keep track of the frequency of each number in the data structure. We will also maintain a separate HashMap to keep track of the frequency of each frequency value. When adding a number, we update both HashMaps accordingly. When deleting one occurrence of a number, we update the HashMaps. To check if a particular frequency exists, we simply query the second HashMap.
Time Complexity
- Adding a number: O(1)
- Deleting one occurrence of a number: O(1)
- Checking if a frequency exists: O(1)
Space Complexity
O(n), where n is the number of unique numbers in the data structure.
import java.util.HashMap;
class FrequencyTracker {
private HashMap<Integer, Integer> numToFreq;
private HashMap<Integer, Integer> freqToCount;
public FrequencyTracker() {
numToFreq = new HashMap<>();
freqToCount = new HashMap<>();
}
public void add(int number) {
int prevFreq = numToFreq.getOrDefault(number, 0);
numToFreq.put(number, prevFreq + 1);
if (prevFreq > 0) {
int prevCount = freqToCount.get(prevFreq);
if (prevCount == 1) {
freqToCount.remove(prevFreq);
} else {
freqToCount.put(prevFreq, prevCount - 1);
}
}
int newFreq = prevFreq + 1;
freqToCount.put(newFreq, freqToCount.getOrDefault(newFreq, 0) + 1);
}
public void deleteOne(int number) {
if (!numToFreq.containsKey(number)) return;
int prevFreq = numToFreq.get(number);
if (prevFreq == 1) {
numToFreq.remove(number);
} else {
numToFreq.put(number, prevFreq - 1);
int prevCount = freqToCount.get(prevFreq);
if (prevCount == 1) {
freqToCount.remove(prevFreq);
} else {
freqToCount.put(prevFreq, prevCount - 1);
}
freqToCount.put(prevFreq - 1, freqToCount.getOrDefault(prevFreq - 1, 0) + 1);
}
}
public boolean hasFrequency(int frequency) {
return freqToCount.containsKey(frequency);
}
}Code Editor (Testing phase)
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