## Uber Interview Question

Software Engineers**Team:**Marketplace

**Country:**United States

Over my interpretation, most frequent is different of longest sequence.

And i have a question, if there is more elements with the same frequency? Should return a list of them?

```
from collections import Iterable
def get_most_frequent(input_list):
max_sequence = 0
most_frequent = None
numbers_frequence = {}
if(not isinstance(input_list, Iterable)):
return most_frequent
for n in input_list:
if(numbers_frequence.get(n, False) is not False):
numbers_frequence[n] += 1
else:
numbers_frequence[n] = 1
if(numbers_frequence[n] > max_sequence):
max_sequence = numbers_frequence[n]
most_frequent = [n]
elif(numbers_frequence[n] is max_sequence):
most_frequent.append(n)
return most_frequent
print(get_most_frequent([0,1,2,3,4,4,4]))
print(get_most_frequent([5,5,5,7,7,7,8,9]))
print(get_most_frequent([0]))
```

Not overly efficient, but this should get the job done (java):

```
public static void findKthMostFrequent(int[] arr, int k) {
Map<Integer, Integer> intToFrequency = new HashMap<>(arr.length);
// O(n)
for (int i=0; i<arr.length; i++) {
int elem = arr[i];
int freq = 0;
if (intToFrequency.containsKey(elem)) {
freq = intToFrequency.get(elem);
}
intToFrequency.put(elem, freq + 1);
}
// O(n)
Map<Integer, List<Integer>> frequencyToInts = new HashMap<>();
for (Integer elem : intToFrequency.keySet()) {
Integer freq = intToFrequency.get(elem);
List<Integer> intsForFrequency = frequencyToInts.getOrDefault(freq, new ArrayList<>());
intsForFrequency.add(elem);
frequencyToInts.put(freq, intsForFrequency);
}
// Sort the keys: O(nlogn) worst case
List<Integer> frequenciesAscending = new ArrayList<>(frequencyToInts.keySet()); // another O(n) to convert to list, there might be some Set sorting function to avoid this
Collections.sort(frequenciesAscending); // O(nlogn), merge sort
int kthFrequency = frequenciesAscending.get(frequenciesAscending.size()-k);
System.out.println("Kth most frequent element(s)" + frequencyToInts.get(kthFrequency));
// time complexity: O(n + n + nlogn) = O(nlogn)
}
```

```
public List<Integer> topKFrequent(int[] nums, int k) {
//k most frequent element
int count = 0;
Map<Integer,Integer> map = new HashMap<>();
List<Integer> result = new ArrayList<>();
for(int i = 0; i < nums.length;i++){
if(map.containsKey(nums[i])){
map.put(nums[i],map.get(nums[i])+1);
}else
map.put(nums[i],1);
}
while(count < k){
int max = 0;
int value = 0;
for(Map.Entry<Integer, Integer> entry: map.entrySet()){
max = Math.max(max, entry.getValue());
if(entry.getValue() >= max){
max = entry.getValue();
value = entry.getKey();
}
}
map.remove(value);
result.add(value);
count++;
}
return result;
}
```

Count each number, and put into hashmap. Dump the hashmap contents into an auxiliary array. Perform quickselect on that array. Total O(n) time and O(n) space.

```
from collections import defaultdict
def helper(aux, lo, hi, c):
pivot_count, _ = aux[hi]
i = lo # i represents the exclusive upper bound on count < pivot_count
k = hi # k represents the exclusive lower bound on count > pivot_count
j = lo # i < j < k; (j, k) is the interval where count == pivot_count
while j <= k:
count, num = aux[j]
if count < pivot_count:
aux[i], aux[j] = aux[j], aux[i]
i += 1
j += 1
elif count > pivot_count:
aux[j], aux[k] = aux[k], aux[j]
k -= 1
else:
j += 1
aux[k + 1], aux[hi] = aux[hi], aux[k + 1]
left, right = i, k + 1
if c - 1 < left:
return helper(aux, lo, left - 1, c)
elif left <= c - 1 <= right:
return aux[left][1]
else:
return helper(aux, right + 1, hi, c)
def kthMostFrequentNumber(nums, k):
assert k > 0, f'k must be greater than 0: {k}'
assert len(nums) > 0, f'nums must be non-empty'
counts = defaultdict(int)
for num in nums:
counts[num] += 1
assert k <= len(counts), f'k must be between 1 and {len(counts)}: {k}'
aux = [(counts[num], num) for num in nums]
return helper(aux, 0, len(nums) - 1, k)
print(kthMostFrequentNumber([1, 1, 1, 1, 2, 2, 2, 2, 3, 3], 1))
print(kthMostFrequentNumber([1, 2, 3, 2, 1, 2, 2, 2, 3], 2))
print(kthMostFrequentNumber([1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 5, 5], 4))
```

I'll give you a hint. You can use hash tables for this.

- utkarsh.bhatt12 January 04, 2018Simply iterate over the array and store the element and the number of occurrence in that array.

Now return the element which has the highest number of occurrences in your hash table. This should be O(n) solution.

Please correct me if I've made any mistakes.