medium Top 'K' Frequent Numbers
Problem Statement
Given an unsorted array of numbers, find the top ‘K’ frequently occurring numbers in it.
Example 1:
Input: [1, 3, 5, 12, 11, 12, 11], K = 2
Output: [12, 11]
Explanation: Both '11' and '12' appeared twice.
Example 2:
Input: [5, 12, 11, 3, 11], K = 2
Output: [11, 5] or [11, 12] or [11, 3]
Explanation: Only '11' appeared twice; all other numbers appeared once.
Constraints:
- 1 <= nums.length <= 105
- -105 <= nums[i] <= 105
- k is in the range [1, the number of unique elements in the array].
- It is guaranteed that the answer is unique.
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✅ Solution Top 'K' Frequent Numbers
Problem Statement
Given an unsorted array of numbers, find the top ‘K’ frequently occurring numbers in it.
Example 1:
Input: [1, 3, 5, 12, 11, 12, 11], K = 2
Output: [12, 11]
Explanation: Both '11' and '12' appeared twice.
Example 2:
Input: [5, 12, 11, 3, 11], K = 2
Output: [11, 5] or [11, 12] or [11, 3]
Explanation: Only '11' appeared twice; all other numbers appeared once.
Constraints:
- 1 <= nums.length <= 105
- -105 <= nums[i] <= 105
- k is in the range [1, the number of unique elements in the array].
- It is guaranteed that the answer is unique.
Solution
This problem follows Top ‘K’ Numbers. The only difference is that in this problem, we need to find the most frequently occurring number compared to finding the largest numbers.
We can follow the same approach as discussed in the Top K Elements problem. However, in this problem, we first need to know the frequency of each number, for which we can use a HashMap. Once we have the frequency map, we can use a Min Heap to find the ‘K’ most frequently occurring number. In the Min Heap, instead of comparing numbers we will compare their frequencies in order to get frequently occurring numbers
Code
Here is what our algorithm will look like:
import java.util.*;
class Solution {
public List<Integer> findTopKFrequentNumbers(int[] nums, int k) {
// find the frequency of each number
Map<Integer, Integer> numFrequencyMap = new HashMap<>();
for (int n : nums)
numFrequencyMap.put(n, numFrequencyMap.getOrDefault(n, 0) + 1);
PriorityQueue<Map.Entry<Integer, Integer>> minHeap = new PriorityQueue<
Map.Entry<Integer, Integer>
>((e1, e2) -> e1.getValue() - e2.getValue());
// go through all numbers of the numFrequencyMap and push them in the minHeap, which
// will have top k frequent numbers.
// If the heap size is more than k, we remove the smallest (top) number
for (Map.Entry<Integer, Integer> entry : numFrequencyMap.entrySet()) {
minHeap.add(entry);
if (minHeap.size() > k) {
minHeap.poll();
}
}
// create a list of top k numbers
List<Integer> topNumbers = new ArrayList<>(k);
while (!minHeap.isEmpty()) {
topNumbers.add(minHeap.poll().getKey());
}
return topNumbers;
}
public static void main(String[] args) {
Solution sol = new Solution();
List<Integer> result = sol.findTopKFrequentNumbers(
new int[] { 1, 3, 5, 12, 11, 12, 11 },
2
);
System.out.println("Here are the K frequent numbers: " + result);
result = sol.findTopKFrequentNumbers(new int[] { 5, 12, 11, 3, 11 }, 2);
System.out.println("Here are the K frequent numbers: " + result);
}
}
Time Complexity
The time complexity of the above algorithm is
Space Complexity
The space complexity will be O(N). Even though we are storing only ‘K’ numbers in the heap. For the frequency map, however, we need to store all the ‘N’ numbers.
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