Unique Frequency Values — Problem Statement & Solution Guide
Problem Description
You are provided with an integer array nums representing a set of observed signal intensities. Your objective is to identify the distinct intensity values that occur exactly once within the dataset. Specifically, you must return a list of all unique integers from nums whose frequency of appearance is precisely 1. If no such values exist, return an empty list. The order of elements in the returned list does not matter, but each value must appear only once in the result.
This problem requires you to analyze the distribution of values in the array and filter out those that are repeated. It is a common pattern in data analysis where identifying singular occurrences is critical for anomaly detection or baseline establishment. You should aim for an efficient solution that avoids unnecessary nested loops, leveraging hash-based data structures for optimal performance.
Input: An array of integers nums.
Output: An array of integers containing all values from nums that appear exactly once.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Unique Frequency Values"
WHY DOES IT MATTER?
Frequency counting is a fundamental pattern for problems where the relationship between elements matters more than their values alone. Recognizing this pattern lets you replace nested loops with hash‑based aggregations, dramatically improving scalability.
OPTIMIZATION CHALLENGE
The key insight is to separate counting from selection, using a constant‑time update structure (hash map) to achieve linear time overall, rather than repeatedly scanning the array for each element's count.
REAL-WORLD CONNECTION
In distributed logging systems, you often need to identify log messages that appear only once (e.g., rare error codes) among billions of entries; a hash map or count‑min sketch serves the same purpose as the frequency map in this problem.
During an interview, first state the O(n²) brute‑force idea, then immediately propose the hash‑map as the natural optimization; this shows you can spot the inefficiency and apply the right data structure without over‑engineering.
COMPLEXITY AT A GLANCE
O(n)O(k)Core Theory — Why This Approach?
The problem reduces to counting the frequency of each integer in the input array and then selecting those with a count of exactly one. A naive double‑loop that compares each element with every other element runs in O(n²) time, which quickly becomes infeasible for large n because the number of comparisons grows quadratically. The optimal paradigm leverages a hash map (or dictionary) to record frequencies in a single linear pass, turning the counting step into O(1) average‑time updates per element. After the frequency map is built, a second linear scan extracts the keys whose associated value equals one, preserving the required O(n) overall time complexity while using O(k) extra space, where k is the number of distinct values.
This approach exemplifies the "frequency counting" pattern, a staple in array and string problems such as finding the first non‑repeating character or determining if two arrays are anagrams. By decoupling the counting from the selection phase, we avoid repeated scans of the same data and keep both time and space usage predictable. Moreover, the hash‑based solution gracefully handles negative numbers and large integer ranges without needing auxiliary arrays sized to the maximum possible value.
Interview Questions on This Problem
Q1How would you modify the solution if the output must preserve the original order of appearance of the unique elements?
Maintain a list (or array) to record the order of first occurrence while populating the frequency map. After counting, iterate over the original array and add an element to the result list only if its frequency is one; this guarantees order preservation with O(n) time and O(k) extra space.
Q2Can you solve the problem in O(n) time and O(1) extra space assuming the input array contains only integers in the range [0, n‑1]?
Yes. Use the array itself as a frequency counter by incrementing the value at index nums[i] % n by n for each element, then a second pass identifies indices whose value is less than 2*n, indicating a single occurrence. This in‑place counting works because the range constraint guarantees indices are valid and avoids extra storage.
Q3What would be the impact on time and space complexity if the input were a read‑only stream rather than a mutable array?
With a read‑only stream you cannot revisit elements, so you must store frequencies as you go, still requiring O(k) space for the hash map. The time remains O(n) because each element is processed once, but you lose the ability to perform a second pass for order preservation unless you also store the stream elements, which would increase space to O(n).
Examples
Input
nums = [4, 2, 4, 7, 2, 9]
Output
[7, 9]
Explanation: First, count the frequency of each element: 4 appears 2 times, 2 appears 2 times, 7 appears 1 time, and 9 appears 1 time. The values with a frequency of exactly 1 are 7 and 9. Thus, the output is [7, 9].
Input
nums = [1, 1, 1, 1]
Output
[]
Explanation: Count the frequencies: 1 appears 4 times. Since no value appears exactly once, the result is an empty list.
Input
nums = [5, 3, 5, 8, 3, 8, 2]
Output
[2]
Explanation: Frequencies: 5 appears 2 times, 3 appears 2 times, 8 appears 2 times, and 2 appears 1 time. Only 2 has a frequency of 1. Therefore, the output is [2].
Input
nums = [10, 20, 30, 40]
Output
[10, 20, 30, 40]
Explanation: Each element (10, 20, 30, 40) appears exactly once. Since all values are unique in their frequency (count = 1), all of them are included in the result.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- The input array may contain duplicate values.
- The order of elements in the output array is not specified.
Optimal Approach & Strategy
Use a hash map to record frequencies in a single pass, then iterate again to collect keys with a count of one. This runs in O(n) time with O(k) additional space, where k is the number of distinct values.
Brute Force Approach
Loop over each element and for each one scan the entire array to count its occurrences; if the count is one, add it to the result. This requires O(n²) time and O(1) extra space.
Verified Code Solutions
function solution(nums) {
const frequencyMap = {};
for (let num of nums) {
frequencyMap[num] = (frequencyMap[num] || 0) + 1;
}
const result = [];
for (let num in frequencyMap) {
if (frequencyMap[num] === 1) {
result.push(parseInt(num));
}
}
return result;
}class Solution {
public:
vector<int> solution(vector<int>& nums) {
unordered_map<int, int> frequencyMap;
for (int num : nums) {
frequencyMap[num]++;
}
vector<int> result;
for (auto& pair : frequencyMap) {
if (pair.second == 1) {
result.push_back(pair.first);
}
}
return result;
}
}class Solution {
public int[] solution(int[] nums) {
Map<Integer, Integer> frequencyMap = new HashMap<>();
for (int num : nums) {
frequencyMap.put(num, frequencyMap.getOrDefault(num, 0) + 1);
}
List<Integer> result = new ArrayList<>();
for (int num : frequencyMap.keySet()) {
if (frequencyMap.get(num) == 1) {
result.add(num);
}
}
return result.stream().mapToInt(Integer::intValue).toArray();
}
}def solution(nums):
frequency_map = {}
for num in nums:
frequency_map[num] = frequency_map.get(num, 0) + 1
result = []
for num in frequency_map:
if frequency_map[num] == 1:
result.append(int(num))
return resultfunction solution(nums) {
const frequencyMap = {};
for (let num of nums) {
frequencyMap[num] = (frequencyMap[num] || 0) + 1;
}
const result = [];
for (let num in frequencyMap) {
if (frequencyMap[num] === 1) {
result.push(parseInt(num));
}
}
return result;
}Asked in Top Tech Interviews
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