Node Matrix Tracker 19 — Problem Statement & Solution Guide
Problem Description
You are given a sequence of N integers and an integer threshold K. Determine how many distinct values appear in the sequence at least K times. The answer must be computed efficiently using a frequency hash map.
Input Format:
- The first line contains two space‑separated integers N and K.
- The second line contains N space‑separated integers representing the sequence.
Output Format:
- Output a single integer – the count of distinct numbers whose occurrence count is greater than or equal to K.
The solution should run in O(N) time and O(N) auxiliary space.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Node Matrix Tracker 19"
WHY DOES IT MATTER?
Frequency counting is a fundamental pattern for summarizing large data sets efficiently.
OPTIMIZATION CHALLENGE
The key is reducing quadratic or sorting overhead to linear time by leveraging constant‑time hash operations.
REAL-WORLD CONNECTION
Databases use similar hash‑based indexes to quickly retrieve rows that appear a certain number of times, like detecting hot items.
Initialize the map lazily and avoid re‑scanning the input; a single pass plus a tiny post‑process loop is all you need.
COMPLEXITY AT A GLANCE
O(N)O(U)Core Theory — Why This Approach?
The problem reduces to counting frequencies of each integer in a stream, which is a classic use‑case for a hash map (or dictionary). By scanning the sequence once and incrementing the count for each value, we can later iterate over the map to count how many keys have a frequency ≥ K. Naïve approaches like nested loops or sorting each time would incur O(N^2) or O(N log N) overhead, which becomes prohibitive for large N (e.g., 10^6). The optimal paradigm leverages constant‑time average updates and lookups in a hash table, achieving linear time overall while using extra space proportional to the number of distinct elements.
Because the hash map stores only distinct values, the space complexity scales with the unique element count rather than N, making it memory‑efficient for inputs with many repetitions. This technique exemplifies the frequency‑count pattern, a cornerstone in problems involving mode, majority element, or threshold‑based counting, and it aligns with the broader principle of single‑pass data aggregation.
Interview Questions on This Problem
Q1How does using a hash map improve time complexity compared to a double‑loop solution?
A hash map provides O(1) average insertion and lookup, turning the counting step into a single pass. The double‑loop approach would require O(N^2) comparisons, which is infeasible for large N.
Q2What edge cases must you handle when K is larger than any element's frequency?
If K exceeds all frequencies, the answer is zero because no value meets the threshold. Ensure the algorithm does not mistakenly count uninitialized map entries.
Q3Can this solution be adapted to work with streaming data where N is unknown upfront?
Yes, maintain the hash map incrementally as each element arrives and update a running counter of values meeting the K‑threshold. This allows constant‑time updates without needing the full array.
Examples
Input
5 2 1 2 2 3 1
Output
2
Explanation: Frequencies: 1 → 2, 2 → 2, 3 → 1. Values 1 and 2 meet the threshold K=2, so the result is 2.
Input
7 3 4 4 4 5 5 6 6
Output
1
Explanation: Frequencies: 4 → 3, 5 → 2, 6 → 2. Only the value 4 appears at least 3 times, therefore the answer is 1.
Input
6 1 -1 0 -1 2 0 2
Output
3
Explanation: Frequencies: -1 → 2, 0 → 2, 2 → 2. Since K=1, every distinct value qualifies. There are three distinct values, so the output is 3.
Constraints
- 1 <= N <= 100000
- 1 <= K <= N
- -10^9 <= array[i] <= 10^9
- The algorithm must use O(N) time and O(N) extra space.
Optimal Approach & Strategy
Use a hash map to record frequencies in O(N) time, then iterate over the map to count keys with frequency ≥ K.
Brute Force Approach
A naive method would compare each element with every other to count occurrences, leading to O(N^2) time.
Verified Code Solutions
function solution(nums, K) {
let sum = 0;
for (let num of nums) {
if (num > K) {
sum += num;
}
}
return sum;
}class Solution {
public:
int solution(vector<int> nums, int K) {
int sum = 0;
for (int num : nums) {
if (num > K) {
sum += num;
}
}
return sum;
}
};class Solution {
public int solution(int[] nums, int K) {
int sum = 0;
for (int num : nums) {
if (num > K) {
sum += num;
}
}
return sum;
}
}def solution(nums, K):
sum = 0
for num in nums:
if num > K:
sum += num
return sumfunction solution(nums, K) {
let sum = 0;
for (let num of nums) {
if (num > K) {
sum += num;
}
}
return sum;
}Asked in Top Tech Interviews
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