Tome Cache Aligner 11 — Problem Statement & Solution Guide
Problem Description
You are given a sequence of integers that represent performance metrics collected from a tome cache system. Your task is to identify the metric value that occurs most frequently in the sequence. If several values share the highest frequency, you must return the smallest of those values. The sequence is provided as a queue of integers, meaning the order of arrival is preserved, but the final answer depends only on the frequency of each distinct value.
Input format:
- The first line contains a single integer n (1 ≤ n ≤ 10^5), the number of metrics.
- The second line contains n space‑separated integers a1, a2, …, an (−10^9 ≤ ai ≤ 10^9), the metrics in the order they were recorded.
Output format:
- Output a single integer: the metric value that appears most frequently, breaking ties by choosing the smallest value.
The problem requires building a frequency hash map to count occurrences and then selecting the appropriate value. The solution should run in O(n) time and use O(k) additional space, where k is the number of distinct metrics.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Tome Cache Aligner 11"
WHY DOES IT MATTER?
Frequency counting is a foundational pattern in data analysis and system monitoring. It is essential for identifying outliers, popular items, and bottlenecks in performance metrics. Mastering this pattern allows engineers to efficiently process large datasets and make data-driven decisions.
OPTIMIZATION CHALLENGE
The key insight is to decouple the counting phase from the comparison phase. By using a hash map to store frequencies in O(1) average time, you avoid the O(n) scan for each element. The tie-breaking logic (smallest value) is handled during the single pass or a final scan of the map, ensuring the overall complexity remains linear.
REAL-WORLD CONNECTION
This pattern is directly analogous to cache hit/miss analysis in distributed systems. Identifying the most frequently accessed cache key helps in optimizing cache eviction policies (e.g., LFU - Least Frequently Used) or pre-fetching strategies to improve system latency.
In interviews, explicitly state that the 'queue' aspect is a red herring for the core algorithm. The order of arrival does not affect the frequency count or the smallest value tie-breaker. Focus on the hash map implementation and the logic for updating the maximum frequency and the corresponding minimum value.
COMPLEXITY AT A GLANCE
O(n)O(k)Core Theory — Why This Approach?
The problem of finding the most frequent element in a sequence, with a tie-breaker for the smallest value, is fundamentally a frequency counting problem. While the input is presented as a queue to emphasize the preservation of arrival order, the core task requires global knowledge of the entire dataset to determine frequencies. Naive approaches, such as iterating through the queue for each unique element to count its occurrences, result in O(n^2) time complexity, which is infeasible for large-scale performance metrics where n can reach millions. This quadratic behavior fails because it repeatedly scans the data structure without leveraging the associative properties of hash maps.
Interview Questions on This Problem
Q1At a fintech platform processing millions of transaction metrics per second, how would you modify this solution to handle a streaming input where the entire queue is not available in memory at once?
For streaming data, you cannot store the entire queue. Instead, you would maintain a hash map of frequencies and a min-heap (or a sorted structure) to track the top candidates. However, since the tie-breaker is the smallest value, you can maintain a variable for the current max frequency and the smallest value associated with it. As new elements arrive, update the frequency map. If a new frequency exceeds the current max, update the answer. If it equals the max, compare the values and keep the smaller one. This allows O(1) amortized time per element and O(k) space where k is the number of unique metrics.
Q2In a high-growth startup's logging system, if the 'smallest value' tie-breaker was replaced with 'most recently seen value', how would your data structure choice change?
If the tie-breaker is 'most recently seen', a simple hash map is no longer sufficient for O(1) retrieval of the most recent among ties. You would need to store the last seen index for each value in the hash map. When a tie in frequency occurs, you compare the last seen indices. Alternatively, if the input is a stream, you can simply keep track of the last value that achieved the current maximum frequency, as the 'most recent' property is naturally handled by the order of processing.
Q3At a global product company, if the metric values are bounded (e.g., 0 to 1000), how can you optimize space complexity compared to using a hash map?
If the range of values is small and known, you can use a fixed-size array (counting sort approach) instead of a hash map. This reduces the space complexity from O(k) to O(M), where M is the range of values, and eliminates the overhead of hash collisions and pointer chasing. The time complexity remains O(n) for counting, but the constant factors are significantly lower, making it more cache-friendly and faster in practice.
Examples
Input
7 1 2 2 3 3 3 4
Output
3
Explanation: Frequencies: 1→1, 2→2, 3→3, 4→1. The maximum frequency is 3 for value 3, so the answer is 3.
Input
5 5 5 4 4 3
Output
4
Explanation: Frequencies: 5→2, 4→2, 3→1. Both 5 and 4 have the highest frequency of 2; the smaller value is 4, so the answer is 4.
Input
4 10 20 30 40
Output
10
Explanation: All values appear once. The smallest value among them is 10, which is returned.
Input
6 0 0 0 0 0 0
Output
0
Explanation: The only value 0 appears six times, so the answer is 0.
Constraints
- 1 <= n <= 10^5
- -10^9 <= ai <= 10^9
Optimal Approach & Strategy
Use a hash map to count the frequency of each element in a single pass through the queue. Track the maximum frequency and the smallest value associated with it during the iteration or in a final pass over the map entries.
Brute Force Approach
Iterate through each unique element in the queue and count its occurrences by scanning the entire queue again. This results in O(n^2) time complexity, which is inefficient for large inputs.
Verified Code Solutions
function solution(nums) {
let sum = 0;
for (let i = 0; i < nums.length; i++) {
sum += nums[i];
}
if (nums.length % 2 === 1) {
sum += nums[Math.floor(nums.length / 2)];
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums) {
int sum = 0;
for (int i = 0; i < nums.size(); i++) {
sum += nums[i];
}
if (nums.size() % 2 === 1) {
sum += nums[nums.size() / 2];
}
return sum;
}
};class Solution {
public int solution(int[] nums) {
int sum = 0;
for (int i = 0; i < nums.length; i++) {
sum += nums[i];
}
if (nums.length % 2 === 1) {
sum += nums[nums.length / 2];
}
return sum;
}
}def solution(nums):
sum = 0
for i in range(len(nums)):
sum += nums[i]
if len(nums) % 2 === 1:
sum += nums[len(nums) // 2]
return sumfunction solution(nums) {
let sum = 0;
for (let i = 0; i < nums.length; i++) {
sum += nums[i];
}
if (nums.length % 2 === 1) {
sum += nums[Math.floor(nums.length / 2)];
}
return sum;
}Asked in Top Tech Interviews
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