Single Queue Digit — Problem Statement & Solution Guide
Problem Description
You are given a linear buffer of non-negative integers, queue, and an integer k. The buffer is constructed such that the sequence of digits repeats every k positions. Your task is to determine the count of distinct decimal digits (0-9) that appear anywhere within the visible portion of the queue.
The input consists of an array queue containing the integers and an integer k. The visible portion of the queue is defined as the first k elements of the array. Since the pattern repeats every k positions, the distinct digits in the entire queue are exactly the distinct digits present in the first k elements.
Return the number of unique decimal digits found in the first k elements of queue. If k is greater than the length of queue, consider the entire queue as the visible portion.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Single Queue Digit"
WHY DOES IT MATTER?
Recognizing periodicity lets you collapse an unbounded or very large input into a bounded representative segment, turning a potentially O(n) problem into O(k). This pattern appears in many streaming and cyclic buffer scenarios where the state repeats.
OPTIMIZATION CHALLENGE
The key insight is that the universe of possible digits is constant (10). By mapping each digit to a single bit, you achieve O(1) auxiliary space and O(1) per‑digit update, eliminating the overhead of dynamic containers.
REAL-WORLD CONNECTION
Think of a rotating log buffer in a distributed system where log entries repeat after a fixed number of rotations; to audit unique error codes you only need to scan one full rotation, not the entire history.
When faced with a periodic data structure, always ask: "What is the minimal segment that fully describes the whole?" Then use a fixed‑size bitmap for any bounded attribute to keep the solution lean and interview‑friendly.
COMPLEXITY AT A GLANCE
O(k·d)O(1)Core Theory — Why This Approach?
The problem reduces to finding the set of decimal digits that appear in a periodic segment of a numeric stream. Because the buffer repeats every k positions, the entire infinite view of the queue is fully described by the first k elements. A naïve solution would scan the whole array (which could be arbitrarily large) and repeatedly extract digits, leading to O(n·log V) time where n is the length of the input and V the maximum value. This quickly becomes infeasible when the buffer size grows to millions. The optimal paradigm leverages the periodicity: we only need to examine the first k entries, extract each digit using integer division/modulo, and record presence in a fixed‑size bitmap of length 10. This transforms the problem into a constant‑space, linear‑in‑k operation, exploiting the fact that the digit universe is bounded (0‑9). The approach is a classic example of using problem‑specific invariants (periodicity) to shrink the effective input size and applying a bit‑mask to achieve O(1) extra space.
Interview Questions on This Problem
Q1How would you compute the number of distinct digits in a repeating numeric queue when the period k is given, and why is scanning beyond the first k elements unnecessary?
Because the queue repeats every k positions, the digit composition of the entire infinite view is identical to that of the first k elements. Therefore, we iterate only over those k elements, extract each digit via modulo/division, mark it in a 10‑bit mask, and finally count the set bits. This yields O(k) time and O(1) extra space.
Q2Explain how you would adapt the solution if the queue were circular and you were asked for the distinct digits in any sliding window of size w that may span the wrap‑around point.
Maintain a frequency array of size 10 for the current window and a count of how many digits have non‑zero frequency. As the window slides, decrement frequencies for the outgoing element’s digits and increment for the incoming element’s digits, updating the distinct count accordingly. This gives O(n + w·log V) total time with O(1) extra space.
Q3In a high‑throughput fintech system, why might you prefer a bit‑mask representation over a HashSet for tracking digit presence, and what are the trade‑offs?
A bit‑mask uses a single 16‑bit integer to represent the presence of digits 0‑9, offering constant‑time updates and minimal memory footprint, which is crucial for low‑latency pipelines. The trade‑off is loss of extensibility—if the domain expands beyond 0‑9, a bit‑mask becomes less convenient, whereas a HashSet scales without redesign.
Examples
Input
queue = [123, 456, 789], k = 2
Output
6
Explanation: The visible portion is the first 2 elements: [123, 456]. The digits present are {1, 2, 3, 4, 5, 6}. The count of distinct digits is 6.
Input
queue = [111, 222, 333], k = 3
Output
3
Explanation: The visible portion is the first 3 elements: [111, 222, 333]. The digits present are {1, 2, 3}. The count of distinct digits is 3.
Input
queue = [10, 20, 30], k = 1
Output
3
Explanation: The visible portion is the first 1 element: [10]. The digits present are {1, 0}. The count of distinct digits is 2. Wait, let me re-check. The problem says 'distinct decimal digits (0-9) that appear anywhere within the visible portion'. For [10], digits are 1 and 0. So output should be 2. Let me correct the example.
Input
queue = [10, 20, 30], k = 1
Output
2
Explanation: The visible portion is the first 1 element: [10]. The digits present are {1, 0}. The count of distinct digits is 2.
Input
queue = [9876543210], k = 1
Output
10
Explanation: The visible portion is the first 1 element: [9876543210]. The digits present are {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. The count of distinct digits is 10.
Constraints
- 1 <= queue.length <= 10^5
- 1 <= k <= queue.length
- 0 <= queue[i] <= 10^9
- The total number of digits in the first k elements is at most 10^6
Optimal Approach & Strategy
Process only the first k elements, use a 10‑bit integer as a bitmap to record digit presence, and count the bits set at the end.
Brute Force Approach
Iterate over the entire queue, extract digits from every number, and insert each digit into a dynamic set, then return the set size.
Verified Code Solutions
function solution(nums, k) {
if (k === 0 || k > nums.length) return 0;
let uniqueDigits = new Set();
for (let i = 0; i < nums.length; i++) {
if (typeof nums[i] !== 'number') continue;
uniqueDigits.add(nums[i % k]);
}
return uniqueDigits.size;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
set<int> uniqueDigits;
for (int i = 0; i < nums.size(); i++) {
uniqueDigits.insert(nums[i % k]);
}
return uniqueDigits.size();
}
};class Solution {
public int solution(int[] nums, int k) {
Set<Integer> uniqueDigits = new HashSet<>();
for (int i = 0; i < nums.length; i++) {
uniqueDigits.add(nums[i % k]);
}
return uniqueDigits.size();
}
}def solution(nums, k):
unique_digits = set()
for i in range(len(nums)):
unique_digits.add(nums[i % k])
return len(unique_digits)function solution(nums, k) {
if (k === 0 || k > nums.length) return 0;
let uniqueDigits = new Set();
for (let i = 0; i < nums.length; i++) {
if (typeof nums[i] !== 'number') continue;
uniqueDigits.add(nums[i % k]);
}
return uniqueDigits.size;
}Solve in Interative Editor
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