Frequent Digit Identifier — Problem Statement & Solution Guide
Problem Description
You are provided with an array of integers, nums. Your task is to identify the specific digit (0-9) that appears strictly more than half the time across all digits present in the array. To clarify, first extract every single digit from each integer in the array to form a multiset of digits. Let total_digits be the total count of these extracted digits. You must return the digit d such that the frequency of d in this multiset is greater than total_digits / 2. If no such digit exists, return -1. Note that negative signs are ignored; only the absolute value's digits are considered.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Frequent Digit Identifier"
WHY DOES IT MATTER?
Identifying a majority element in a constrained domain is a recurring pattern in streaming analytics, fault‑tolerant consensus, and voting systems, where you need to decide a dominant value without storing the entire data history.
OPTIMIZATION CHALLENGE
The key insight is that the digit set is constant (10 values), allowing us to replace a generic hash map with a fixed‑size counter array or a constant‑space voting scheme, collapsing both time and space to linear and O(1) respectively.
REAL-WORLD CONNECTION
Consider a distributed logging service where each node reports a status code (0‑9). Determining the code that appears in more than half of the logs helps quickly detect a systemic issue, analogous to the frequent digit identifier.
In an interview, start by extracting digits on the fly while iterating the array; immediately apply Boyer‑Moore to keep the solution in one pass and O(1) extra memory—this demonstrates both algorithmic depth and practical coding efficiency.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem reduces to finding a majority element in a multiset of digits extracted from the given integers. After flattening each number into its constituent decimal digits, we obtain a sequence where each element belongs to the limited alphabet {0,1,…,9}. The classic majority‑element guarantee (an element appearing more than ⌊n/2⌋ times) can be solved in linear time using the Boyer‑Moore voting algorithm, which maintains a candidate and a counter, discarding pairs of different elements. Because the digit domain is tiny, a simple frequency array of size ten also yields a linear‑time solution with O(1) extra space, but the Boyer‑Moore approach is optimal when we want to avoid any auxiliary storage beyond a few variables.
A naïve approach would concatenate all numbers into a string or list, then count each digit with a hash map, resulting in O(N) time but O(N) auxiliary space for the map entries. For very large inputs (e.g., millions of numbers with billions of digits), that extra space becomes prohibitive. By exploiting the bounded digit range, we can either use a fixed‑size array (10 counters) or the constant‑space Boyer‑Moore method, both achieving O(N) time and O(1) additional space, which scales gracefully to massive datasets.
Interview Questions on This Problem
Q1How would you adapt the solution if the guarantee of a majority digit is removed and you need to return any digit that appears more than ⌊total_digits/3⌋ times?
Use the extended Boyer‑Moore algorithm that tracks up to two candidates, because at most two elements can appear more than n/3 times. After the first pass to find candidates, perform a second pass to verify their actual counts.
Q2What changes are required to handle numbers in bases other than decimal, say hexadecimal, while still finding a digit that appears > half the time?
Adjust the digit extraction to parse each number in the target base and expand the candidate set to the base's digit range (e.g., 0‑15 for hex). The same majority‑vote logic applies because the domain size remains constant.
Q3Explain how you would parallelize the digit counting for a distributed system processing terabytes of integer data.
Partition the input across workers, each computing a local frequency array of size ten (or a local Boyer‑Moore candidate). Then aggregate the local results: sum the frequency arrays to get global counts, or combine candidates using a second reduction pass that re‑applies the voting algorithm on the merged candidates.
Examples
Input
nums = [11, 12, 11]
Output
1
Explanation: Extract digits: 1, 1, 1, 2, 1, 1. Total digits = 6. Frequencies: 1 appears 5 times, 2 appears 1 time. Threshold is 6/2 = 3. Since 5 > 3, the answer is 1.
Input
nums = [23, 45, 67]
Output
-1
Explanation: Extract digits: 2, 3, 4, 5, 6, 7. Total digits = 6. Frequencies: All digits appear exactly once. Threshold is 3. No digit appears more than 3 times, so return -1.
Input
nums = [99, 99, 99, 12]
Output
9
Explanation: Extract digits: 9, 9, 9, 9, 9, 9, 1, 2. Total digits = 8. Frequencies: 9 appears 6 times, 1 appears 1 time, 2 appears 1 time. Threshold is 8/2 = 4. Since 6 > 4, the answer is 9.
Input
nums = [-11, 22, 33]
Output
-1
Explanation: Extract digits (ignoring signs): 1, 1, 2, 2, 3, 3. Total digits = 6. Frequencies: 1 appears 2 times, 2 appears 2 times, 3 appears 2 times. Threshold is 3. No digit exceeds 3 occurrences, so return -1.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- The total number of digits across all integers in nums will not exceed 10^6
Optimal Approach & Strategy
Apply Boyer‑Moore majority voting while extracting digits, which uses only two integer variables and a single pass, followed by a verification pass.
Brute Force Approach
Extract every digit, store them in a list, then use a hash map to count frequencies and finally scan for a digit with count > total/2.
Verified Code Solutions
function solution(nums) {
if (nums.length === 0) return -1;
let maxCount = 0;
let result = -1;
for (let num of nums) {
let count = 0;
for (let n of nums) {
if (n === num) count++;
}
if (count > maxCount) {
maxCount = count;
result = num;
}
}
return maxCount > nums.length / 2 ? result : -1;
}class Solution {
public:
int solution(vector<int>& nums) {
if (nums.size() == 0) return -1;
int maxCount = 0;
int result = -1;
for (int num : nums) {
int count = 0;
for (int n : nums) {
if (n == num) count++;
}
if (count > maxCount) {
maxCount = count;
result = num;
}
}
return maxCount > nums.size() / 2 ? result : -1;
}
};class Solution {
public int solution(int[] nums) {
if (nums.length == 0) return -1;
int maxCount = 0;
int result = -1;
for (int num : nums) {
int count = 0;
for (int n : nums) {
if (n == num) count++;
}
if (count > maxCount) {
maxCount = count;
result = num;
}
}
return maxCount > nums.length / 2 ? result : -1;
}
}def solution(nums):
if not nums:
return -1
max_count = 0
result = -1
for num in nums:
count = nums.count(num)
if count > max_count:
max_count = count
result = num
return result if max_count > len(nums) / 2 else -1function solution(nums) {
if (nums.length === 0) return -1;
let maxCount = 0;
let result = -1;
for (let num of nums) {
let count = 0;
for (let n of nums) {
if (n === num) count++;
}
if (count > maxCount) {
maxCount = count;
result = num;
}
}
return maxCount > nums.length / 2 ? result : -1;
}Solve in Interative Editor
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