Verified Target Index — Problem Statement & Solution Guide
Problem Description
You are provided with an array nums of integers representing a sequence of system metrics. Your objective is to identify the 'Verified Target Index'. An index i is considered verified if the frequency of the value nums[i] in the entire array is strictly greater than the frequency of any other distinct value in the array. If multiple values share the highest frequency, no index is verified. If a unique value holds the maximum frequency, return the smallest index i where nums[i] equals that value. If no such unique maximum frequency exists, return -1.
The core challenge involves efficiently determining the frequency distribution of all elements and identifying the unique mode. This requires a single pass to count occurrences and a subsequent check to ensure uniqueness of the maximum count. The solution must handle large input sizes efficiently, leveraging hash maps for O(1) average-case lookups and updates.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Verified Target Index"
WHY DOES IT MATTER?
Frequency counting with hash maps is a fundamental pattern for any problem that asks about majority, mode, or dominance relationships. Recognizing when a simple count suffices prevents unnecessary sorting or nested loops, directly impacting runtime on large datasets.
OPTIMIZATION CHALLENGE
The key insight is separating counting from decision making. By building a global frequency map in O(n) time, we avoid repeated scans for each element. The second pass merely checks the max frequency and its uniqueness, which is also O(n). This reduces the naive O(n²) to optimal linear complexity.
REAL-WORLD CONNECTION
In distributed logging systems, identifying the most frequent error code (and ensuring it's not tied) determines whether an alert should be raised. The same counting logic applies: aggregate counts across nodes, find the unique top error, and trigger a response.
During an interview, first state the counting‑then‑checking strategy, then write the hash map construction. After that, explicitly verify that the max frequency is unique before returning any index—this extra validation often trips candidates.
COMPLEXITY AT A GLANCE
O(n)O(n)Core Theory — Why This Approach?
The core of this problem lies in frequency analysis, a classic application of hash maps for counting occurrences in linear time. A naive solution would scan the array for each element, counting its occurrences with a nested loop, leading to O(n²) time which quickly becomes infeasible for large inputs (n up to 10⁵ or more). By leveraging a hash map (or dictionary) we can aggregate counts in a single pass, turning the problem into a two‑step linear scan: first to build the frequency table, then to identify the maximum frequency and verify its uniqueness. This paradigm—"count then decide"—is a staple in problems where relative frequencies dictate the answer, and it avoids the combinatorial explosion of repeated scans.
Interview Questions on This Problem
Q1How would you modify the solution if you needed to return all indices of the value with the highest unique frequency instead of just one?
First compute the frequency map, then determine the maximum frequency and ensure it is unique. If unique, iterate the original array a second time collecting every index where the element equals the max‑frequency value; otherwise return an empty list.
Q2Can you solve the problem in O(1) additional space if the input array is mutable and you may reorder it?
Yes. Sort the array in‑place (O(n log n) time) and then scan to find the longest run of equal numbers; track the run length and its value. After confirming the run length is strictly greater than any other run, you can locate an original index via a linear scan or by storing original positions before sorting.
Q3What would be the impact on the algorithm if the array could contain billions of elements that do not fit into memory?
You would need an external‑memory (streaming) approach: use a two‑pass algorithm with a hash‑based counting sketch (e.g., Count‑Min Sketch) to approximate frequencies, then a second pass to verify the candidate with the highest estimated count, ensuring the uniqueness condition.
Examples
Input
nums = [4, 2, 4, 7, 4]
Output
0
Explanation: Step 1: Count frequencies: {4: 3, 2: 1, 7: 1}. Step 2: Identify max frequency: 3. Step 3: Check uniqueness: Only 4 has frequency 3. Step 4: Find smallest index of 4: Index 0. Output: 0.
Input
nums = [1, 2, 2, 1]
Output
-1
Explanation: Step 1: Count frequencies: {1: 2, 2: 2}. Step 2: Identify max frequency: 2. Step 3: Check uniqueness: Both 1 and 2 have frequency 2. Not unique. Output: -1.
Input
nums = [5, 5, 3, 5, 9, 5]
Output
0
Explanation: Step 1: Count frequencies: {5: 4, 3: 1, 9: 1}. Step 2: Identify max frequency: 4. Step 3: Check uniqueness: Only 5 has frequency 4. Step 4: Find smallest index of 5: Index 0. Output: 0.
Input
nums = [7, 8, 9, 10]
Output
-1
Explanation: Step 1: Count frequencies: {7: 1, 8: 1, 9: 1, 10: 1}. Step 2: Identify max frequency: 1. Step 3: Check uniqueness: All values have frequency 1. Not unique. Output: -1.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- The array may contain negative integers and duplicates.
- Time complexity must be O(N) for optimal performance.
- Space complexity must be O(N) to store frequency counts.
Optimal Approach & Strategy
Build a frequency hash map in one pass, then in a second pass find the maximum frequency and ensure it is unique; finally return any index holding that value.
Brute Force Approach
For each index, scan the whole array to count occurrences of its value, then compare that count against counts of all other distinct values.
Verified Code Solutions
function solution(nums) {
let sum = 0;
for (let num of nums) {
sum += num;
}
return sum;
}class Solution {
public:
int solution(vector<int> nums) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
};class Solution {
public int solution(int[] nums) {
int sum = 0;
for (int num : nums) {
sum += num;
}
return sum;
}
}def solution(nums):
sum = 0
for num in nums:
sum += num
return sumfunction solution(nums) {
let sum = 0;
for (let num of nums) {
sum += num;
}
return sum;
}Asked in Top Tech Interviews
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