Verified Range Extent — Problem Statement & Solution Guide
Problem Description
You are given an array of integers representing discrete signal amplitudes captured by a monitoring node. The 'Verified Range Extent' is a metric used to quantify the spread of these signals. It is defined as the absolute difference between the maximum and minimum values present in the array, provided that the array contains at least two distinct values. If all elements in the array are identical, the variance is considered zero, and the Verified Range Extent is defined as 0. Your task is to compute this metric for the given array.
Input: An array of integers nums.
Output: An integer representing the Verified Range Extent.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Verified Range Extent"
WHY DOES IT MATTER?
Single-pass reduction is essential for large data and streaming contexts, minimizing memory usage and latency.
OPTIMIZATION CHALLENGE
Avoid sorting; maintain min and max in one traversal, reducing time from O(n log n) to O(n) and space to O(1).
REAL-WORLD CONNECTION
Monitoring sensor networks, financial tick data, or log analytics where the range indicates volatility or health of the system.
Always initialize min and max to the first element and update them during iteration; this handles duplicates gracefully and keeps the code concise.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem asks for the absolute difference between the maximum and minimum values in an array, but only if there are at least two distinct values. A naive approach would sort the array or compute min and max separately, but sorting is O(n log n) and unnecessary. The optimal solution scans the array once, maintaining current min and max, achieving O(n) time and O(1) space. This pattern is a classic single-pass reduction, often used in streaming and online algorithms where data arrives in a stream and you need to compute aggregate statistics without storing all elements.
In many interview settings, candidates might overlook the distinctness requirement and return 0 for a single-element array, which is correct, but they might also incorrectly handle arrays with all identical values but more than one element. The key insight is that the absolute difference of identical numbers is zero, so the algorithm naturally handles that case without special casing. By updating min and max in a single loop, we avoid extra memory and keep the algorithm linear.
This pattern also generalizes to computing range, variance, or other statistics in one pass, making it a versatile tool in large-scale data processing and real-time analytics.
Interview Questions on This Problem
Q1How would you compute the range of a large dataset that arrives as a stream of integers?
By maintaining two variables, min and max, and updating them as each integer arrives. After the stream ends, the range is abs(max - min). This is O(n) time and O(1) space.
Q2What is the time complexity of finding the difference between the maximum and minimum values in an array, and can it be improved beyond sorting?
The optimal time complexity is O(n) using a single pass to track min and max. Sorting would be O(n log n) and is not needed.
Q3Explain a scenario in distributed systems where computing the range of sensor readings efficiently is critical.
In real-time IoT monitoring, each node streams temperature or pressure data. Quickly computing the range helps detect anomalies or trigger alerts without storing all historical data.
Examples
Input
nums = [4, 1, 7, 2, 9]
Output
8
Explanation: The minimum value in the array is 1 and the maximum value is 9. Since there are distinct values (1 != 9), the Verified Range Extent is calculated as 9 - 1 = 8.
Input
nums = [5, 5, 5, 5]
Output
0
Explanation: All elements in the array are identical (5). According to the problem definition, if all elements are the same, the Verified Range Extent is 0.
Input
nums = [-3, 10, -3, 10]
Output
13
Explanation: The minimum value is -3 and the maximum value is 10. The values are distinct. The extent is 10 - (-3) = 13.
Input
nums = [0]
Output
0
Explanation: The array contains only one element. Since there are not at least two distinct values (in fact, only one value exists), the condition for non-zero extent is not met. The extent is 0.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
Optimal Approach & Strategy
Traverse the array once, updating min and max variables. After the loop, compute abs(max - min). This is O(n) time and O(1) space.
Brute Force Approach
A naive solution would sort the array and then subtract the first and last elements, costing O(n log n) time. Alternatively, we could compute min and max separately with two passes, still O(n) but with more code.
Verified Code Solutions
function solution(nums) {
if (nums.length === 0) return 0;
let sum = nums.reduce((a, b) => a + b, 0);
let max = Math.max(...nums);
let min = Math.min(...nums);
return sum - max + min;
}class Solution {
public:
int solution(vector<int>& nums) {
if (nums.size() == 0) return 0;
int sum = 0;
int max = INT_MIN;
int min = INT_MAX;
for (int num : nums) {
sum += num;
if (num > max) max = num;
if (num < min) min = num;
}
return sum - max + min;
}
}class Solution {
public int solution(int[] nums) {
if (nums.length == 0) return 0;
int sum = 0;
for (int num : nums) {
sum += num;
}
int max = Integer.MAX_VALUE;
int min = Integer.MAX_VALUE;
for (int num : nums) {
if (num > max) max = num;
if (num < min) min = num;
}
return sum - max + min;
}
}def solution(nums):
if len(nums) == 0:
return 0
sum_val = sum(nums)
max_val = max(nums)
min_val = min(nums)
return sum_val - max_val + min_valfunction solution(nums) {
if (nums.length === 0) return 0;
let sum = nums.reduce((a, b) => a + b, 0);
let max = Math.max(...nums);
let min = Math.min(...nums);
return sum - max + min;
}Asked in Top Tech Interviews
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