Range Span of an Integer Sequence — Problem Statement & Solution Guide
Problem Description
Given an array of integers, compute the range span of the sequence. The range span is defined as the arithmetic difference between the maximum value and the minimum value present in the array. This metric represents the total width of the value distribution within the dataset.
If the array contains only one element, the range span is zero, as there is no variation between the maximum and minimum values.
Your task is to implement a function that takes an array of integers as input and returns the computed range span as an integer.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Range Span of an Integer Sequence"
WHY DOES IT MATTER?
The min/max pattern is a foundational technique for many range-related problems, enabling linear-time solutions that are essential for real-time analytics and large-scale data processing.
OPTIMIZATION CHALLENGE
The key insight is that the maximum and minimum can be updated independently in a single pass, eliminating the need for pairwise comparisons or sorting.
REAL-WORLD CONNECTION
Consider a distributed log aggregation system where each node reports its highest and lowest latency; aggregating these extremes across nodes yields the overall latency range without inspecting every individual log entry.
When presenting this pattern, emphasize the importance of initializing min to +∞ and max to -∞ (or the first element) to avoid off-by-one errors, and discuss how this pattern scales to streaming data.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The range span of an integer sequence is simply the difference between the maximum and minimum values in the array. A naive approach would compare every pair of elements to find the extremes, resulting in an O(n^2) time complexity that quickly becomes infeasible for large datasets. The optimal paradigm reduces this to a single linear scan: maintain two variables—currentMin and currentMax—update them as each element is processed, and compute the span as currentMax - currentMin. This O(n) time, O(1) space solution is both elegant and scalable, making it ideal for production systems that handle millions of records.
Interview Questions on This Problem
Q1How would you compute the range span of an array in O(n) time, and why is this approach preferred over a nested loop in a production environment?
By iterating once through the array, tracking the minimum and maximum values seen so far. This linear time complexity avoids the quadratic blow-up of nested loops, ensuring the algorithm remains efficient even for very large inputs, which is critical in high-throughput services.
Q2A fintech company needs to monitor the volatility of transaction amounts in real time. How could the range span algorithm be adapted to handle a sliding window of the last N transactions?
Maintain a deque or balanced BST to store the last N values, updating min and max as elements enter and leave the window. This allows constant-time retrieval of the current span while keeping the window size fixed, enabling real-time volatility metrics.
Q3During a coding interview at a high-growth startup, you are asked to explain why you would choose a single-pass min/max approach over sorting the array to find the extremes. What points would you emphasize?
Sorting is O(n log n) and requires additional memory, whereas the single-pass method is O(n) time and O(1) space. In an interview, highlighting these efficiency gains demonstrates awareness of algorithmic trade-offs and a focus on scalable solutions.
Examples
Input
nums = [3, 1, 4, 1, 5, 9, 2, 6]
Output
8
Explanation: Step 1: Identify the minimum value in the array. The smallest number is 1. Step 2: Identify the maximum value in the array. The largest number is 9. Step 3: Calculate the difference: 9 - 1 = 8. Therefore, the range span is 8.
Input
nums = [42]
Output
0
Explanation: Step 1: The array contains a single element, 42. Step 2: The minimum value is 42. Step 3: The maximum value is 42. Step 4: Calculate the difference: 42 - 42 = 0. Therefore, the range span is 0.
Input
nums = [-5, 10, -3, 7, 2]
Output
15
Explanation: Step 1: Identify the minimum value. The smallest number is -5. Step 2: Identify the maximum value. The largest number is 10. Step 3: Calculate the difference: 10 - (-5) = 10 + 5 = 15. Therefore, the range span is 15.
Input
nums = [100, 100, 100, 100]
Output
0
Explanation: Step 1: Identify the minimum value. All elements are 100, so the minimum is 100. Step 2: Identify the maximum value. All elements are 100, so the maximum is 100. Step 3: Calculate the difference: 100 - 100 = 0. Therefore, the range span is 0.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
Optimal Approach & Strategy
Traverse the array once, updating two variables: currentMin and currentMax. After the loop, compute currentMax - currentMin, achieving O(n) time and O(1) space.
Brute Force Approach
A naive method would compare every pair of elements to find the maximum and minimum, requiring nested loops and O(n^2) time. This quickly becomes impractical as the array size grows.
Verified Code Solutions
function solution(nums) { return Math.max(...nums) - Math.min(...nums); }class Solution { public: int solution(vector<int>& nums) { int max = INT_MIN; int min = INT_MAX; for (int num : nums) { if (num > max) max = num; if (num < min) min = num; } return max - min; } };class Solution { public int solution(int[] nums) { int max = Integer.MIN_VALUE; int min = Integer.MAX_VALUE; for (int num : nums) { if (num > max) max = num; if (num < min) min = num; } return max - min; } }def solution(nums): return max(nums) - min(nums)function solution(nums) { return Math.max(...nums) - Math.min(...nums); }Asked in Top Tech Interviews
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