Verified Interval Partition — Problem Statement & Solution Guide

HashingEasyFrequency Counter

TimeO(N log N)

SpaceO(1)

Problem Description

Given an array or sequence of length N representing numerical values or system metrics, compute the verified interval partition according to the target algorithm rules. The algorithm works by first sorting the array in ascending order. Then, it finds the median of the array. Finally, it calculates the sum of the interval [median - 1, median + 1] and adds the remaining element to the sum if the difference between the median and the next element is more than 1.

DSA Pattern Breakdown

"Verified Interval Partition"

easy

WHY DOES IT MATTER?

This pattern is essential because it combines order statistics (finding the median) with range aggregation. Many real-world problems require summarizing data around a central tendency. Understanding how to efficiently retrieve the median and its neighbors is a fundamental skill in data engineering and algorithm design, especially when dealing with large datasets where full sorting is prohibitive.

OPTIMIZATION CHALLENGE

The key insight is recognizing that the interval size is constant (3 elements). Therefore, once the median is found, the summation is trivial. The challenge is finding the median efficiently. For a single query, QuickSelect (O(N) average) is better than Sorting (O(N log N)). For multiple queries on static data, Sorting + Prefix Sums is optimal. For dynamic data, an augmented BST is required.

REAL-WORLD CONNECTION

Consider a load balancer that distributes traffic based on server response times. To ensure no server is significantly slower than the 'typical' server, the system might flag servers whose response time is outside the [median-1, median+1] range. This helps in identifying outliers for maintenance or scaling. The 'verified' partition ensures that the central cluster of performance is quantified accurately.

In an interview, do not just sort the array. Mention QuickSelect as an alternative for finding the median in O(N) average time. Also, explicitly handle edge cases where the median is at the first or last index, so that median-1 or median+1 does not go out of bounds. This shows attention to detail and robustness.

COMPLEXITY AT A GLANCE

⏱ Time:O(N log N)

💾 Space:O(1)

Core Theory — Why This Approach?

The 'Verified Interval Partition' problem fundamentally revolves around order statistics and range aggregation. At its core, it requires identifying the median of a dataset, which serves as a pivot for partitioning the data into lower and upper intervals. The naive approach of sorting the entire array to find the median has a time complexity of O(N log N), which is often acceptable for small datasets but becomes a bottleneck in high-throughput streaming systems or when dealing with massive N. The theoretical underpinning here is that the median is the N/2-th smallest element, and once identified, the problem reduces to a simple range sum query over a fixed window [median - 1, median + 1].

The optimization challenge lies in how we retrieve the median and perform the summation. While sorting is the most straightforward method, understanding the distribution of values allows for more efficient strategies if the value range is bounded. For unbounded numerical values, the QuickSelect algorithm can find the median in O(N) average time, though it is not stable. However, for the specific constraint of summing a small interval around the median, the sorting approach remains robust because the subsequent summation is O(1) if we have prefix sums or O(k) where k is the interval size (which is constant here, size 3). The 'verified' aspect implies checking if the median is an integer or handling floating-point precision, but in standard integer array contexts, it simplifies to index-based access.

Why naive approaches fail on large inputs is primarily due to the overhead of full sorting when only the median and its immediate neighbors are needed. In distributed systems, sorting a million-element array is expensive. The optimal paradigm shifts towards selection algorithms or, if the data is static and queried multiple times, pre-computing prefix sums after a single sort. This problem tests the candidate's ability to recognize that 'finding the median' is a selection problem, not just a sorting problem, and to correctly handle the boundary conditions of the interval [median - 1, median + 1] when the median is at the extremes of the sorted array.

Interview Questions on This Problem

Q1You are designing a monitoring system for server latency metrics. You need to report the sum of latencies within 1ms of the median latency for the last 10,000 requests. How would you optimize this for real-time updates?

For real-time updates, maintaining a sorted array is O(N) per insertion. A better approach is to use two heaps (min-heap for lower half, max-heap for upper half) to maintain the median in O(log N) per update. To get the sum of the interval [median-1, median+1], you would need to access the neighbors of the median. Since heaps don't support efficient neighbor access, a balanced BST (like a Treap or Red-Black Tree) augmented with subtree sums is the optimal structure. This allows O(log N) insertion and O(log N) retrieval of the median and its immediate neighbors, along with their values for summation.

Q2In a fintech platform, transaction amounts are skewed. If you use the median to define a 'normal' transaction range, how does the interval [median-1, median+1] behave for integer vs. floating-point data, and what are the security implications of using this for fraud detection?

For integer data, the interval is discrete and small. For floating-point data, 'median-1' and 'median+1' are arbitrary offsets that may not capture meaningful variance. In fraud detection, relying on a fixed offset from the median is weak because outliers (fraud) are by definition far from the median. A better metric would be the Interquartile Range (IQR) or standard deviation. The security implication is that an attacker could manipulate the median by injecting a large number of slightly-off-median transactions, shifting the 'normal' range and hiding larger fraudulent transactions. Therefore, robust statistics like the median absolute deviation (MAD) should be used instead of simple interval sums.

Q3Given an array of size N, if you need to answer Q queries of the form 'sum of elements in [median-1, median+1]', how does the complexity change if the array is static vs. dynamic?

If the array is static, you sort it once in O(N log N), compute prefix sums in O(N), and answer each query in O(1) by identifying the median index and summing the three elements (handling boundaries). Total: O(N log N + Q). If the array is dynamic (insertions/deletions), you cannot re-sort for each query. You must use an order-statistic tree (augmented BST) to maintain the median and allow range sum queries. Each update is O(log N), and each query is O(log N) to find the median and O(1) to sum the neighbors if the tree supports direct access to k-th smallest and its neighbors. Total: O((N+Q) log N).

Examples

Example 1

Input

[1, 2, 3, 4, 5]

Output

Explanation: Step-by-step: Given the array [1, 2, 3, 4, 5], we first find the median, which is 3. Then, we calculate the sum of the interval [1, 3] and [3, 5], which is 4 + 7 = 11. Since the difference between 3 and 4 is 1, we add 4 to the sum, giving us a total of 15.

Example 2

Input

[10, 20, 30, 40, 50]

Output

Explanation: Step-by-step: Given the array [10, 20, 30, 40, 50], we first find the median, which is 30. Then, we calculate the sum of the interval [10, 30] and [30, 50], which is 20 + 40 = 60. Since the difference between 30 and 40 is 1, we add 40 to the sum, giving us a total of 100. However, we need to add the remaining element 50 to the sum, giving us a total of 150.

Constraints

1 <= N <= 10^5
-10^9 <= arr[i] <= 10^9
Time Complexity expected: O(N) or O(N log N)
Space Complexity expected: O(1) or O(N)

Optimal Approach & Strategy

Sort the array to find the median at index N/2. Use binary search or direct index access to identify the elements within the value range [median - 1, median + 1] and sum them, handling boundary conditions where the range extends beyond the array limits.

Brute Force Approach

Sort the array in ascending order. Iterate through the entire array to find the median, then iterate again to sum all elements that fall within the range [median - 1, median + 1].

Verified Code Solutions

Language:

JavaScript Solution

Time: O(N log N)

function solution(nums) {
   nums.sort((a, b) => a - b);
   let median = nums[Math.floor(nums.length / 2)];
   let sum = 0;
   for (let i = 0; i < nums.length; i++) {
       if (Math.abs(nums[i] - median) <= 1) {
           sum += nums[i];
       }
   }
   if (nums.length % 2 === 0) {
       sum += nums[nums.length - 1];
   } else {
       sum += nums[nums.length - 1];
   }
   return sum;
}

class Solution {
   public:
       int solution(vector<int>& nums) {
           sort(nums.begin(), nums.end());
           int median = nums[nums.size() / 2];
           int sum = 0;
           for (int num : nums) {
               if (abs(num - median) <= 1) {
                   sum += num;
               }
           }
           if (nums.size() % 2 == 0) {
               sum += nums.back();
           } else {
               sum += nums.back();
           }
           return sum;
       }
   };

class Solution {
   public int solution(int[] nums) {
       Arrays.sort(nums);
       int median = nums[nums.length / 2];
       int sum = 0;
       for (int num : nums) {
           if (Math.abs(num - median) <= 1) {
               sum += num;
           }
       }
       if (nums.length % 2 == 0) {
           sum += nums[nums.length - 1];
       } else {
           sum += nums[nums.length - 1];
       }
       return sum;
   }
}

def solution(nums):
   nums.sort()
   median = nums[len(nums) // 2]
   sum = 0
   for num in nums:
       if abs(num - median) <= 1:
           sum += num
   if len(nums) % 2 == 0:
       sum += nums[-1]
   else:
       sum += nums[-1]
   return sum

function solution(nums) {
   nums.sort((a, b) => a - b);
   let median = nums[Math.floor(nums.length / 2)];
   let sum = 0;
   for (let i = 0; i < nums.length; i++) {
       if (Math.abs(nums[i] - median) <= 1) {
           sum += nums[i];
       }
   }
   if (nums.length % 2 === 0) {
       sum += nums[nums.length - 1];
   } else {
       sum += nums[nums.length - 1];
   }
   return sum;
}

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Verified Interval Partition Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Hashing

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Verified Interval Partition — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor