Bounded Range Segment Evaluator 3 — Problem Statement & Solution Guide
Problem Description
You are provided with an integer array nums of length N and two integer indices left and right (0-indexed). Your task is to compute the sum of all elements in the subarray defined by the inclusive range [left, right]. This operation is fundamental in segment tree implementations and prefix sum optimizations, where efficient range queries are required. The function must return the exact arithmetic sum of the elements from index left to index right, inclusive. If the range is invalid (e.g., left > right), return 0. This problem serves as a baseline for understanding state transitions in dynamic programming contexts where cumulative sums over bounded intervals are prerequisites for more complex knapsack-style optimizations.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Bounded Range Segment Evaluator 3"
WHY DOES IT MATTER?
Range‑sum queries appear in virtually every data‑intensive system—analytics, monitoring, and finance—where fast aggregation over sliding windows is critical. Mastering prefix sums or segment trees demonstrates an ability to convert a seemingly expensive operation into a constant‑time lookup, a skill valued across all engineering tiers.
OPTIMIZATION CHALLENGE
The key insight is that the sum of a segment can be expressed as the difference of two cumulative sums. By precomputing these cumulative values once, we eliminate the need for repeated iteration.
REAL-WORLD CONNECTION
Think of a streaming service that needs to compute total watch time for any user‑defined interval. By storing cumulative watch minutes up to each timestamp, the service can instantly report any interval without re‑scanning the entire log.
During an interview, write the prefix array first, verify it with a small example, and then show the O(1) query formula. If the problem mentions updates, pivot to a Fenwick Tree and explain its log‑N update/query trade‑off.
COMPLEXITY AT A GLANCE
O(N) preprocessing + O(1) per queryO(N) auxiliary space for prefix arrayCore Theory — Why This Approach?
The bounded range segment evaluator is a classic example of range query problems where we need to compute the sum of elements between two indices repeatedly. A naive solution iterates over the subarray for each query, leading to O(N) time per query, which becomes prohibitive when N and the number of queries are large. The optimal paradigm leverages prefix sums (or a segment tree for mutable arrays) to preprocess cumulative sums in O(N) time, allowing any range sum to be answered in O(1) by subtracting two prefix values. This approach transforms the problem from linear per‑query work to constant‑time lookups, dramatically improving scalability for real‑world workloads such as analytics dashboards or financial time‑series aggregations.
Interview Questions on This Problem
Q1How would you answer multiple range sum queries on a static array in O(1) time per query?
Pre‑compute a prefix sum array where prefix[i] = sum of nums[0..i]. Then answer a query [l, r] as prefix[r] - (l > 0 ? prefix[l‑1] : 0).
Q2If the array were mutable (updates allowed), which data structure would you choose to support both point updates and range sum queries efficiently?
A Fenwick Tree (Binary Indexed Tree) or a Segment Tree can handle point updates in O(log N) and range sum queries in O(log N) while using O(N) space.
Q3Why might a naïve O(N) per‑query solution still be acceptable in some production systems?
When the number of queries is extremely low, the array size is small, or the latency budget permits linear scans, the overhead of building and maintaining extra structures may outweigh the benefits.
Examples
Input
nums = [4, 2, 7, 1, 9], left = 1, right = 3
Output
10
Explanation: The subarray from index 1 to 3 is [2, 7, 1]. The sum is 2 + 7 + 1 = 10.
Input
nums = [10, -5, 3, 8, 2], left = 0, right = 4
Output
18
Explanation: The entire array is considered. The sum is 10 + (-5) + 3 + 8 + 2 = 18.
Input
nums = [1, 2, 3, 4, 5], left = 2, right = 2
Output
3
Explanation: The range is a single element at index 2. The value is 3.
Input
nums = [7, 7, 7], left = 2, right = 1
Output
0
Explanation: The left index (2) is greater than the right index (1), indicating an invalid range. By definition, the sum is 0.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
- 0 <= left <= right < nums.length
- The sum of all elements in any valid range will fit within a 64-bit integer.
Optimal Approach & Strategy
Build a prefix sum array in O(N) once, then answer each query with a single subtraction in O(1).
Brute Force Approach
Iterate from left to right, adding each element to a running total for every query, which costs O(right‑left+1) per query.
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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