Balanced Tree Span Calculator 9 — Problem Statement & Solution Guide
Problem Description
You are given a static array of integers. Your task is to answer a series of range‑sum queries. Each query specifies two indices, L and R (0‑based), and you must output the sum of all elements from position L to position R inclusive. The array does not change between queries.
Input format:
- The first line contains an integer N, the number of elements in the array.
- The second line contains N space‑separated integers, the array values.
- The third line contains an integer Q, the number of queries.
- Each of the following Q lines contains two integers L and R, describing a query.
Output format:
- For every query, output a single line containing the computed sum.
The goal is to process all queries efficiently, taking advantage of a segment tree or similar data structure to achieve logarithmic time per query.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Balanced Tree Span Calculator 9"
WHY DOES IT MATTER?
Precomputation patterns like prefix sums are essential because they convert expensive per‑query work into a one‑time cost, enabling real‑time responses in systems where latency is critical.
OPTIMIZATION CHALLENGE
The key insight is that the sum of a subarray can be expressed as the difference of two cumulative sums, eliminating the need to iterate over the subarray each time.
REAL-WORLD CONNECTION
Think of a bank’s transaction ledger: a daily balance sheet is essentially a prefix sum of all deposits and withdrawals, allowing instant balance checks for any date range without re‑scanning the entire history.
When explaining this to an interviewer, emphasize the O(N) preprocessing and O(1) query trade‑off, and be ready to discuss edge cases like empty ranges or negative numbers.
COMPLEXITY AT A GLANCE
O(N + Q) overall, O(1) per queryO(N)Core Theory — Why This Approach?
Range sum queries on a static array can be answered efficiently by precomputing a prefix sum array. The prefix sum at index i stores the sum of all elements from the start of the array up to i, inclusive. With this auxiliary array, the sum of any subarray [L,R] is simply prefix[R] - prefix[L-1] (or prefix[R] if L==0). This transforms each query from O(N) time to O(1) time after an O(N) preprocessing step.
A naive approach recomputes the sum for each query by iterating over the requested range, leading to O(Q·N) time for Q queries, which quickly becomes infeasible for large N and Q (e.g., 10^5 each). The prefix sum method reduces the overall complexity to O(N + Q) while using only O(N) additional space, making it ideal for read‑only or static data scenarios.
The underlying algorithmic pattern is a classic example of *precomputation* to trade space for time. By storing cumulative information, we avoid redundant work during query processing. This pattern is widely used in competitive programming, database indexing, and real‑time analytics where fast read access is critical.
Interview Questions on This Problem
Q1How would you answer a range sum query on a static array in O(1) time?
I would build a prefix sum array in O(N) time and answer each query by computing prefix[R] - prefix[L-1] (or prefix[R] if L==0). This gives O(1) per query and O(N) preprocessing.
Q2What are the trade‑offs of using a segment tree versus a prefix sum array for this problem?
A segment tree supports updates in O(log N) and queries in O(log N), using O(N) space. A prefix sum array is simpler, uses O(N) space, and gives O(1) queries but cannot handle updates efficiently. For static arrays, prefix sums are preferable.
Q3In a distributed system, how might you handle range sum queries across multiple shards?
Each shard can maintain its own prefix sums for its local data. To answer a global query, aggregate the relevant prefix sums from shards, adjusting for boundaries, which reduces inter‑shard communication and keeps query time near O(1) per shard.
Examples
Input
5 1 2 3 4 5 3 0 2 1 3 2 4
Output
6 9 12
Explanation: Query 1: sum of indices 0 to 2 → 1+2+3 = 6. Query 2: sum of indices 1 to 3 → 2+3+4 = 9. Query 3: sum of indices 2 to 4 → 3+4+5 = 12.
Input
4 -1 0 1 2 2 0 3 2 2
Output
2 1
Explanation: Query 1: sum of indices 0 to 3 → -1+0+1+2 = 2. Query 2: sum of index 2 only → 1.
Input
6 5 -3 7 0 -2 4 4 0 5 1 4 3 3 2 2
Output
11 2 0 7
Explanation: Query 1: 5-3+7+0-2+4 = 11. Query 2: -3+7+0-2 = 2. Query 3: element at index 3 is 0. Query 4: element at index 2 is 7.
Constraints
- 1 <= N <= 100000
- 1 <= Q <= 100000
- 0 <= L <= R < N
- -1000000000 <= nums[i] <= 1000000000
- The absolute value of any query result will fit within a 64‑bit signed integer.
Optimal Approach & Strategy
Precompute a prefix sum array in O(N) time, then answer each query in O(1) by subtracting two prefix sums.
Brute Force Approach
Iterate over each element from L to R and add it to a running sum, resulting in O(N) time 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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