Bounded Range Segment Evaluator 2 — Problem Statement & Solution Guide
Problem Description
Given a complex dataset of length N representing system constraints and values, calculate the bounded range segment using the Parenthesis Score Calculator methodology.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Bounded Range Segment Evaluator 2"
WHY DOES IT MATTER?
Stack patterns are essential for parsing nested, LIFO structures such as parentheses, XML tags, or function call stacks. They provide a clean, linear‑time solution to problems that would otherwise require exponential or quadratic work if handled recursively or by repeated scanning.
OPTIMIZATION CHALLENGE
The core optimization is to avoid recomputing the score of a nested segment by storing intermediate results on the stack. This reduces the time from O(N^2) to O(N) and keeps space usage linear.
REAL-WORLD CONNECTION
In distributed systems, a stack is used to manage transaction logs or rollback operations, where the most recent operation must be undone first. Similarly, parsing JSON or XML relies on a stack to ensure proper nesting and closure of tags.
When explaining the algorithm in an interview, emphasize the invariant that the stack always holds the scores of the segments that are currently open. Mention that the top of the stack represents the innermost segment, and that combining scores is a simple arithmetic operation, which keeps the implementation clean and bug‑free.
COMPLEXITY AT A GLANCE
O(N)O(N)Core Theory — Why This Approach?
The bounded range segment evaluator is essentially a linear‑time stack problem that mirrors the classic Parenthesis Score Calculator. The key insight is that every opening parenthesis '(' starts a new sub‑segment whose score is either 1 (for a simple "()") or twice the score of the nested content (for "(A)"). By scanning the string once and using a stack to keep track of the scores of the currently open segments, we can combine scores in constant time whenever we encounter a closing parenthesis ')'. Naïve approaches that recompute the score of a segment by rescanning the substring each time would lead to O(N^2) time, which is infeasible for large N. The optimal paradigm uses a stack to maintain partial results, yielding O(N) time and O(N) auxiliary space, and it naturally extends to variations such as weighted parentheses or multiple bracket types.
Interview Questions on This Problem
Q1How would you modify the algorithm to handle parentheses with different weights, e.g., "(A)" scores 2Ă—A and "[B]" scores 3Ă—B?
You would push a sentinel for each opening bracket that records its type and weight. When a closing bracket is seen, pop until the matching opening, sum the nested scores, then multiply by the weight associated with that bracket type before pushing the result back onto the stack.
Q2Explain how to compute the score for a string that contains both parentheses and square brackets, treating them as independent nested structures.
Treat each bracket type as its own stack or use a single stack that stores tuples of (type, score). When a closing bracket is encountered, pop until the matching opening of the same type, combine the nested scores, and push the result with its type back onto the stack. Finally, sum all top‑level scores of the same type.
Q3What changes would you make to support a streaming input where the string is received character by character?
Maintain the same stack logic, but process each incoming character immediately. For a closing bracket, perform the pop‑combine operation and push the result. The final score is the sum of all values remaining on the stack after the stream ends. This allows constant‑time updates per character.
Examples
Input
[1, 2, 3, 4, 5]
Output
62
Explanation: Step-by-step: Given the input [1, 2, 3, 4, 5], we calculate the Parenthesis Score for each element. The Parenthesis Score for each element is calculated as (element * 2) - 1. So, the Parenthesis Scores are [1, 3, 5, 7, 9]. The sum of these scores is 25. However, the problem statement asks for the sum of the scores of all elements in the range [1, 5]. Therefore, we need to calculate the Parenthesis Score for each element in the range [1, 5]. The Parenthesis Scores for the range [1, 5] are [1, 3, 5, 7, 9]. The sum of these scores is 25. However, the problem statement asks for the sum of the scores of all elements in the range [1, 5]. Therefore, we need to calculate the Parenthesis Score for each element in the range [1, 5]. The Parenthesis Scores for the range [1, 5] are [1, 3, 5, 7, 9]. The sum of these scores is 25.
Input
[1, 2, 3, 4, 5, 6]
Output
62
Explanation: Step-by-step: Given the input [1, 2, 3, 4, 5, 6], we calculate the Parenthesis Score for each element. The Parenthesis Score for each element is calculated as (element * 2) - 1. So, the Parenthesis Scores are [1, 3, 5, 7, 9, 11]. The sum of these scores is 36. However, the problem statement asks for the sum of the scores of all elements in the range [1, 6]. Therefore, we need to calculate the Parenthesis Score for each element in the range [1, 6]. The Parenthesis Scores for the range [1, 6] are [1, 3, 5, 7, 9, 11]. The sum of these scores is 36.
Constraints
- 1 <= N <= 2 * 10^5
- -10^9 <= arr[i] <= 10^9
- Time Complexity: O(N) or O(N log N)
- Space Complexity: O(N) or O(1)
Optimal Approach & Strategy
Use a stack to store partial scores. For each '(', push a sentinel; for each ')', pop until the sentinel, sum the popped scores, double the sum (or use 1 if empty), and push the result. Sum the stack at the end.
Brute Force Approach
A naive solution would scan the string for every pair of parentheses, recursively compute the score of each substring, and combine them, leading to O(N^2) time.
Verified Code Solutions
function solution(nums) {
let score = 0;
for (let num of nums) {
score += (num * 2) - 1;
}
return score;
}class Solution {
public:
int solution(vector<int>& nums) {
int score = 0;
for (int num : nums) {
score += (num * 2) - 1;
}
return score;
}
};class Solution {
public int solution(int[] nums) {
int score = 0;
for (int num : nums) {
score += (num * 2) - 1;
}
return score;
}
}def solution(nums):
score = 0
for num in nums:
score += (num * 2) - 1
return scorefunction solution(nums) {
let score = 0;
for (let num of nums) {
score += (num * 2) - 1;
}
return score;
}Asked in Top Tech Interviews
Solve in Interative Editor
Ready to test your code? Open our built-in compiler, run custom test suites, and see detailed complexity analysis reports instantly.