Bounded Range Segment Evaluator 6 — Problem Statement & Solution Guide

StackMediumParenthesis Score Calculator

TimeO(N)

SpaceO(N)

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

"Bounded Range Segment Evaluator 6"

medium

WHY DOES IT MATTER?

Understanding how to collapse nested structures with a stack is fundamental for many parsing, evaluation, and range‑query problems; it turns exponential‑looking work into linear work by exploiting the inherent hierarchy of the input.

OPTIMIZATION CHALLENGE

The breakthrough is recognizing that the score of a segment can be computed solely from its immediate children, allowing us to store only the current depth’s partial result on the stack and combine it instantly at each closing token.

REAL-WORLD CONNECTION

Think of a cloud resource manager that tracks quota limits for nested projects: each project inherits limits from its parent, and when a sub‑project finishes, its usage is rolled up to the parent—exactly the push/pop semantics of a stack.

During an interview, write the stack‑based solution first, then immediately walk through a small example on the whiteboard; this demonstrates both correctness and your mental model before you dive into code.

COMPLEXITY AT A GLANCE

⏱ Time:O(N)

💾 Space:O(N)

Core Theory — Why This Approach?

The Bounded Range Segment Evaluator problem can be modeled as a variant of the classic Parenthesis Score Calculator. Each opening constraint (e.g., ‘[‘) introduces a new bounded segment, and each closing constraint (e.g., ‘]’) finalizes the segment, possibly nesting multiple layers. A stack naturally captures this nesting: pushing the current segment’s partial result when we encounter an opening token and popping to combine results when we see a closing token. Naïve solutions enumerate every possible sub‑array, compute its bounds, and aggregate scores, leading to O(N²) time and quickly exhausting time limits for large N. The optimal paradigm leverages the LIFO property of a stack to process the input in a single left‑to‑right pass, maintaining the cumulative score of the current depth and merging scores in constant time at each closing token, thus achieving linear complexity.

The key insight is that the score of a bounded segment depends only on the scores of its immediate inner segments, not on the entire history of the array. By storing intermediate scores on the stack, we avoid recomputation and can instantly combine nested results using the problem‑specific scoring rule (often a multiplication by 2 or addition of a base value). This mirrors how parentheses scoring works: "()" yields 1, and "(A)" yields 2·score(A). Translating that to the bounded range context yields an O(N) time, O(N) space algorithm that scales to the maximum input sizes typical in interview settings.

Interview Questions on This Problem

Q1How does a stack help compute the score of nested bounded ranges in O(N) time?

A stack maintains the partial scores of open segments; when we encounter a closing token we pop the top score, apply the scoring rule (e.g., double it or add a base), and add it to the new top, thereby collapsing nested structures in constant time per character.

Q2What would be the time and space complexity if you used a two‑pointer sliding window instead of a stack?

A sliding window would still need to recompute scores for overlapping windows, leading to O(N²) time in the worst case, while space remains O(1); the stack approach is superior because it avoids redundant recomputation.

Q3In a real system, how would you modify the algorithm to handle dynamic updates (insertions/deletions) to the dataset?

You could augment the stack with a segment tree or binary indexed tree that stores scores for each segment, allowing O(log N) updates and queries while preserving the nesting logic through lazy propagation of score changes.

Examples

Example 1

Input

[5, 9, 13, 17, -1, 8, 6]

Output

Explanation: Step-by-step: Given the input [5, 9, 13, 17, -1, 8, 6], we calculate the score as follows: (5 + 9) - 13 + 17 - (-1) + 8 - 6 = 34

Example 2

Input

[8, 6]

Output

Explanation: Step-by-step: Given the input [8, 6], we calculate the score as follows: (8 + 6) - 8 + 6 = 14

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

Traverse the array once, using a stack to store partial scores of open segments; on each closing token pop, compute the segment’s final score, and merge it with the previous level, achieving O(N) time.

Brute Force Approach

Enumerate every possible sub‑segment, compute its bounds and score independently, and keep the maximum; this requires O(N²) time.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(N)

function solution(nums) {
   let score = 0;
   for (let i = 0; i < nums.length; i++) {
       if (nums[i] === -1) {
           if (i % 2 === 0) {
               if (i + 1 < nums.length) {
                   score += nums[i + 1];
               }
           } else {
               if (i + 1 < nums.length) {
                   score -= nums[i + 1];
               }
           }
       } else {
           if (i % 2 === 0) {
               score += nums[i];
           } else {
               score -= nums[i];
           }
       }
   }
   return score;
}

class Solution {
   public:
       int solution(vector<int>& nums) {
           int score = 0;
           for (int i = 0; i < nums.size(); i++) {
               if (nums[i] == -1) {
                   if (i % 2 == 0) {
                       if (i + 1 < nums.size()) {
                           score += nums[i + 1];
                       }
                   } else {
                       if (i + 1 < nums.size()) {
                           score -= nums[i + 1];
                       }
                   }
               } else {
                   if (i % 2 == 0) {
                       score += nums[i];
                   } else {
                       score -= nums[i];
                   }
               }
           }
           return score;
       }
   };

class Solution {
   public int solution(int[] nums) {
       int score = 0;
       for (int i = 0; i < nums.length; i++) {
           if (nums[i] == -1) {
               if (i % 2 == 0) {
                   if (i + 1 < nums.length) {
                       score += nums[i + 1];
                   }
               } else {
                   if (i + 1 < nums.length) {
                       score -= nums[i + 1];
                   }
               }
           } else {
               if (i % 2 == 0) {
                   score += nums[i];
               } else {
                   score -= nums[i];
               }
           }
       }
       return score;
   }
}

def solution(nums):
   score = 0
   for i in range(len(nums)):
       if nums[i] == -1:
           if i % 2 == 0:
               if i + 1 < len(nums):
                   score += nums[i + 1]
           else:
               if i + 1 < len(nums):
                   score -= nums[i + 1]
       else:
           if i % 2 == 0:
               score += nums[i]
           else:
               score -= nums[i]
   return score

function solution(nums) {
   let score = 0;
   for (let i = 0; i < nums.length; i++) {
       if (nums[i] === -1) {
           if (i % 2 === 0) {
               if (i + 1 < nums.length) {
                   score += nums[i + 1];
               }
           } else {
               if (i + 1 < nums.length) {
                   score -= nums[i + 1];
               }
           }
       } else {
           if (i % 2 === 0) {
               score += nums[i];
           } else {
               score -= nums[i];
           }
       }
   }
   return score;
}

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Bounded Range Segment Evaluator 6 Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Stack

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Bounded Range Segment Evaluator 6 — 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