Bounded Range Segment Evaluator 5 — Problem Statement & Solution Guide

Dynamic ProgrammingEasyKnapsack State Optimization

TimeO(N * C)

SpaceO(C)

Problem Description

Given a complex dataset of length N representing system constraints and values, calculate the bounded range segment using the Knapsack State Optimization methodology.

DSA Pattern Breakdown

"Bounded Range Segment Evaluator 5"

easy

WHY DOES IT MATTER?

The pattern demonstrates how to turn an apparently quadratic sliding‑window problem into a linear‑time solution by reusing DP states, a skill that appears in many resource‑allocation and budgeting questions across large‑scale systems.

OPTIMIZATION CHALLENGE

The breakthrough is recognizing that the knapsack transition is reversible; by storing enough auxiliary information (e.g., previous dp values or using a monotonic queue), we can subtract the contribution of the element leaving the window in O(1) time, eliminating the need for full recomputation.

REAL-WORLD CONNECTION

Think of a distributed cache that must keep the top‑k most valuable items within a moving time window; instead of recomputing the top‑k from scratch each second, the system incrementally evicts expired entries and inserts new ones, mirroring the sliding‑window DP updates.

When coding, first implement the classic 0/1 knapsack DP, then wrap it in a sliding‑window loop. Use a separate array to snapshot dp before each insertion so you can roll back the outgoing element efficiently—this avoids hidden O(N) penalties.

COMPLEXITY AT A GLANCE

⏱ Time:O(N * C)

💾 Space:O(C)

Core Theory — Why This Approach?

The Bounded Range Segment Evaluator can be modeled as a variant of the classic 0/1 knapsack where each item (constraint) may be taken at most once and we are interested in the best achievable value within a specific weight (or index) interval. A naive DP that iterates over every possible sub‑array and recomputes the knapsack state for each start point leads to O(N^2 * W) time, which quickly becomes infeasible for N ≈ 10^5 and weight limits in the thousands. The optimal paradigm leverages the fact that the knapsack transition is linear and can be reused across overlapping segments; by maintaining a rolling DP table and applying the “bounded knapsack state compression” (also known as the “modulo‑class optimization”), we update the DP in O(1) amortized per element while preserving correctness. This reduces the overall complexity to O(N · R) where R is the size of the bounded range, and with further monotonic‑queue tricks it can be brought down to O(N · log R) or even O(N) for fixed‑size ranges.

In practice the algorithm stores a one‑dimensional DP array dp[w] = maximum value achievable with total weight w for the current sliding window. When the window slides forward, the contribution of the element leaving the window is subtracted using a pre‑computed “inverse transition”, and the new element is incorporated via the standard knapsack update (dp[w] = max(dp[w], dp[w‑weight_i] + value_i)). Because each weight class is processed independently modulo the item weight, the state space does not explode, and the DP remains compact. This state‑reuse technique is the essence of knapsack state optimization and is the key to handling large N with bounded ranges efficiently.

Interview Questions on This Problem

Q1How would you adapt the classic 0/1 knapsack DP to answer queries for the maximum value in any sub‑array of length L?

Maintain a sliding DP array for the current window of length L. When the window moves, remove the effect of the outgoing element by reversing its DP update (using stored previous states) and then apply the knapsack transition for the incoming element. This yields O(N · L) total time, which is optimal for fixed L.

Q2Explain why the modulo‑class optimization works for bounded knapsack when the weight limit is large but the number of distinct weights is small.

Items with the same weight belong to the same residue class modulo that weight. By processing each class separately, we can treat the DP transition as a series of independent 1‑dimensional convolutions, allowing us to update dp[w] using only dp[w‑k·weight] values. This reduces redundant work and bounds the complexity by the number of classes rather than the raw weight limit.

Q3A fintech platform needs to evaluate risk scores over rolling windows of transaction amounts. Which aspects of the bounded range segment evaluator make it suitable for this task?

The evaluator’s sliding‑window DP reuses previously computed states, giving O(1) amortized update per new transaction. It also respects a hard bound on the total transaction amount (the knapsack capacity), ensuring the risk score stays within regulatory limits while efficiently handling high‑throughput streams.

Examples

Example 1

Input

[1, 2, 3, 4, 5], W = 5

Output

Explanation: Step-by-step: Given the input array [1, 2, 3, 4, 5] and W = 5, we first calculate the sum of the array, which is 15. Since the sum exceeds W, we need to find the maximum sum of a subarray that does not exceed W. We can achieve this by using a sliding window approach, where we maintain a window of elements that sum up to W. In this case, the maximum sum of a subarray that does not exceed W is 0.

Example 2

Input

[]

Output

Explanation: Step-by-step: Given an empty input array, we return 0 as the maximum sum of a subarray that does not exceed W is 0.

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

Maintain a single DP array for the sliding window, updating it incrementally as the window moves forward and undoing the effect of the element that leaves the window. This yields linear time overall.

Brute Force Approach

For every possible start index, recompute a full knapsack DP over the next L elements, tracking the best value that fits the capacity. This repeats the DP N times, leading to quadratic time.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(N * C)

function solution(nums, W) {
      if (nums.length === 0) return 0;
      let sum = 0;
      let maxSum = 0;
      for (let i = 0; i < nums.length; i++) {
         sum += nums[i];
         if (sum > W) {
            sum = nums[i];
         }
         maxSum = Math.max(maxSum, sum);
      }
      return maxSum;
   }

class Solution {
   public:
      int solution(vector<int>& nums, int W) {
         if (nums.size() == 0) return 0;
         int sum = 0;
         int maxSum = 0;
         for (int num : nums) {
            sum += num;
            if (sum > W) {
               sum = num;
            }
            maxSum = max(maxSum, sum);
         }
         return maxSum;
      }
   };

class Solution {
      public int solution(int[] nums, int W) {
         if (nums.length == 0) return 0;
         int sum = 0;
         int maxSum = 0;
         for (int num : nums) {
            sum += num;
            if (sum > W) {
               sum = num;
            }
            maxSum = Math.max(maxSum, sum);
         }
         return maxSum;
      }
   }

def solution(nums, W):
      if not nums:
         return 0
      total_sum = 0
      max_sum = 0
      for num in nums:
         total_sum += num
         if total_sum > W:
            total_sum = num
         max_sum = max(max_sum, total_sum)
      return max_sum

function solution(nums, W) {
      if (nums.length === 0) return 0;
      let sum = 0;
      let maxSum = 0;
      for (let i = 0; i < nums.length; i++) {
         sum += nums[i];
         if (sum > W) {
            sum = nums[i];
         }
         maxSum = Math.max(maxSum, sum);
      }
      return maxSum;
   }

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Bounded Range Segment Evaluator 5 Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Dynamic Programming

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Bounded Range Segment Evaluator 5 — 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