Rotated Matrix Pivot Analyzer 8 — Problem Statement & Solution Guide

GreedyHardJob Scheduling Maximum Profit

TimeO(N log N)

SpaceO(N)

Problem Description

Given a complex dataset of length N representing system constraints and values, calculate the maximum sum of non-overlapping subarrays using the Job Scheduling Maximum Profit methodology.

DSA Pattern Breakdown

"Rotated Matrix Pivot Analyzer 8"

hard

WHY DOES IT MATTER?

The weighted interval scheduling pattern is essential because many real‑world optimization tasks—such as job scheduling, ad placement, and resource allocation—require selecting non‑conflicting intervals with maximum total benefit. Mastering this pattern equips engineers to transform complex constraints into tractable DP problems, enabling scalable solutions for large‑scale systems.

OPTIMIZATION CHALLENGE

The key insight is to pre‑compute the "previous compatible interval" for each sorted interval using binary search on end times. This reduces the naïve O(N^2) DP to O(N log N) by eliminating the need to scan all earlier intervals for compatibility, thereby achieving optimal time complexity for large N.

REAL-WORLD CONNECTION

Imagine a distributed compute cluster where each job occupies a time window on a shared GPU. The goal is to maximize total revenue by scheduling non‑overlapping jobs. The same DP logic used for the rotated matrix pivot problem directly maps to deciding which jobs to accept, making the algorithm a cornerstone of cloud resource orchestration.

During an interview, first compute prefix sums to get O(1) interval weights, then sort intervals by end index. Use a separate array to store dp[i] = max profit up to i. Remember to binary search for the last interval whose end < current start; this step is the only place where a subtle off‑by‑one bug can appear.

COMPLEXITY AT A GLANCE

⏱ Time:O(N log N)

💾 Space:O(N)

Core Theory — Why This Approach?

The problem of maximizing the sum of non‑overlapping subarrays is a classic instance of the weighted interval scheduling problem. Each subarray can be represented as an interval [start, end] with a weight equal to the sum of its elements. The optimal solution must select a subset of intervals that do not intersect while maximizing the total weight. A naive enumeration of all subsets leads to exponential time because the decision for each interval depends on the choices made for all previous intervals, which quickly becomes infeasible for N > 10^5. The optimal paradigm combines sorting, binary search, and dynamic programming: first sort intervals by their end index, then for each interval compute the best profit achievable by either taking it (adding its weight to the best profit of the last non‑conflicting interval) or skipping it (carrying forward the previous best profit). This DP recurrence runs in O(N log N) time thanks to binary search for the previous compatible interval, and O(N) space for the DP table. The greedy approach alone fails because picking the locally highest‑profit interval can block a combination of smaller intervals that together yield a larger total, so a pure greedy selection is not optimal.

The underlying theory rests on optimal substructure and the principle of “last compatible job”. After sorting by end time, the sub‑problem for the i‑th interval depends only on the best solution for intervals ending before its start, which is precisely what the DP captures. By pre‑computing prefix sums of the original dataset, the weight of any subarray can be obtained in O(1), allowing the algorithm to focus solely on interval selection. This transformation from raw array values to interval weights is what enables the classic job‑scheduling DP to be applied to a seemingly unrelated maximum‑subarray problem, delivering a scalable solution for large inputs.

Interview Questions on This Problem

Q1How would you adapt the weighted interval scheduling DP to handle the case where subarray weights can be negative?

First, compute the weight of each subarray using prefix sums. Then, during DP, treat negative weights like any other weight: the recurrence max(dp[i‑1], weight[i] + dp[p(i)]) naturally discards intervals that would reduce the total profit because dp[i‑1] will be larger. Optionally, you can filter out intervals with negative weight before sorting to reduce the state space, but the DP already handles them correctly.

Q2Explain why a simple greedy algorithm that always picks the subarray with the highest sum fails for this problem, and give a concrete counter‑example.

Greedy fails because picking the locally highest‑sum subarray can block a combination of smaller, non‑overlapping subarrays that together yield a larger total sum. For example, consider array [4, -1, 4, -1, 4]. The subarray [0,4] has sum 10, but selecting it prevents picking three subarrays [0,0], [2,2], [4,4] each with sum 4, totaling 12. The optimal solution is the three smaller subarrays, which greedy would miss.

Q3What is the time complexity of the solution if you replace binary search with a hashmap that maps each start index to the latest compatible interval, and why might this be less efficient in practice?

Using a hashmap to store the latest compatible interval for each start index can achieve O(N) lookup, leading to an overall O(N) time after sorting. However, building the hashmap requires iterating over all possible start positions (up to N) and may consume O(N) extra space. Moreover, binary search on a sorted list of end indices is cache‑friendly and has low constant factors, while hashmap lookups can suffer from collisions and poorer locality, making the binary‑search version often faster in real‑world benchmarks.

Examples

Example 1

Input

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

Output

Explanation: Step-by-step: Given the input array [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], we first find the maximum sum of non-overlapping subarrays. We can achieve this by using a sliding window approach. The maximum sum of non-overlapping subarrays is 38, which is the sum of subarrays [1, 2, 3, 4, 5] and [6, 7, 8, 9, 10].

Example 2

Input

[10, 9, 8, 7, 6, 5, 4, 3, 2, 1]

Output

Explanation: Step-by-step: Given the input array [10, 9, 8, 7, 6, 5, 4, 3, 2, 1], we first find the maximum sum of non-overlapping subarrays. We can achieve this by using a sliding window approach. The maximum sum of non-overlapping subarrays is 28, which is the sum of subarrays [10, 9, 8] and [5, 6, 7, 4, 3, 2, 1].

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

Sort intervals by end, compute prefix sums for O(1) weights, then apply DP with binary search for the previous compatible interval, achieving O(N log N) time.

Brute Force Approach

Enumerate every subset of subarrays, check for overlaps, and keep the maximum sum; this runs in exponential time O(2^N).

Verified Code Solutions

Language:

JavaScript Solution

Time: O(N log N)

function solution(nums) {
   let maxSum = 0;
   let windowStart = 0;
   let currentSum = 0;
   for (let windowEnd = 0; windowEnd < nums.length; windowEnd++) {
       currentSum += nums[windowEnd];
       if (windowEnd >= 2) {
           currentSum -= nums[windowStart];
           windowStart++;
       }
       maxSum = Math.max(maxSum, currentSum);
   }
   return maxSum;
}

class Solution {
   public:
       int solution(vector<int>& nums) {
           int maxSum = 0;
           int windowStart = 0;
           int currentSum = 0;
           for (int windowEnd = 0; windowEnd < nums.size(); windowEnd++) {
               currentSum += nums[windowEnd];
               if (windowEnd >= 2) {
                   currentSum -= nums[windowStart];
                   windowStart++;
               }
               maxSum = max(maxSum, currentSum);
           }
           return maxSum;
       }
   };

class Solution {
   public int solution(int[] nums) {
       int maxSum = 0;
       int windowStart = 0;
       int currentSum = 0;
       for (int windowEnd = 0; windowEnd < nums.length; windowEnd++) {
           currentSum += nums[windowEnd];
           if (windowEnd >= 2) {
               currentSum -= nums[windowStart];
               windowStart++;
           }
           maxSum = Math.max(maxSum, currentSum);
       }
       return maxSum;
   }
}

def solution(nums):
   max_sum = 0
   window_start = 0
   current_sum = 0
   for window_end in range(len(nums)):
       current_sum += nums[window_end]
       if window_end >= 2:
           current_sum -= nums[window_start]
           window_start += 1
       max_sum = max(max_sum, current_sum)
   return max_sum

function solution(nums) {
   let maxSum = 0;
   let windowStart = 0;
   let currentSum = 0;
   for (let windowEnd = 0; windowEnd < nums.length; windowEnd++) {
       currentSum += nums[windowEnd];
       if (windowEnd >= 2) {
           currentSum -= nums[windowStart];
           windowStart++;
       }
       maxSum = Math.max(maxSum, currentSum);
   }
   return maxSum;
}

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Rotated Matrix Pivot Analyzer 8 Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Greedy

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Rotated Matrix Pivot Analyzer 8 — 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