Start by understanding that each conflicting pair limits subarrays containing both elements. By removing one carefully chosen pair, you maximize subarrays that remain valid. Using array traversal combined with segment tree updates helps compute the longest valid subarray starting at each index efficiently.
Problem Statement
You are given an integer n representing an array nums with elements from 1 to n in order, and a list of conflictingPairs where each pair [a, b] indicates elements that cannot coexist in a subarray. Your task is to remove exactly one conflicting pair and calculate how many non-empty subarrays do not contain both elements of any remaining conflicting pair.
Return the maximum number of valid subarrays achievable after removing exactly one conflicting pair. Use efficient strategies such as prefix sums, enumeration, and segment tree updates to handle large n and multiple conflicting pairs within the constraints.
Examples
Example 1
Input: n = 4, conflictingPairs = [[2,3],[1,4]]
Output: 9
Example 2
Input: n = 5, conflictingPairs = [[1,2],[2,5],[3,5]]
Output: 12
Constraints
- 2 <= n <= 105
- 1 <= conflictingPairs.length <= 2 * n
- conflictingPairs[i].length == 2
- 1 <= conflictingPairs[i][j] <= n
- conflictingPairs[i][0] != conflictingPairs[i][1]
Solution Approach
Preprocess conflicting pairs
Map each number to the indices of pairs it participates in. Track the earliest ending index of a valid subarray starting from each position to quickly identify invalid segments.
Simulate removal of each pair
Iterate through each conflicting pair and temporarily ignore it. Update the segment tree or prefix structures to compute the longest valid subarrays for remaining pairs, keeping a running count of valid subarrays to determine the maximum.
Compute maximum subarrays
After evaluating each pair removal, combine results using prefix sums to efficiently count all non-empty valid subarrays. Return the highest count found, ensuring array and segment tree updates are applied correctly.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | O(n) |
| Space | O(n) |
Time complexity is O(n) by processing each element and conflicting pair once using segment tree or prefix logic. Space complexity is O(n) to store mappings and segment structures for quick subarray evaluation.
What Interviewers Usually Probe
- Check if candidates efficiently handle overlapping conflicting pairs using arrays and segment tree reasoning.
- Look for correct calculation of longest valid subarrays starting from each index.
- Assess whether the candidate correctly simulates removal of each conflicting pair for maximal subarrays.
Common Pitfalls or Variants
Common pitfalls
- Forgetting to remove only one conflicting pair and recalculating subarrays for remaining pairs.
- Inefficiently iterating over all subarrays instead of using segment tree or prefix sum optimizations.
- Mishandling edge cases where conflicting pairs overlap at the start or end of the array.
Follow-up variants
- Change the problem to allow removal of k conflicting pairs instead of one.
- Compute maximum subarrays when conflicting pairs are dynamic and updated online.
- Extend to arrays with repeated numbers instead of 1 to n sequential integers.
How GhostInterview Helps
- GhostInterview highlights which conflicting pair removal maximizes subarrays using simulation plus segment tree logic.
- It visualizes invalid segments and tracks longest valid subarrays efficiently for each start index.
- It provides step-by-step computation to avoid double-counting and miscalculating overlapping subarrays.
Topic Pages
Related GhostInterview Pages
- LeetCode Interview Copilot - Use GhostInterview as a live solver when you want direct help with LeetCode-style coding questions.
- Coding Interview Assistant - See how GhostInterview supports array, string, linked list, graph, and tree interview workflows.
- How GhostInterview Works - Review the screenshot, reasoning, and answer flow before using the solver in a live interview.
FAQ
What is the main strategy to solve Maximize Subarrays After Removing One Conflicting Pair?
Focus on removing one conflicting pair at a time and use segment tree or prefix sum logic to calculate the longest valid subarrays efficiently.
Can this problem be solved without segment trees?
Yes, with careful array traversal and prefix sums, but segment trees improve efficiency for larger n and overlapping conflicts.
How do conflicting pairs affect subarray counts?
Each conflicting pair prevents subarrays containing both elements, so identifying which pair to remove maximizes valid subarrays.
What is the time complexity for large n?
With proper preprocessing and segment tree updates, the problem can be solved in O(n) time for all pairs.
Is enumeration of subarrays necessary?
Direct enumeration is inefficient; instead, simulate pair removal and count valid subarrays using prefix sums or segment tree tracking.
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