Problem Description

You have an integer n that represents an array nums = [1, 2, 3, ..., n] containing numbers from 1 to n in order.

You're also given a 2D array conflictingPairs where each conflictingPairs[i] = [a, b] indicates that elements a and b form a conflicting pair.

Your task is to:

1. Remove exactly one element from conflictingPairs (delete one conflicting pair)
2. After removal, count all possible non-empty subarrays of nums that are valid

A subarray is considered valid if it doesn't contain both elements from any remaining conflicting pair. In other words, for each remaining pair [a, b], a valid subarray cannot have both a and b present at the same time.

The goal is to find which single conflicting pair to remove such that the number of valid subarrays is maximized, and return this maximum count.

Example to clarify:

• If n = 3, then nums = [1, 2, 3]
• If conflictingPairs = [[1, 3], [2, 3]]
• If we remove [1, 3], only [2, 3] remains as a conflict
• Valid subarrays would be: [1], [2], [3], [1, 2] (since 1 and 2 don't conflict after removing [1,3])
• Invalid subarrays would be: [2, 3], [1, 2, 3] (both contain 2 and 3 which still conflict)

The function should return the maximum possible number of valid subarrays after optimally choosing which conflicting pair to remove.

Intuition

Let's think about how conflicting pairs restrict our subarrays. For any subarray starting at position a, it can extend rightward until it hits a position b where (a, b) forms a conflicting pair. The key insight is that for each starting position, we need to find the earliest (minimum) position where a conflict occurs.

Consider iterating through starting positions from right to left (from n down to 1). For each starting position a, we need to track all positions b > a that conflict with a. The rightmost valid ending position for subarrays starting at a is one less than the minimum such b.

Now, what happens when we remove one conflicting pair? If we remove a pair that contains this minimum b value (let's call it b1), then the new limit becomes the second minimum value (call it b2). This gives us additional valid subarrays.

The clever observation is that we don't need to try removing every possible pair individually. Instead, we can:

1. Calculate the base count of valid subarrays assuming no pairs are removed (using the minimum b values)
2. Track how much benefit we'd get by removing pairs that involve each minimum b1

For each starting position a, we maintain:

• b1: the minimum conflicting position (this determines our base count)
• b2: the second minimum conflicting position (this determines the benefit if we remove a pair involving b1)

The benefit of removing a pair involving b1 is b2 - b1 additional subarrays. We accumulate these potential benefits in a counter array cnt[b1].

Finally, the answer is the base count plus the maximum benefit we can get from removing one pair (which is the maximum value in our cnt array).

This approach is efficient because instead of trying all possible removals, we identify which removal would give us the maximum benefit by tracking the gap between the first and second minimum conflicting positions.

Learn more about Segment Tree and Prefix Sum patterns.

Solution Approach

The implementation follows the intuition by processing conflicting pairs and tracking minimum values efficiently:

1. Data Structure Setup:

• Create an adjacency list g where g[a] stores all values b that conflict with a (where a < b)
• Normalize pairs so that the smaller value is always first: if a > b, swap them
• Initialize a counter array cnt of size n + 2 to track potential benefits

2. Main Algorithm:

Process starting positions from right to left (n down to 1):

• Maintain two variables:
  • b1: minimum conflicting position (initially n + 1)
  • b2: second minimum conflicting position (initially n + 1)

For each starting position a:

• Update minimums: Iterate through all b values in g[a] (positions that conflict with a)

  • If b < b1: Update both values - b2 = b1, b1 = b
  • Else if b < b2: Update only b2 = b

• Calculate contributions:

  • Base contribution: ans += b1 - a (number of valid subarrays starting at a without removing any pair)
  • Potential benefit: cnt[b1] += b2 - b1 (additional subarrays if we remove a pair involving b1)

3. Find Maximum Benefit:

• Track the maximum value in cnt array as add
• This represents the maximum additional subarrays we can get by removing one conflicting pair

4. Return Result:

• Final answer = ans + add
• Where ans is the base count and add is the maximum benefit from removal

Key Pattern: The solution uses a greedy approach combined with dynamic tracking of minimum values. By processing positions in reverse order, we can efficiently maintain the two smallest conflicting positions for each starting point without recomputing them repeatedly.

Time Complexity: O(n + m) where m is the number of conflicting pairs Space Complexity: O(n + m) for storing the adjacency list and counter array

Example Walkthrough

Let's walk through a concrete example with n = 4 and conflictingPairs = [[1, 3], [2, 4], [1, 4]].

This means nums = [1, 2, 3, 4] and we have three conflicting pairs.

Step 1: Build adjacency list After normalizing pairs (smaller value first), we get:

• g[1] = [3, 4] (1 conflicts with 3 and 4)
• g[2] = [4] (2 conflicts with 4)
• g[3] = [] (3 has no conflicts with higher numbers)
• g[4] = [] (4 is the last element)

Step 2: Process each starting position from right to left

Initialize: ans = 0, cnt = [0, 0, 0, 0, 0, 0] (size n+2)

Starting at position 4:

• b1 = 5, b2 = 5 (initialized to n+1)
• No conflicts in g[4]
• Base contribution: ans += 5 - 4 = 1 (subarray [4])
• Benefit if we remove pair with b1=5: cnt[5] += 5 - 5 = 0
• Running ans = 1

Starting at position 3:

• b1 = 5, b2 = 5
• No conflicts in g[3]
• Base contribution: ans += 5 - 3 = 2 (subarrays [3] and [3,4])
• Benefit: cnt[5] += 5 - 5 = 0
• Running ans = 3

Starting at position 2:

• b1 = 5, b2 = 5
• Process conflict with 4: b1 = 4, b2 = 5
• Base contribution: ans += 4 - 2 = 2 (subarrays [2] and [2,3])
• Benefit if we remove pair with b1=4: cnt[4] += 5 - 4 = 1
• Running ans = 5

Starting at position 1:

• b1 = 5, b2 = 5
• Process conflict with 3: b1 = 3, b2 = 5
• Process conflict with 4: Since 4 > 3 but 4 < 5, update b2 = 4
• Final: b1 = 3, b2 = 4
• Base contribution: ans += 3 - 1 = 2 (subarrays [1] and [1,2])
• Benefit if we remove pair with b1=3: cnt[3] += 4 - 3 = 1
• Running ans = 7

Step 3: Find maximum benefit

• cnt = [0, 0, 0, 1, 1, 0]
• Maximum value in cnt is 1

Step 4: Return result

• Final answer = 7 + 1 = 8

Verification: The 8 valid subarrays after optimal removal are:

• If we remove [1,3]: [1], [2], [3], [4], [1,2], [2,3], [3,4], [1,2,3]
• If we remove [2,4]: [1], [2], [3], [4], [1,2], [2,3], [3,4], [2,3,4]

Both removals give us 8 valid subarrays, which matches our calculated answer.

Solution Implementation

Time and Space Complexity

The time complexity is O(n + m), where n is the input parameter n and m is the number of conflicting pairs. The algorithm iterates through all values from n down to 1 (taking O(n) time), and for each value a, it processes all edges in g[a]. Since each conflicting pair is stored exactly once in the adjacency list, the total time to process all edges across all iterations is O(m). Therefore, the overall time complexity is O(n + m).

The space complexity is O(n + m). The adjacency list g requires O(n + m) space (array of size n+1 plus storage for all m edges), and the cnt array requires O(n) space. Since O(n + m) dominates, the overall space complexity is O(n + m).

Note: The reference answer states O(n) for both complexities where n is the number of conflicting pairs, which appears to use different notation. In the reference, n seems to refer to the number of pairs (m in our analysis), making the reference answer O(m) in our notation. However, this would be incorrect for the actual implementation since the algorithm must iterate through all values from n to 1, making the time complexity dependent on both the parameter n and the number of pairs m.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Tracking of Minimum Conflicting Positions

The Pitfall: The current implementation has a critical bug in how it tracks the smallest and second smallest conflicting positions. The variables smallest_conflict and second_smallest_conflict are initialized once outside the loop and never reset for each position. This causes them to accumulate values from previous iterations, leading to incorrect calculations.

Why This Happens: When processing position current_node, we need to find the two smallest positions that conflict with it. However, the code maintains these values across all iterations, meaning conflicts from position n are still considered when processing position n-1, n-2, etc.

The Fix: Reset the tracking variables for each position:

for current_node in range(n, 0, -1):
    # Reset for each position - THIS IS THE FIX
    smallest_conflict = n + 1
    second_smallest_conflict = n + 1
  
    # Update smallest and second smallest conflicting nodes
    for conflict_node in graph[current_node]:
        if conflict_node < smallest_conflict:
            second_smallest_conflict = smallest_conflict
            smallest_conflict = conflict_node
        elif conflict_node < second_smallest_conflict:
            second_smallest_conflict = conflict_node
  
    # Rest of the logic remains the same

2. Misunderstanding the Problem's Counting Logic

The Pitfall: Developers might incorrectly assume that removing a conflicting pair [a, b] only affects subarrays containing both a and b. The actual impact is more nuanced - it affects all subarrays starting from any position up to a that would extend to or past b.

Correct Understanding:

• When we remove a pair involving position b1, we extend the valid range from [current_node, b1-1] to [current_node, b2-1]
• This gives us b2 - b1 additional valid subarrays starting at current_node
• The count[b1] accumulates all such benefits across different starting positions

3. Edge Case: No Conflicts for a Position

The Pitfall: If a position has no conflicts (empty graph[current_node]), the smallest_conflict remains as n + 1, which is correct. However, developers might try to "optimize" by skipping such positions, which would miss counting valid subarrays.

Why It Matters: Even positions with no conflicts contribute to the answer. A position with no conflicts can form subarrays all the way to position n, contributing (n + 1) - current_node valid subarrays.