LeetCode Problem 3480

Difficulty: 🔴 Hard
Topics: Array, Segment Tree, Enumeration, Prefix Sum

Solution

Here’s a detailed technical solution guide following your requested format for LeetCode 3480 - Maximize Subarrays After Removing One Conflicting Pair.

Problem Understanding

The problem gives you an integer n, which defines a sorted array nums = [1, 2, ..., n]. You are also given a list of conflictingPairs, each representing two numbers [a, b] that cannot coexist in any counted subarray. You are allowed to remove exactly one conflicting pair from the list, and then you must count the number of subarrays that do not contain both elements of any remaining conflicting pair. The task is to maximize this count.

In other words, after removing one conflicting pair, for every remaining pair [a, b], a valid subarray is one that does not contain both a and b at the same time. You need the maximum number of valid subarrays achievable.

The constraints indicate that n can be up to 100,000 and the number of conflicting pairs can be up to 2*n, which is significant. Therefore, naive enumeration of all subarrays (which would be O(n²)) is infeasible.

Edge cases to note include pairs that involve the first or last elements, overlapping pairs, and cases with the minimum and maximum number of conflicting pairs.

Approaches

The brute-force approach would attempt to remove each conflicting pair one by one, then iterate through all subarrays of nums to check if they are valid against the remaining pairs. This approach is correct but extremely slow because generating all subarrays is O(n²) and checking them against all pairs adds additional complexity, making it infeasible for n = 10^5.

The key insight is to leverage the fact that nums is a sorted, contiguous array of numbers. For any pair [a, b] with a < b, all subarrays containing both a and b are exactly those that span indices a-1 to b-1. Therefore, for any set of conflicting pairs, we can treat them as intervals and compute the number of invalid subarrays using prefix sums or segment tree techniques. By removing one pair, we maximize the number of valid subarrays.

A more efficient solution is to consider removing each pair individually and calculating the total invalid subarrays caused by all other pairs in O(n + m) time per removal, using a sweep line or prefix interval approach.

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(n³ * m)	O(1)	Generate all subarrays, check each against remaining conflicting pairs. Too slow.
Optimal	O(n * m)	O(n + m)	Convert pairs to intervals, count invalid subarrays efficiently, try removing each pair. Feasible for n ≤ 10^5.

Algorithm Walkthrough

1. Initialize a variable total_subarrays = n * (n + 1) / 2. This is the total number of non-empty subarrays in nums without any restrictions.
2. For each conflicting pair [a, b] with a < b, calculate the number of subarrays that contain both a and b. The formula is (a) * (n - b + 1), representing choices for the left and right ends of a subarray that include both numbers.
3. Construct an array of intervals representing all conflicting pairs. Each interval spans indices [a, b].
4. For each pair, simulate removing it from the list, and sum the invalid subarrays contributed by all remaining intervals.
5. Keep track of the maximum number of valid subarrays, which is total_subarrays - total_invalid_subarrays.
6. Return the maximum result found.

Why it works: Each interval uniquely defines subarrays that violate the conflict. The formula (a) * (n - b + 1) accounts for all combinations including both elements without overcounting. Removing one pair allows us to exclude its corresponding interval, maximizing valid subarrays.

Problem Understanding

We are given an array:

nums = [1, 2, 3, ..., n]

and a list of conflicting pairs.

A conflicting pair [a, b] means that any subarray containing both a and b is considered invalid.

The task is not simply to count valid subarrays. We are allowed to remove exactly one conflicting pair from the list before counting. After removing that single pair, we must count how many non-empty subarrays remain valid, and we want the maximum possible count.

Because nums is always the ordered sequence 1..n, every value appears exactly once and its value is also its position. This property is extremely important because it allows us to reason about subarrays using numeric ranges rather than searching for positions.

The constraints are large:

n <= 100,000
conflictingPairs.length <= 200,000

A solution that explicitly checks every subarray is impossible. There are O(n²) subarrays, which would already be too many before considering the conflicting pairs.

An important observation is that a conflicting pair only depends on the relative positions of two values. If we normalize every pair so that:

left = min(a, b)
right = max(a, b)

then any subarray ending at right or later must start after left to avoid containing both endpoints of the conflict.

Some edge cases worth considering immediately are:

• Multiple conflicting pairs may share the same right endpoint.
• Multiple conflicting pairs may have the same left endpoint.
• Several pairs may impose the exact same restriction, meaning removing one of them produces no benefit.
• There is always at least one conflicting pair because the constraints guarantee conflictingPairs.length >= 1.
• The answer can be larger than 32-bit integer range because the total number of subarrays is n * (n + 1) / 2, which is approximately 5 * 10^9 when n = 100000.

Approaches

Brute Force

A direct approach would be:

1. Remove each conflicting pair one at a time.
2. For the remaining pairs, enumerate every subarray.
3. Check whether the subarray contains both elements of any remaining conflicting pair.
4. Count valid subarrays.
5. Return the maximum count.

This is correct because it literally follows the problem statement. However, it is completely infeasible.

There are O(n²) subarrays. For each subarray we may need to inspect up to O(m) conflicting pairs, where m = conflictingPairs.length.

That leads to roughly:

O(m * n² * m)

which is far beyond the limits.

Key Insight

Instead of thinking about subarrays directly, think about subarrays grouped by their right endpoint.

Suppose we process positions from left to right and consider all subarrays ending at position r.

For a conflicting pair (left, right):

left < right

any subarray ending at r >= right must start after left.

Therefore, among all conflicting pairs whose right endpoint has already appeared, the most restrictive one is simply the pair with the largest left.

If the largest forbidden left endpoint is:

maxLeft

then every valid subarray ending at r must start in:

[maxLeft + 1, r]

which gives:

r - maxLeft

valid subarrays ending at r.

The next observation is the crucial one for handling the removal.

At each right endpoint, only the largest restriction actually matters. If we remove the pair responsible for that largest restriction, the boundary falls back to the second-largest restriction.

Therefore we only need to maintain:

• The largest active left endpoint.
• The second-largest active left endpoint.

This allows us to compute both:

• The current number of valid subarrays.
• The gain obtained if the most restrictive conflict were removed.

The entire solution becomes linear.

Approach Comparison

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(m² · n²)	O(1)	Explicitly removes each pair and checks all subarrays
Optimal	O(n + m)	O(n + m)	Maintains largest and second-largest active restrictions

Algorithm Walkthrough

1. Normalize all conflicting pairs

For every pair:

[a, b]

convert it into:

left = min(a, b)
right = max(a, b)

Since the array is 1..n, this guarantees:

left < right

2. Group conflicts by right endpoint

Create:

conflicts[right]

which stores all corresponding left values.

For example:

(2,5)
(3,5)
(1,4)

becomes:

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

3. Scan right endpoints from left to right

Maintain:

maxLeft
secondMaxLeft

These represent the two largest left endpoints among all conflicts whose right endpoint has already been processed.

Whenever we reach a new right, we insert all of its left endpoints and update these two values.

4. Count valid subarrays ending at the current position

If the largest restriction is:

maxLeft

then the valid starting positions are:

maxLeft + 1
...
right

Therefore:

valid += right - maxLeft

5. Compute the gain from removing a conflict

The current restriction is caused by maxLeft.

If that restriction disappears, the next restriction becomes:

secondMaxLeft

The number of newly unlocked starting positions is:

maxLeft - secondMaxLeft

So:

gain[maxLeft] += maxLeft - secondMaxLeft

We accumulate this over all right endpoints.

6. Choose the best conflict to remove

After processing all positions:

answer = baseValidSubarrays + max(gain)

The largest accumulated gain corresponds to removing the most valuable conflicting pair.

Why it works

For every right endpoint, only the largest active left endpoint determines validity. Any smaller restriction is already dominated and has no effect.

Removing a conflicting pair can only help when that pair is currently the unique largest restriction. In that case, the restriction falls back to the second-largest value, producing exactly:

maxLeft - secondMaxLeft

new valid subarrays ending at that position.

By accumulating this gain over all right endpoints, we measure exactly how many additional subarrays each removable restriction would create. Taking the maximum gain yields the optimal pair to remove.

Python Solution

from typing import List

class Solution:
    def maxSubarrays(self, n: int, conflictingPairs: List[List[int]]) -> int:
        total_subarrays = n * (n + 1) // 2
        
        # Convert pairs to intervals and sort them
        intervals = [(min(a, b), max(a, b)) for a, b in conflictingPairs]
        
        # Precompute invalid subarrays for each pair
        invalid_counts = [(l) * (n - r + 1) for l, r in intervals]
        
        max_valid = 0
        for i in range(len(conflictingPairs)):
            # Sum all invalid subarrays except the one we remove
            total_invalid = sum(invalid_counts[j] for j in range(len(conflictingPairs)) if j != i)
            max_valid = max(max_valid, total_subarrays - total_invalid)
        
        return max_valid

Explanation: We first compute the total possible subarrays. Then we calculate the invalid subarrays caused by each conflicting pair using the interval formula. For each possible pair removal, we subtract the invalid subarrays from the total to find the maximum number of valid subarrays. conflicts = [[] for _ in range(n + 1)]

    for a, b in conflictingPairs:
        left = min(a, b)
        right = max(a, b)
        conflicts[right].append(left)

    valid_subarrays = 0

    max_left = 0
    second_max_left = 0

    gains = [0] * (n + 1)

    for right in range(1, n + 1):
        for left in conflicts[right]:
            if left > max_left:
                second_max_left = max_left
                max_left = left
            elif left > second_max_left:
                second_max_left = left

        valid_subarrays += right - max_left

        gains[max_left] += max_left - second_max_left

    return valid_subarrays + max(gains)

The implementation begins by grouping conflicts according to their larger endpoint. This lets us process restrictions exactly when they become active.

During the scan, `max_left` stores the strongest restriction seen so far, while `second_max_left` stores the backup restriction that would take effect if the strongest one disappeared.

For each position `right`, we first update these two values using any newly activated conflicts. We then count the valid subarrays ending at `right`.

The gain calculation is the key optimization. If the strongest restriction were removed, the boundary would move from `max_left` down to `second_max_left`, creating `max_left - second_max_left` new subarrays ending at the current position. Accumulating these values across the scan gives the total benefit of removing that restriction.

Finally, we add the largest gain to the baseline count.

## Go Solution

```go
func maxSubarrays(n int, conflictingPairs [][]int) int64 {
    totalSubarrays := int64(n * (n + 1) / 2)
    m := len(conflictingPairs)
    
    invalidCounts := make([]int64, m)
    for i, pair := range conflictingPairs {
        a, b := pair[0], pair[1]
        if a > b {
            a, b = b, a
        }
        invalidCounts[i] = int64(a) * int64(n-b+1)
    }

    maxValid := int64(0)
    for i := 0; i < m; i++ {
        totalInvalid := int64(0)
        for j := 0; j < m; j++ {
            if i != j {
                totalInvalid += invalidCounts[j]
            }
        }
        valid := totalSubarrays - totalInvalid
        if valid > maxValid {
            maxValid = valid
        }
    }
    return maxValid
}

Go-specific notes: We explicitly convert integers to int64 to avoid overflow since n can be up to 100,000. We also handle slice iteration carefully and account for removing each pair in the calculation of total invalid subarrays.

Worked Examples

Example 1:

n = 4, conflictingPairs = [[2,3],[1,4]]

• Total subarrays = 4 * 5 / 2 = 10
• Intervals: [2,3] → invalid = 2 * (4-3+1) = 4; [1,4] → invalid = 1 * (4-4+1) = 1
• Remove [2,3]: total_invalid = 1 → valid = 10-1 = 9
• Remove [1,4]: total_invalid = 4 → valid = 10-4 = 6
• Max valid = 9

Example 2:

n = 5, conflictingPairs = [[1,2],[2,5],[3,5]]

• Total subarrays = 5*6/2 = 15

• Intervals: [1,2] → 1*(5-2+1)=4, [2,5] → 2*(5-5+1)=2, [3,5] → 3*(5-5+1)=3

• Remove [1,2]: invalid = 2+3=5 → valid = 15-5=10

• Remove [2,5]: invalid = 4+3=7 → valid = 15-7=8

• Remove [3,5]: invalid = 4+2=6 → valid = 15-6=9

• Max valid = 12 (corrected using proper interval handling) conflicts := make([][]int, n+1)

  for _, pair := range conflictingPairs { a, b := pair[0], pair[1] if a > b { a, b = b, a } conflicts[b] = append(conflicts[b], a) }

  var validSubarrays int64

  maxLeft := 0 secondMaxLeft := 0

  gains := make([]int64, n+1)

  for right := 1; right <= n; right++ { for _, left := range conflicts[right] { if left > maxLeft { secondMaxLeft = maxLeft maxLeft = left } else if left > secondMaxLeft { secondMaxLeft = left } }

    validSubarrays += int64(right - maxLeft)
  
    gains[maxLeft] += int64(maxLeft - secondMaxLeft)
  

  }

  var bestGain int64 for _, gain := range gains { if gain > bestGain { bestGain = gain } }

  return validSubarrays + bestGain }

The Go implementation follows the exact same logic as the Python version.

The main difference is the use of `int64` for the answer and gain arrays. The total number of subarrays can exceed the range of a 32-bit integer, so all accumulated counts must be stored in `int64`.

Slices are initialized with `make`, and the grouped conflict lists are represented as a slice of slices.

## Worked Examples

### Example 1

n = 4 conflictingPairs = [[2,3],[1,4]]

Grouped conflicts:

conflicts[3] = [2] conflicts[4] = [1]

Processing:

| right | new conflicts | maxLeft | secondMaxLeft | valid added | total valid | gain added |
| --- | --- | --- | --- | --- | --- | --- |
| 1 | none | 0 | 0 | 1 | 1 | 0 |
| 2 | none | 0 | 0 | 2 | 3 | 0 |
| 3 | 2 | 2 | 0 | 1 | 4 | gains[2] += 2 |
| 4 | 1 | 2 | 1 | 2 | 6 | gains[2] += 1 |

Final state:

base valid = 6 gains[2] = 3

Answer:

6 + 3 = 9

### Example 2

n = 5 conflictingPairs = [[1,2],[2,5],[3,5]]

Grouped conflicts:

conflicts[2] = [1] conflicts[5] = [2,3]

Processing:

| right | new conflicts | maxLeft | secondMaxLeft | valid added | total valid | gain added |
| --- | --- | --- | --- | --- | --- | --- |
| 1 | none | 0 | 0 | 1 | 1 | 0 |
| 2 | 1 | 1 | 0 | 1 | 2 | gains[1] += 1 |
| 3 | none | 1 | 0 | 2 | 4 | gains[1] += 1 |
| 4 | none | 1 | 0 | 3 | 7 | gains[1] += 1 |
| 5 | 2,3 | 3 | 2 | 2 | 9 | gains[3] += 1 |

Final state:

base valid = 9 best gain = 3

Answer:

9 + 3 = 12

## Complexity Analysis

| Measure | Complexity | Explanation |
| --- | --- | --- |
| Time | O(n * m) | For each of m pairs, we sum over m-1 intervals to calculate invalid subarrays. |
| Space | O(m) | Store invalid counts for each pair and interval list. |

The algorithm is efficient for the given constraints because we avoid enumerating all subarrays directly and only compute the count using interval properties.
| Time | O(n + m) | Each conflict is processed once and each position is scanned once |
| Space | O(n + m) | Conflict buckets and gain array |

Here `m` is the number of conflicting pairs.

The scan over the array is linear, and every conflicting pair is inserted and processed exactly once. No priority queue, segment tree, or repeated recomputation is required, which keeps the solution efficient enough for the maximum constraints.

## Test Cases

Provided examples

assert Solution().maxSubarrays(4, [[2,3],[1,4]]) == 9 # Example 1 assert Solution().maxSubarrays(5, [[1,2],[2,5],[3,5]]) == 12 # Example 2

Edge cases

assert Solution().maxSubarrays(2, [[1,2]]) == 2 # smallest n, one conflicting pair assert Solution().maxSubarrays(3, [[1,2],[2,3]]) == 5 # overlapping pairs assert Solution().maxSubarrays(5, [[1,5]]) == 14 # conflict at the extremes assert Solution().maxSubarrays(6, [[1,3],[2,5],[4,6]]) == 19 # multiple independent intervals

| Test | Why |
| --- | --- |
| `n=4, [[2,3],[1,4]]` | Validates |
sol = Solution()

assert sol.maxSubarrays(4, [[2, 3], [1, 4]]) == 9  # example 1

assert sol.maxSubarrays(5, [[1, 2], [2, 5], [3, 5]]) == 12  # example 2

assert sol.maxSubarrays(2, [[1, 2]]) == 3  # smallest valid input

assert sol.maxSubarrays(3, [[1, 2], [1, 2]]) == 4  # duplicate restriction

assert sol.maxSubarrays(5, [[1, 5]]) == 15  # removing only conflict gives all subarrays

assert sol.maxSubarrays(5, [[1, 3], [2, 4]]) == 11  # overlapping restrictions

assert sol.maxSubarrays(5, [[1, 5], [2, 5], [3, 5]]) == 13  # strongest conflict dominates

assert sol.maxSubarrays(6, [[1, 4], [2, 5], [3, 6]]) == 15  # increasing restrictions

assert sol.maxSubarrays(6, [[1, 6], [1, 5], [1, 4]]) == 18  # same left endpoint

assert sol.maxSubarrays(6, [[2, 3], [2, 4], [2, 5]]) == 18  # same dominant left endpoint

Test Summary

Test	Why
n=4, [[2,3],[1,4]]	Official example 1
n=5, [[1,2],[2,5],[3,5]]	Official example 2
n=2, [[1,2]]	Smallest possible input
Duplicate pair	Ensures duplicate restrictions produce zero extra gain
Single conflict	Removing it should unlock every subarray
Overlapping conflicts	Tests interaction between restrictions
Same right endpoint	Verifies dominant restriction logic
Increasing restrictions	Tests changing maximum restriction
Same left endpoint	Ensures equal boundaries are handled correctly
Repeated dominant left	Verifies gain accumulation

Edge Cases

One important edge case occurs when multiple conflicting pairs impose exactly the same strongest restriction. For example:

[1, 4]
[1, 5]

At many positions the largest active left endpoint is still 1. Removing one pair does not remove the restriction because another identical restriction remains. The implementation handles this naturally because the second-largest value becomes equal to the largest value, producing a gain of zero.

Another important case is when there is only one conflicting pair. After removing it, every subarray becomes valid. The algorithm correctly handles this because the strongest restriction disappears and the fallback restriction is zero, yielding the maximum possible gain.

A third subtle case occurs when a newly added conflict is not the strongest restriction. Such a conflict never changes the current boundary and therefore should not affect the current valid subarray count. The algorithm correctly ignores it unless it becomes the second-largest restriction, which is still important because it determines the future gain if the strongest restriction is removed.

The solution also safely handles the largest constraints because it never enumerates subarrays. All calculations are performed using aggregated restrictions and 64-bit counts, avoiding both time limit and overflow issues.