Tejav
#3480

Maximize Subarrays After Removing One Conflicting Pair

Hard
ArraySegment TreeEnumerationPrefix SumHash MapArray
LeetCode ↗

Approaches

Brute ForceOptimal
Complexity Comparison
Brute ForceOptimal Solution★
Time
O(n²)
O(n)
Space
O(1)
O(n)
💡

Intuition

Time O(n)Space O(n)

Use prefix sums to efficiently calculate valid subarrays by tracking the longest valid segment for each starting index, adjusting for conflicting pairs.

⚙️

Algorithm

3 steps
  1. 1Step 1: Initialize an array f where f[i] represents the end index of the longest valid subarray starting at index i.
  2. 2Step 2: For each conflicting pair, adjust the f array to mark invalid ranges.
  3. 3Step 3: Calculate the total valid subarrays using the formula: sigma(f[i]) - n * (n + 1) / 2 + n.
solution.py8 lines
1def maxSubarrays(n, conflictingPairs):
2    f = [0] * (n + 1)
3    for i in range(1, n + 1):
4        f[i] = i
5    for a, b in conflictingPairs:
6        f[max(a, b)] = min(f[max(a, b)], min(a, b) - 1)
7    total = sum(f[i] - i + 1 for i in range(1, n + 1))
8    return total - n * (n + 1) // 2 + n
ℹ

Complexity note: The algorithm processes each element and conflicting pair linearly, leading to linear time complexity.

  • 1Removing one conflicting pair can significantly increase valid subarrays.
  • 2Using prefix sums allows efficient counting without generating all subarrays.

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