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3480. Maximize Subarrays After Removing One Conflicting Pair

  • Time: $O(n + |\texttt{conflictPairs}|)$
  • Space: $O(n)$
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class Solution {
 public:
  long long maxSubarrays(int n, vector<vector<int>>& conflictingPairs) {
    long validSubarrays = 0;
    int maxLeft = 0;
    int secondMaxLeft = 0;
    // gains[i] := the number of additional valid subarrays we can gain if the
    // restriction at index `i` is removed
    vector<long> gains(n + 1);
    // conflicts[r] := left endpoints that conflict with subarrays ending in r
    vector<vector<int>> conflicts(n + 1);

    for (const vector<int>& pair : conflictingPairs) {
      const int a = pair[0];
      const int b = pair[1];
      conflicts[max(a, b)].push_back(min(a, b));
    }

    for (int right = 1; right <= n; ++right) {
      for (const int left : conflicts[right]) {
        if (left > maxLeft) {
          secondMaxLeft = maxLeft;
          maxLeft = left;
        } else if (left > secondMaxLeft) {
          secondMaxLeft = left;
        }
      }
      // Subarrays [maxLeft + 1..right],
      //           [maxLeft + 2..right],
      //           ...
      //           [right..right] are valid.
      validSubarrays += right - maxLeft;
      // If we remove `maxLeft` (the most restrictive conflict), we gain
      // `maxLeft - secondMaxLeft` new subarrays:
      // [secondMaxLeft + 1..right],
      // [secondMaxLeft + 2..right],
      // ...
      // [maxLeft..right].
      gains[maxLeft] += maxLeft - secondMaxLeft;
    }

    return validSubarrays + ranges::max(gains);
  }
};

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class Solution {
  public long maxSubarrays(int n, int[][] conflictingPairs) {
    long validSubarrays = 0;
    int maxLeft = 0;
    int secondMaxLeft = 0;
    // gains[i] := the number of additional valid subarrays we can gain if the
    // restriction at index `i` is removed
    long[] gains = new long[n + 1];
    // conflicts[r] := left endpoints that conflict with subarrays ending in r
    List<Integer>[] conflicts = new List[n + 1];

    for (int i = 0; i <= n; ++i)
      conflicts[i] = new ArrayList<>();

    for (int[] pair : conflictingPairs) {
      final int a = pair[0];
      final int b = pair[1];
      conflicts[Math.max(a, b)].add(Math.min(a, b));
    }

    for (int right = 1; right <= n; ++right) {
      for (int left : conflicts[right]) {
        if (left > maxLeft) {
          secondMaxLeft = maxLeft;
          maxLeft = left;
        } else if (left > secondMaxLeft) {
          secondMaxLeft = left;
        }
      }
      // Subarrays [maxLeft + 1..right],
      //           [maxLeft + 2..right],
      //           ...
      //           [right..right] are valid.
      validSubarrays += right - maxLeft;
      // If we remove `maxLeft` (the most restrictive conflict), we gain
      // `maxLeft - secondMaxLeft` new subarrays:
      // [secondMaxLeft + 1..right],
      // [secondMaxLeft + 2..right],
      // ...
      // [maxLeft..right].
      gains[maxLeft] += maxLeft - secondMaxLeft;
    }

    return validSubarrays + Arrays.stream(gains).max().getAsLong();
  }
}

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class Solution:
  def maxSubarrays(self, n: int, conflictingPairs: list[list[int]]) -> int:
    validSubarrays = 0
    maxLeft = 0
    secondMaxLeft = 0
    # gains[i] := the number of additional valid subarrays we can gain if the
    # restriction at index `i` is removed
    gains = [0] * (n + 1)
    # conflicts[r] := left endpoints that conflict with subarrays ending in r
    conflicts = [[] for _ in range(n + 1)]

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

    for right in range(1, n + 1):
      for left in conflicts[right]:
        if left > maxLeft:
          secondMaxLeft = maxLeft
          maxLeft = left
        elif left > secondMaxLeft:
          secondMaxLeft = left
      # Subarrays [maxLeft + 1..right],
      #           [maxLeft + 2..right],
      #           ...
      #           [right..right] are valid.
      validSubarrays += right - maxLeft
      # If we remove `maxLeft` (the most restrictive conflict), we gain
      # `maxLeft - secondMaxLeft` new subarrays:
      # [secondMaxLeft + 1..right],
      # [secondMaxLeft + 2..right],
      # ...
      # [maxLeft..right].
      gains[maxLeft] += maxLeft - secondMaxLeft

    return validSubarrays + max(gains)