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1964. Find the Longest Valid Obstacle Course at Each Position 👍

  • Time: $O(n\log n)$
  • Space: $O(n)$
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class Solution {
 public:
  vector<int> longestObstacleCourseAtEachPosition(vector<int>& obstacles) {
    vector<int> ans;
    // tail[i] := the minimum tail of all increasing subseqs having length i + 1
    // it's easy to see that tail must be an increasing array
    vector<int> tail;

    for (const int obstacle : obstacles)
      if (tail.empty() || obstacle >= tail.back()) {
        tail.push_back(obstacle);
        ans.push_back(tail.size());
      } else {
        const int index = firstGreater(tail, obstacle);
        tail[index] = obstacle;
        ans.push_back(index + 1);
      }

    return ans;
  }

 private:
  // Find the first index l s.t A[l] > target
  // Returns A.size() if can't find
  int firstGreater(const vector<int>& A, int target) {
    return upper_bound(begin(A), end(A), target) - begin(A);
  }
};

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class Solution {
  public int[] longestObstacleCourseAtEachPosition(int[] obstacles) {
    List<Integer> ans = new ArrayList<>();
    // tail[i] := the minimum tail of all increasing subseqs with length i + 1
    // it's easy to see that tail must be an increasing array
    List<Integer> tail = new ArrayList<>();

    for (final int obstacle : obstacles)
      if (tail.isEmpty() || obstacle >= tail.get(tail.size() - 1)) {
        tail.add(obstacle);
        ans.add(tail.size());
      } else {
        final int index = firstGreater(tail, obstacle);
        tail.set(index, obstacle);
        ans.add(index + 1);
      }

    return ans.stream().mapToInt(Integer::intValue).toArray();
  }

  // Find the first index l s.t A.get(l) > target
  // Returns obstacles.length if can't find
  private int firstGreater(List<Integer> A, int target) {
    int l = 0;
    int r = A.size();

    while (l < r) {
      final int m = (l + r) / 2;
      if (A.get(m) > target)
        r = m;
      else
        l = m + 1;
    }

    return l;
  }
}

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class Solution:
  def longestObstacleCourseAtEachPosition(self, obstacles: List[int]) -> List[int]:
    ans = []
    # tail[i] := the minimum tail of all increasing subseqs having length i + 1
    # it's easy to see that tail must be an increasing array
    tail = []

    for obstacle in obstacles:
      if not tail or obstacle >= tail[-1]:
        tail.append(obstacle)
        ans.append(len(tail))
      else:
        index = bisect_right(tail, obstacle)
        tail[index] = obstacle
        ans.append(index + 1)

    return ans