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1713. Minimum Operations to Make a Subsequence 👍

  • Time: $O(n\log n)$
  • Space: $O(n)$
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class Solution {
 public:
  int minOperations(vector<int>& target, vector<int>& arr) {
    vector<int> indices;
    unordered_map<int, int> numToIndex;

    for (int i = 0; i < target.size(); ++i)
      numToIndex[target[i]] = i;

    for (const int a : arr)
      if (const auto it = numToIndex.find(a); it != numToIndex.end())
        indices.push_back(it->second);

    return target.size() - lengthOfLIS(indices);
  }

 private:
  // Same as 300. Longest Increasing Subsequence
  int lengthOfLIS(vector<int>& nums) {
    // tail[i] := the min 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 num : nums)
      if (tail.empty() || num > tail.back())
        tail.push_back(num);
      else
        tail[firstGreaterEqual(tail, num)] = num;

    return tail.size();
  }

 private:
  int firstGreaterEqual(const vector<int>& A, int target) {
    return lower_bound(A.begin(), A.end(), target) - A.begin();
  }
};

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class Solution {
  public int minOperations(int[] target, int[] arr) {
    List<Integer> indices = new ArrayList<>();
    Map<Integer, Integer> numToIndex = new HashMap<>();

    for (int i = 0; i < target.length; ++i)
      numToIndex.put(target[i], i);

    for (final int a : arr)
      if (numToIndex.containsKey(a))
        indices.add(numToIndex.get(a));

    return target.length - lengthOfLIS(indices);
  }

  // Same as 300. Longest Increasing Subsequence
  private int lengthOfLIS(List<Integer> nums) {
    // tail[i] := the min 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 num : nums)
      if (tail.isEmpty() || num > tail.get(tail.size() - 1))
        tail.add(num);
      else
        tail.set(firstGreaterEqual(tail, num), num);

    return tail.size();
  }

  private int firstGreaterEqual(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 minOperations(self, target: List[int], arr: List[int]) -> int:
    indices = []
    numToIndex = {num: i for i, num in enumerate(target)}

    for a in arr:
      if a in numToIndex:
        indices.append(numToIndex[a])

    return len(target) - self._lengthOfLIS(indices)

  # Same as 300. Longest Increasing Subsequence
  def _lengthOfLIS(self, nums: List[int]) -> int:
    # tail[i] := the min tail of all increasing subseqs having length i + 1
    # It's easy to see that tail must be an increasing array.
    tail = []

    for num in nums:
      if not tail or num > tail[-1]:
        tail.append(num)
      else:
        tail[bisect.bisect_left(tail, num)] = num

    return len(tail)