Skip to content

1713. Minimum Operations to Make a Subsequence 👍

  • Time: $O(n\log n)$
  • Space: $O(n)$
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
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) {
    // tails[i] := the minimum tail of all the increasing subsequences having
    // length i + 1
    vector<int> tails;
    for (const int num : nums)
      if (tails.empty() || num > tails.back())
        tails.push_back(num);
      else
        tails[firstGreaterEqual(tails, num)] = num;
    return tails.size();
  }

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

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
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) {
    // tails[i] := the minimum tail of all the increasing subsequences with
    // length i + 1
    List<Integer> tails = new ArrayList<>();
    for (final int num : nums)
      if (tails.isEmpty() || num > tails.get(tails.size() - 1))
        tails.add(num);
      else
        tails.set(firstGreaterEqual(tails, num), num);
    return tails.size();
  }

  private int firstGreaterEqual(List<Integer> arr, int target) {
    final int i = Collections.binarySearch(arr, target);
    return i < 0 ? -i - 1 : i;
  }
}

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
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:
    # tails[i] := the minimum tail of all the increasing subsequences having
    # length i + 1
    tails = []
    for num in nums:
      if not tails or num > tails[-1]:
        tails.append(num)
      else:
        tails[bisect.bisect_left(tails, num)] = num
    return len(tails)