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300. Longest Increasing Subsequence 👍

Approach 1: 2D DP

  • Time: $O(n^2)$
  • Space: $O(n)$
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class Solution {
 public:
  int lengthOfLIS(vector<int>& nums) {
    if (nums.empty())
      return 0;

    // dp[i] := Length of LIS ending at nums[i]
    vector<int> dp(nums.size(), 1);

    for (int i = 1; i < nums.size(); ++i)
      for (int j = 0; j < i; ++j)
        if (nums[j] < nums[i])
          dp[i] = max(dp[i], dp[j] + 1);

    return *max_element(begin(dp), end(dp));
  }
};
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class Solution {
  public int lengthOfLIS(int[] nums) {
    if (nums.length == 0)
      return 0;

    // dp[i] := Length of LIS ending at nums[i]
    int[] dp = new int[nums.length];
    Arrays.fill(dp, 1);

    for (int i = 1; i < nums.length; ++i)
      for (int j = 0; j < i; ++j)
        if (nums[j] < nums[i])
          dp[i] = Math.max(dp[i], dp[j] + 1);

    return Arrays.stream(dp).max().getAsInt();
  }
}
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class Solution:
  def lengthOfLIS(self, nums: List[int]) -> int:
    if not nums:
      return 0

    # dp[i] := LIS ending at nums[i]
    dp = [1] * len(nums)

    for i in range(1, len(nums)):
      for j in range(i):
        if nums[j] < nums[i]:
          dp[i] = max(dp[i], dp[j] + 1)

    return max(dp)
  • Time: $O(n\log n)$
  • Space: $O(n)$
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class Solution {
 public:
  int lengthOfLIS(vector<int>& nums) {
    // 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 num : nums)
      if (tail.empty() || num > tail.back())
        tail.push_back(num);
      else
        tail[firstGreaterEqual(tail, num)] = num;

    return tail.size();
  }

 private:
  // Find the first index l s.t A[l] >= target
  // Returns A.size() if can't find
  int firstGreaterEqual(const vector<int>& A, int target) {
    return lower_bound(begin(A), end(A), target) - begin(A);
  }
};
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class Solution {
  public int lengthOfLIS(int[] nums) {
    // 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 num : nums)
      if (tail.isEmpty() || num > tail.get(tail.size() - 1))
        tail.add(num);
      else
        tail.set(firstGreaterEqual(tail, num), num);

    return tail.size();
  }

  // Find the first index l s.t A.get(l) >= target
  // Returns nums.size() if can't find
  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 lengthOfLIS(self, nums: List[int]) -> int:
    # 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 num in nums:
      if not tail or num > tail[-1]:
        tail.append(num)
      else:
        tail[bisect_left(tail, num)] = num

    return len(tail)