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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] := the length of LIS ending in 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 ranges::max(dp);
  }
};

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class Solution {
  public int lengthOfLIS(int[] nums) {
    if (nums.length == 0)
      return 0;

    // dp[i] := the length of LIS ending in 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] the length of LIS ending in 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) {
    // 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>& A, int target) {
    return ranges::lower_bound(A, target) - A.begin();
  }
};

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class Solution {
  public int lengthOfLIS(int[] nums) {
    // tails[i] := the minimum tail of all the increasing subsequences having
    // 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> A, int target) {
    final int i = Collections.binarySearch(A, target);
    return i < 0 ? -i - 1 : i;
  }
}

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class Solution:
  def lengthOfLIS(self, nums: list[int]) -> int:
    # tails[i] := the minimum tails 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)