2407. Longest Increasing Subsequence II LeetCode Solution

Last updated: February 27, 2025

In this guide, you will get 2407. Longest Increasing Subsequence II LeetCode Solution with the best time and space complexity. The solution to Longest Increasing Subsequence II problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Longest Increasing Subsequence II solution in C++
Longest Increasing Subsequence II solution in Java
Longest Increasing Subsequence II solution in Python
Additional Resources

2407. Longest Increasing Subsequence II LeetCode Solution image

Problem Statement of Longest Increasing Subsequence II

You are given an integer array nums and an integer k.
Find the longest subsequence of nums that meets the following requirements:

The subsequence is strictly increasing and
The difference between adjacent elements in the subsequence is at most k.

Return the length of the longest subsequence that meets the requirements.
A subsequence is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements.

Example 1:

Input: nums = [4,2,1,4,3,4,5,8,15], k = 3
Output: 5
Explanation:
The longest subsequence that meets the requirements is [1,3,4,5,8].
The subsequence has a length of 5, so we return 5.
Note that the subsequence [1,3,4,5,8,15] does not meet the requirements because 15 – 8 = 7 is larger than 3.

Example 2:

Input: nums = [7,4,5,1,8,12,4,7], k = 5
Output: 4
Explanation:
The longest subsequence that meets the requirements is [4,5,8,12].
The subsequence has a length of 4, so we return 4.

Example 3:

Input: nums = [1,5], k = 1
Output: 1
Explanation:
The longest subsequence that meets the requirements is [1].
The subsequence has a length of 1, so we return 1.

Constraints:

1 <= nums.length <= 105
1 <= nums[i], k <= 105

Complexity Analysis

Time Complexity: O(n\log 10^5)
Space Complexity: O(10^5)

2407. Longest Increasing Subsequence II LeetCode Solution in C++

struct SegmentTreeNode {
  int lo;
  int hi;
  int maxLength;
  std::unique_ptr<SegmentTreeNode> left;
  std::unique_ptr<SegmentTreeNode> right;
  // maxLength := the maximum length of LIS ending in [lo..hi]
  SegmentTreeNode(int lo, int hi, int maxLength,
                  std::unique_ptr<SegmentTreeNode> left = nullptr,
                  std::unique_ptr<SegmentTreeNode> right = nullptr)
      : lo(lo),
        hi(hi),
        maxLength(maxLength),
        left(std::move(left)),
        right(std::move(right)) {}
};

class SegmentTree {
 public:
  explicit SegmentTree() : root(make_unique<SegmentTreeNode>(0, 1e5 + 1, 0)) {}

  void updateRange(int i, int j, int maxLength) {
    update(root, i, j, maxLength);
  }

  // Returns the maximum length of LIS ending in [i..j].
  int queryRange(int i, int j) {
    return query(root, i, j);
  }

 private:
  std::unique_ptr<SegmentTreeNode> root;

  void update(std::unique_ptr<SegmentTreeNode>& root, int i, int j,
              int maxLength) {
    if (root->lo == i && root->hi == j) {
      root->maxLength = maxLength;
      root->left = nullptr;
      root->right = nullptr;
      return;
    }
    const int mid = root->lo + (root->hi - root->lo) / 2;
    if (root->left == nullptr) {
      root->left = make_unique<SegmentTreeNode>(root->lo, mid, root->maxLength);
      root->right =
          make_unique<SegmentTreeNode>(mid + 1, root->hi, root->maxLength);
    }
    if (j <= mid)
      update(root->left, i, j, maxLength);
    else if (i > mid)
      update(root->right, i, j, maxLength);
    else {
      update(root->left, i, mid, maxLength);
      update(root->right, mid + 1, j, maxLength);
    }
    root->maxLength = merge(root->left->maxLength, root->right->maxLength);
  }

  int query(std::unique_ptr<SegmentTreeNode>& root, int i, int j) {
    if (root->left == nullptr)
      return root->maxLength;
    if (root->lo == i && root->hi == j)
      return root->maxLength;
    const int mid = root->lo + (root->hi - root->lo) / 2;
    if (j <= mid)
      return query(root->left, i, j);
    if (i > mid)
      return query(root->right, i, j);
    return merge(query(root->left, i, mid), query(root->right, mid + 1, j));
  }

  int merge(int left, int right) const {
    return max(left, right);
  };
};

class Solution {
 public:
  int lengthOfLIS(vector<int>& nums, int k) {
    int ans = 1;
    SegmentTree tree;

    for (const int num : nums) {
      const int left = max(1, num - k);
      const int right = num - 1;
      // the maximum length of LIS ending in [left..right] + the current number
      const int maxLength = tree.queryRange(left, right) + 1;
      ans = max(ans, maxLength);
      tree.updateRange(num, num, maxLength);
    }

    return ans;
  }
};
/* code provided by PROGIEZ */

2407. Longest Increasing Subsequence II LeetCode Solution in Java

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// code provided by PROGIEZ

2407. Longest Increasing Subsequence II LeetCode Solution in Python

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 # code by PROGIEZ

Additional Resources

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