Longest Increasing Subsequence II LeetCode Solution

Problem – Longest Increasing Subsequence II LeetCode Solution

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.

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

Longest Increasing Subsequence II LeetCode Solution in Python

class SEG:
    def __init__(self, n):
        self.n = n
        self.tree = [0] * 2 * self.n
       
    def query(self, l, r):
        l += self.n
        r += self.n
        ans = 0
        while l < r:
            if l & 1:
                ans = max(ans, self.tree[l])
                l += 1
            if r & 1:
                r -= 1
                ans = max(ans, self.tree[r])
            l >>= 1
            r >>= 1
        return ans
    
    def update(self, i, val):
        i += self.n
        self.tree[i] = val
        while i > 1:
            i >>= 1
            self.tree[i] = max(self.tree[i * 2], self.tree[i * 2 + 1])

class Solution:
    def lengthOfLIS(self, A: List[int], k: int) -> int:
        n, ans = max(A), 1
        seg = SEG(n)
        for a in A:
            a -= 1
            premax = seg.query(max(0, a - k), a)
            ans = max(ans, premax + 1)
            seg.update(a, premax + 1)
        return ans

Longest Increasing Subsequence II LeetCode Solution in C++

class Solution {
public: 
    const static int N = 100001;
    array<int, 2*N> seg{};
    
    void update(int pos, int val){ // update max
        pos += N;
        seg[pos] = val;
 
        while (pos > 1) {
            pos >>= 1;
            seg[pos] = max(seg[2*pos], seg[2*pos+1]);
        }
    }
 
    int query(int lo, int hi){ // query max [lo, hi)
        lo += N;
        hi += N;
        int res = 0;
 
        while (lo < hi) {
            if (lo & 1) {
                res = max(res, seg[lo++]);
            }
            if (hi & 1) {
                res = max(res, seg[--hi]);
            }
            lo >>= 1;
            hi >>= 1;
        }
        return res;
    }
    
    int lengthOfLIS(vector<int>& A, int k) {
        int ans = 0;
        for (int i = 0; i < size(A); ++i){
            int l = max(0, A[i]-k);
            int r = A[i];
            int res = query(l, r) + 1; // best res for the current element
            ans = max(res, ans);
            update(A[i], res); // and update it here
        }
        
        return ans;
    }
};

Longest Increasing Subsequence II LeetCode Solution in Java

class Solution {

    int N = 100001;
    int[] seg = new int[2*N];
    
    void update(int pos, int val){  // update max
        pos += N;
        seg[pos] = val;
 
        while (pos > 1) {
            pos >>= 1;
            seg[pos] = Math.max(seg[2*pos], seg[2*pos+1]);
        }
    }
 
    int query(int lo, int hi){ // query max [lo, hi)
        lo += N;
        hi += N;
        int res = 0;
 
        while (lo < hi) {
            if ((lo & 1)==1) {
                res = Math.max(res, seg[lo++]);
            }
            if ((hi & 1)==1) {
                res = Math.max(res, seg[--hi]);
            }
            lo >>= 1;
            hi >>= 1;
        }
        return res;
    }
    
    public int lengthOfLIS(int[] A, int k) {
        int ans = 0;
        for (int i = 0; i < A.length; ++i){
            int l = Math.max(0, A[i]-k);
            int r = A[i];
            int res = query(l, r) + 1; // best res for the current element
            ans = Math.max(res, ans);
            update(A[i], res); // and update it here
        }
        return ans;
    }
}
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